LMR's blackboard
The next best thing to the greatest sports, and other, gematria decode ever (as the divine code of the universe, itself) lies right below in the Reaper's final anagram-with-gematria.
---> Nice try, but the divine code wasn't included. Not quite yet. Ha.
Well, now, an update ... the solution to the fine structure (constants) will be out, in a couple of weeks, at the VegasCasinoTalk.com forum, where I will put up the mock-up solution, to be expanded upon as I go along.
NOTE-1 THAT THIS PART, FOR SOME TIME, WILL BE A WORK IN PROGRESS ... . Slow going from the back end of this bit, to organizing and explaining it in a way to prove out the theory of everything, beyond a reasonable doubt. Won't be an explanation per se, for some time.
NOTE-2 THAT THE FINE STRUCTURE OF THE UNIVERSE HAS BEEN DETERMINED ... . Ie, the electromagnetic, and gravitational, fine-structure constants. I'm in the process now of finalizing the analyses to present to the professional physics community.
Okay, a chance to revisit the above. Instead of a more detailed thus answer, say, with taking the ratio r as a fraction, or, say, thus getting into strictly the cubes instead of the squares , I will try to back things up to form a different perspective. Hopefully, a perspective more in line with the title. After all, I did write something about a "mythical connection" between the numerals, 411, and, 137, as having something to do with the fine structure of the universe. Nor can it hurt, a lot, especially if there's a good reason or two, to blur the lines a bit between the regular simple operators, and, say, their analogous forms of concatenation. Let alone when the question was consciously meant to be about supposedly useless numerals. However, such numerals, which go out of their way to have as little ostensive purpose as possible, just might turn out to be the most meaningful of them all. I mean in the sense that by sticking to themselves, they join up with the other such numerals in an absolutely fixed, or universal, way that is the way of the self. As I go along, I will try to include a few other ways that numerals stick to themselves by their properties that just are. My point now is to continue on from Gerry's answer, by associating the numerals, 411/137 with the ones I think are their alternate-universe analogs in the sense that 316 abstractly becomes twice 142, and the cubes of its parts lead back to it in the way that the squares of particular parts of 411 led back to 411, which is three times 137. But, the thus cubes of the parts of 316 can't thus directly lead back to 316, which is where the blurring above, of the operations, allows for some common solutions. And a way to move beyond the 3 X's, squared 411/137 stuff, to the 2 X's, cubed 316/142 stuff, and, beyond. What if the electromagnet fine structure is only one part of the above, specific question, where the exponent is 2, and, the multiplier is 3, as switched for the alternate thus structure, all within a broader range of other thus dimensionless, universal constants? Perhaps, the thus gravitation constants involve the exponents/multipliers of 4/5, and, 5/4, given that gravity is more of a "neutralized charge", and, that it wouldn't make much sense to try it with 0/1, and, 1/0, because thus multiplying by 0 would be pointless, as would be exponents of 1 on the thus parts of the numerals. On the end, I included several numerical and other examples for taking the numeral 142 as the alternate-universe analog to 137 for the averaged-out reciprocal of the constant. But, no, I didn't blindly do a lot of the straightforward calculations above that Gerry M. briefly elaborated on. The point of the self-similar square sums concatenation was only one way, I think, to prove out a bit of physics by the numerals alone. That it's pointless to debate, then, necessarily unendingly, the physics of a theory of every thing. Such a theory, if right, should be thus self-evident. Incidentally, I don't honestly know the reason that the question was initially closed. Nor that there was some further resistance, given that, at its core, it was merely another math question, but with a funny title. Regardless, no one controls, let alone accurately predicts, the long-term course of science. All that we can do, with certainty, is to close ourselves off from it. So, if you're reading this, I suggest that you copy and paste it to read it at your leisure, later on, before its summarily deleted. I've posted only four things, before, of which two were deleted. The last time, to do with the New Year's fun question about rewriting the numeral, 2023, which I put in terms of 2's that were upside-down 7's. But, supposedly, I "didn't answer" the question. Nothing like a good effort to render an exercise's outcome moot. If it is 142, then 316 is the numeral which goes with it in the way that 411 goes with 137. 411 = [3 X (4^2 + 11^2)] = (3 X 137), and, [-(4^2) + 11^2] = {[70 / (3 - 1)] X 3} --->to (3 X 137) = 411. And, so, 316 --->to (300 - 16) = 284 = 2 X 142 --->to 142..142 --->to, to say, {[1 / (-4 + 2 - 1)] + 42]} = [(3.333333333333333333333333333... )^3 + (1.666666666666666666666666666... )^3] --->to 316. As well, (3.333333333333333333333333... )^3 = 37.0370370370370370... , with the 7 = [1 + 6]; and, (1.666666666666666666666666... )^3 = 4.629629629629... , with the 2 = [1 + 1], and, the 9 = [3^2]. Again, it's the 3 X's with squares for the 411 to get back to itself, and, the 2 X's with cubes for the 316 to get back to itself. Furthermore, 316 = (2 X 158), but, (2 X 185) = 370 = 10 X 37 --->to 137 by reversing the last two digits; and, 411 = (3 X 137), but, (3 X 371) = 1113 = (114 + 1000 - 1) --->to 114..411 --->to [204 / (4 + 2)] = 34 = (2 X 17), but, (2 X 71) = 142. As if 411/137, and, 316/142, are thus direct opposites. As well, 142 --->to (100 X 4 + 2) = 402 = (3 X 134), but, (3 X 431) = 1293 = (1300 - 7) --->to 137. When I first realized that there must be a numeral like 411, with 137, to go with 142, I added a zero to 42 to get 420 = (410 + 10) --->to 411, and, then, a zero to 37 to get 370 = (310 + 60) --->to 316 --->to (300 - 16) = 284 = (2 X 142). The idea was to to have 410, and, 310, plus something, and, then, I noticed that the 16 had to be subracted to lead to (2 X 142) = 284. It could be that there is the entire range of alternate thus numerals as 316 for 284, on the 316-type sides of things. (Such may be imaged as extending down to -9, or -10, at which point the regular, staid version of numerals take over again.) Next, I noticed that 1 added to 316 forms 137, but with its digits out of order, and, that 1 subtracted from 142 forms 411, but with its digits, too, out of order. And, that, by adding the 1 from the front to the end, 137 --->to 308 = (2^2 X 7 X 11) --->to 2711 --->to [(2 X 7)^2 + 11^2] = 317, so that 308 = [300 + (1 + 7)] --->to 317 is, in an abstract sense, half way between 316, and, 137. Similarly, 142 --->to 403, by shifting a 1 from the 3, is (13 X 31) = [(-1 + 14) X (41 - 10)] --->to 114..411 is half way between 411, and, 142. Other ways to write things from the 2 X's, cubed perspective of 316 to itself by way of 142, are 316, to (300 - 16) = 284 = 2 X 142 = [142 + (02 + 040 + 010^2)] --->to [(**10^2** + 04020) + (-2 + 4 + 1)] = (3^3 + 16^3) = 4123, back to 316; and, (2 X 142) = [(0100 + 40 + 02) + (002 + 040 + 100)] --->to {(001 + 04020) + [020^3 + (04 X 0010)]} = 4069 = [(-3^3) + 16^3], back to 316. I phrased the above in symmetrical fashion about the arrows, with the same number of zeroes on their corresponding numerals. Interestingly, the exponents of 2 on the 10's, in the 4123-part, and, the exponent of 3 on 20, in the 4069-part, may have to do with these numerals being in the middle of their sections. The 2 of the 20 is in the middle of 142..241, and, the 1 of the 10 is in the middle about the first arrow above. In the line of (41^2 - 1^2) = 1680 = [(01040 + 200) + (002^2 + 04X010)] --->to 142..241, the 002 is to the exponent of 2 is in the middle of the 142..241 part on the left side of the arrow. But, then where is the corresponding exponent of 3, given that it makes sense to have an exponent of each of 2, and 3, for both the 3 X's, squared stuff, and, the 2 X's, cubed stuff, to mesh? There is no exponent on either of the digits of 137, in the lines, 411 = [3 X (4^2 + 11^2)] = (3 X 137) = 411, and, [-(4^2) + 11^2] = {[70 / (3 - 1)] X 3} --->to (3 X 137) = 411. Ah, but both exponents on the digits of 142 that are the 10's, and the 20, on the 316-side, of thus things above, were in its two lines that worked out, which went from 316, and, back to 316, and, so, the similar sort and degree of exponents for the 411-side should be in its two lines that didn't work out except as expressed, in a cross-over sense, by the 2 X's 142 stuff. As with the exponent of 3 on the 10 in the part about the arrow in {[(00100 + 40) X 2] + ( 2 + 0400 + 10^3)} --->to [(100 + 40 + 2) +(2 + 040 + 0100)] = 1682 = (41^2 + 1^2). The 1680-part goes with the 1682-part, the same as the 4123-part goes with the 4069-part, and, the two thus pairs of parts yield the exponents of 2, and 3, only once each in the same way. Seems like a good pattern to hang onto, that it's the smaller sums, and differences, of the thus squares, and cubes, that naturally work out in terms of their own 411/137, or 316/142, stuff, but, with the larger ones crossing over into the other stuff's territory. And, that the natural flow or pattern for the exponents to match up as above, if possible, happens by the smaller sums, and differences, of the cubes, but, by the larger sums, and differences, of the squares. And, to let the smaller sums, and differences, of the squares to the form 3 X's one 137 rather than by combinations of the digits of 137..137..137, analogous to (2 X 142) going by the combinations of its digits in two separate 142's to add up to its sums, and differences, of cubes. In the lines, 411 = [3 X (4^2 + 11^2)] = (3 X 137), and, [-(4^2) + 11^2] = {[70 / (3 - 1)] X 3} --->to (3 X 137) = 411, there was only one form of 137 allowed, which was directly multiplied by 3. Another way is 316 --->to (31^3 + 6^3) = 30007 = (30001 + 6) --->to 316, but it doesn't thus naturally go back to 316 by way of 2 X's the 142. More abstractly, [2 X (30001 + 6))] = 60014 = (60013 + 1) --->to 1/316, and, that (2 X 316) = 632 = [(601 + 30) + 1] --->to 1/316, where (2 X 316) --->to 1/316. Similarly, 411 --->to (41^2 - 1^2) = 1680 = {40^2 + [10 X (1 + 0^0)^3]} --->to 411, and, 316 --->to [-(3^3) + 16^3] = 4069 = [(4 X 10^3) + 69] --->to [4 X (69 + 10)] = 316. And, that [(1 + 4 + 2)^3 + (-2 + 4 + 1)^3] = (7^3 + 3^3) = 10 X 37 --->to 137, but, [(1 + 4 + 2)^3 - (-2 + 4 + 1)^3] = (7^3 - 3^3) = 316. From the other way around, [(7 - 3)^2 + (-1 - 1 - 3 + 7)^2] = 20 = 10X(4 - 2) --->to 142, but, [(7 - 3)^2 - (-1 - 1 - 3 + 7)^2] = 12 = [(1 + 1)^4 - 4)] --->to 114, or 411. It would be nice, in a way, if every thing matched up, but, then there wouldn't be any rules for, or, then, value in, choosing the values of a, and b, in the question. Say, by blindly putting 316 --->to (31^3 + 6^3) = 30007 = [(1 + 40^2 + 2) + (1 X 4) + (2 X 14200)] --->to (142 + 142 + 142) = 426 = [(3/2) X 284] = [(3/2) X (300 - 16)], back to 3/2 of 316. Or, by putting 316 --->to (30007 - 1) = [(3 X 10000) + 6] = {(-1000 + 4X20^3) - [10^3 - (4 + 2)]} --->to (2 X 142) = 284 = (300 - 16) /to 316. Even something as seemingly straightforward as (41^2 - 1^2) = 1680 = [(1370 + 137 + (103 + 70)] --->to 137..137..137, now with the digits of three 137's to work with, is just too much of a stretch on the 3 X's, 411/137 stuff to be a significant relative pattern by a simple compromise between it, and, the 316/142 stuff. Then how easy it would be to merely substitute 41 = (1 - 30 + 70), or (2 + 40 - 1), and, 1 = [√(7 - 3) - 1], or (-1 + 4 - 2) into (41^2 +/- 1^2) to obtain either of those two sets of 137, or 142, from those sums, and differences, of squares, respectively. Or, from 31 = [(2 X 4^2) - 1], and, 6 = [2 + (4 X 1)] substituted into (31^3 + 6^3). Again, the idea is to find out how to have particularly the 2 X's, cubed stuff of 316/142 actually work at all, and, in a manner similar to the 3 X's, cubed stuff of 411/137. And, next, the other way around for the latter to take on the natural rules of the former. The latter stuff forces the former stuff into a 2 X's 142-way of making the former sums, and differences of, cubes get back to 316, but, the former stuff, which isn't inherently about sums, or differences, of squares, forces the latter stuff to lighten its rules a bit to allow for more of its own thus solutions, as with also [-(4^2) + 11^2] = {[70 / (3 - 1)] X 3} --->to (3 X 137) = 411. In specific, the primary roles of the values of 2, and 3, are reversed from the former to the latter. In general, the rules (for almost randomly, recreationally working with the digits of numerals) may vary according to the values of the multipliers, and exponents, here as 2, and 3, for the squares, and, 3, and, 2, for the cubes. It looks like the exponents may allow for also their reciprocals. More importantly, just as the rules for sums, and differences, for squares are inherently firmer than for cubes, it's harder to define the 142-side of things, to begin with. 137 as a sum of squares comes before thus 153, and, after thus 125, but, where does 142 naturally, let alone inherently, fit into any sequence of sums, or differences, of squares, or cubes? It doesn't, and, so, it follows an entirely different pattern of its own, in the sense that it works with digits of groups of two, possibly more, 142's instead of a multiple of 3 of one 137 properly formed. But, it very much does appear that 316 goes with 142, as does 411 with 137, and, that the former goes with the latter. But, what happens at the in-between cases, when the result of squares, or cubes, isn't a thus multiple, or divisor, of 316, or 411? 411--->to (41^2 + 1^2) = 1682 = (2 X 29^2) = {2 X [-4 X (1 + 1) + 37)^2} --->to 241..137; (41^2 - 1^2) = 1680 = 16 X 105 = (80 X 21) = [(2 X 4 X 10) X (1 X 3 X 7)] --->to 241..137. Similarly, 316 --->to (31^3 - 6^3) = [(2 + 4 - 1)^2 X (13^2 X 7)] --->to 241..137; and (31^3 + 6^3) = (1 X 30007) X (-2 + 4 - 1) --->to 137..241, and, (31^3 + 6^3) = (1 X 30007) = [1 X (37 X 811)] = {1 X [37 X (411 + 400)]} --->to 137....411..114. And, there is (3^3 + 1.6^3) = [(7 X 3 + 10) + {1 / [(3 + 7) + {[.02 / (4 - 1)] - .01 + .42}]}] --->to 731..137....241..142. And, (3^3 - 1.6^3) = [(-7 + 3 X 10) - {1 / [(3 + 7) + {[.02 / (4 - 1)] - .01 + .42}]}] --->to 731..137....241..142. [-(3^3) + 16^3] = 4069 = (-1 + 4070) --->to 147 = (1 X 3 X 7^2) --->to 137; with (41^2 - 1) = 1680 = [(10 X 4)^2 + 2 X 4 X 10] = (1700 - 20) --->to 172 --->to (2 X 71) = 142. And, [(6.666666666666666666666666 ...)^3 + (1.333333333333333333333333... )^3] = {371 - (1 X 73) + [(1 X 2) / (4 - 4 + 2 + 1)]} --->to 371..173....124..421. Now to get back to what the 2 X's, cubed stuff of 316/142 directly has to do with the 3 X's, squared stuff of 411/137. The further idea is to find values a, and b, such that, for example, 411 = (3 X 137) in terms of things reducing to this in the same way that they did with the 316/142 stuff, but with the latter expressed in terms of the 142's written out twice instead of once . But, for the sake of the one stuff crossing over into the other stuff, to learn more about the rules for (almost randomly) throwing these digits together to make the sums, etc, of squares, etc, how can three of the 137's as 137..137..137 thus become the 2 X's of the 316-stuff? First up is [(4.444444444444444444444444... )^2 + (1.111111111111111111111111... )^2] = {(7 X 3) - [1 / (-1 + 3 + 7)^2]} --->to 731..137(..137), from 3 X's, to 2 X's. The 4, and 11, were written in the doubled sense of the 142's of 3, and 16, that was by extending the 3, and 6, by repetition, out from the 1 in the middles. That shows that to make it work by the 3 X's, squared stuff brought into 2 X's, cubed stuff, then one of the three terms in the 3 X's, squared stuff must be squared. Which seems to be the only natural way, if any other way, to express the 411 stuff that takes on also the various characteristics of the 316-stuff. Perhaps the best example to show that the 3 X's, squared stuff of 411/137 may be rewritten as the 2 X's, cubed stuff of 316/142, in two parts, is 411 = {(1 + 3 + 7) + [(7 X 3) -1]^2} --->to 137..731. And, yet another of the solutions provided, by Gerry, to the question, which becomes 803 = [11 X (08^2 + 03^2)] = (11 X 73) = ((1 + 3 + 7) X {[7 X (3 - 1)] + (1^2 + 3^2 + 7^2)}, or, 803 = [(-1 + 3 + 7)^3 + (73 + 1)] --->to 137..731, with only two 137's, which, again, comes to involve either a set of squares, and/or another exponent, so that the 3 X's, squared stuff of 411/137 may be written in terms of the 2 X's, cubed stuff of 316/142, with two parts such as 137..731. For one more example, I guess, of many, there is 411 = (70^2 - 67^2) = ({[(-1 + 3 + 7)^2 - (7 + 3 + 1)]^2 - {(-1 + 3 + 7)^2 - [7 X (3 - 1)]^2}) --->to 137..731 and 137..731. It goes with 316 = (80^2 - 78^2) = {[(13 - 7) + (73 + 1)]^2 - [(-1 + 3 +7)^2 - (7 - 3 - 1)]^2} --->to 137..731....137..731, as 4 = 2^2, instead of 3 sets of 137. So, it looks like also the exponents on themselves in tower form is acceptable. In general, I guess, (-1 + 3 + 7)^2 = (1 + 4 - 2)^(2^2) = 81. There is also 316 = {[(1 + 4 - 2)^(2^2) - (-2 + 4 - 1)]^2 - [(1 + 4 - 2)^(2^2) - (-2 + 4 + 1)]^2}, and, 316 = {[(1^3 + 4^3 + 2^3) + (2 + 4 + 1)]^2 - [(1^3 + 4^3 + 2^3) + (2 + 4 - 1)]^2}, which tend to further blur things by the exponents of 2, instead of 3 for the 2 X's, cubed stuff of 316/142. Totally crossing over is 411 --->to (41^2 + 1^2) = 1682 = [1402 + 2 X (40 + 100)] --->to 142..241; and (41^2 - 1^2) = 1680 = [(1040 + 200) + (20^2 + 4 X 10)] --->to 142..241. 316 --->to (31^3 + 6^3) = (1 X 30007) --->to 137, or 731; and (31^3 - 6^3) = [(13^2 X 7) X (7 - 3 + 1)^2 --->to 137..731. How about 316 --->to 4123 = (3^3 + 16^3) = [19 X (7 X 31)] = {√(301 + 6) X [(6 + 1) X (1 + 30)]} --->to 316..613, but, 4123 = [4110 + (-1 + 14)] --->to 411..114? Or, (310 + 1 + 6) --->to 316 = [(73 + 100) - (73 - 1) - (73 - 1)] --->to 731..731..731? Something else, {2 X [(4.44444444444444444444444444... )^2 - (1.111111111111111111111111... )^2]} = 37.037037037037043703... . Of course, the 241 --->to 142, and, 37 = (10 X 3 + 7) --->to 137, directly. I substituted the 4 and 1.1, with each squared, for the 3 and 1.6 cubed, the latter which gave way to 142. As I got into it a bit earlier, I think that the numerals, for example, 308, 403, 704, and 407 are "gateway" numerals "between" 411, and, 316, and, so, also the two electromagnetic fine-structure constants. 407 = (4^3 + 0^3 + 7^3), like 371 = (3 ^3 + 7^3 + 1^3). 407 sheds a 1 from the 4, and, like 308 sheds 1 from the 8, to become 137. 407 --->to (4 X 7) = 28 = (14 X 2) --->to 142; 308 --->to (3 X 8) = (24 X 1) = [(3 X 10) - 6] --->to 241, and 316. 407 = (11 X 37) = [(1 + 3 + 7) X (7 + 3 X 10)] --->to 137..731; and 403 sheds a 1 from the 3 to become 421, or 124, and, 403 = (13 X 31) = [(-1 + 4^2 - 2) X (2 X 4^2 - 1)] --->to 142..241. As well, 704 = (2^2 + 7 X 100) --->to (2 X 71) = 142, and, 704 = (4^3 X 11)--->to 411. Furthermore, as I noted already, "... 1 added to 316 forms 137, but with its digits out of order, and, that 1 subtracted from 142 forms 411, but with its digits, too, out of order. And, that, by adding the 1 from the front, to the end, 137 --->to 308 = (2^2 X 7 X 11) --->to 2711 --->to [(2 X 7)^2 + 11^2] = 317, so that 308 = [300 + (1 + 7)] --->to 317 is, in an abstract sense, half way between 316, and, 137. Similarly, 142 --->to 403 = (13 X 31) = [(-1 +14) X (41 - 10)] --->to 114..411 is half way between 411, and, 142." Moreover, as demonstrated below to show how, say, thus things revolve around 704, there are the fractional equations to relate most of these numerals, in most ways, when not by the simple exponents of, say, 2, or 3. So, let's take a quick look at one case of many about the numeral, 704. In general it very much seems that thus things become focused about the numerals like 704, in terms of the further "inter-coincidences" of 411/316, and 137/142. The further away from, say, 704, the less well-defined thus things become between 411/316, and 137/142, and, the other thus relevant numerals about 704. With 704 = (1 / 85282283970828229709329922167760553035055088562322322426444162632986043716444498839169349906741804649545178312629555397**72727272727272727272727272727272727272727272727272**684155963862177533049550295036764939039364194800866931790398339627598892817711104) X (70^137 - 4^137), and 704 = (1 / 1433339346697710056724708001873551614860170873466951273021247041372596436742282691989919263882609510744905811900364937569602**2727272727272727272727272727272727272727272727272727**228575706994869793842739502117647297576308935476087738153367899778661266245336170496) X (70^142 - 4^142). A set of twenty-five 72's within the denominator's expression with exponents, 137, and, a set of twenty-six 27's within the denominator's expression with exponents, 142. But from where come the 72's, and, 27's, respectively? Perhaps, from 72 = (70 + √4) --->to 704. Or, another guess, say, an artificial "island of stability", in the sense that the first regular, "atomic magic numbers" add to (2 + 10 + 18 + 36 + 54 + 86 + 118) = 324 = 18^2 = [4^2 + (1 + 1)]^2 --->to 411, and, that the first regular, "nuclear magic numbers" add to (2 + 8 + 20 + 28 + 50 + 82 + 126) = 316, of which the 126 is replaced by 114 in the case of protons (instead of neutrons). Insofar as up to 118 = (4 + 114) --->to 411..114 elements of the periodic table, the regular elements. The first denominator has 250 digits, and, the second denominator, 260. Perhaps, because 250 = (25 X 10), and, 260 = (26 X 10), with the 10's --->to **1**'s as the first digits of 137, and 142. The four **3**'s, and, four **4**'s, together, in the denominators above of 142, and 137, as the middle digits of 137, and 142, respectively. The chance of these, and only these, two different digits as less than 1/100, combined with the chance of two, and only two, thus four-of-a-kind's of, which is less than 1/4^2 in strings of digits of that length, to an overall chance less than 1/1600, let alone the chance of also both of the 72's, 27's, as the end digits of 137, and 142, and, of the 1's from the 10's. As well, the 137-denominator's smallest factor was 2^131, and, its largest was 204 = (204 X 1) digits --->to 142. 131 = (-11 + 142) --->to 1/142, and 1/142 by breaking the two 1's into separate numerators. The 137-denominator was the one to contain the 4444-run for the middle digit 4 of 142. The 142-denominator's smallest factor was 2^136, and, its largest was 69 = {[(7 + 30 + 100) + 1] / √(-3 + 7) digits --->to 731..137. 136 = (-1 + 137) --->to 1/137. The 142-denominator was the one to contain the 3333-run for the middle digit 3 of 137. Incidentally, the largest factors are: 514326796135736895120460142673051831066041223265200309685158844534359838150610822900408943344936522280963286550169127595496011065898376945285661212034176736287165772490986373978333211257192096042560572551 (204 digits), and 376142321484418303677130304058222651666489639564016682778979735427557 (69 digits), respectively. There may not even be a requirement for the larger exponents. When thus working with the digits of those, and other, numerals, it may be, say, that **142** = {(1 / 2159016100347461905325223736872322283514248209961272895096224) X [**61^(30 - 1 + 6) + 3^(30 - 1 + 6)**], and, **137** = [**4^(-1 - 1 + 4) + 11^(-1 - 1 + 4)**]. Interestingly, the above doesn't seem so easy for, say, 316 written as (3^X + 16^X). Perhaps, for larger thus rearrangements of the digits of 316 as exponents on 3, and 16. But, for sure, 411, and 316, are even thus further interrelated. 411 = (-1 / 292581382785534914866756968188193808434984757840316737585126468568361231825543392617790166789265384274056504397949185481590598192266373914749525582968549546007958668898660328722294272154336738011923528548159705325848996390576069031829067457519218795662621818783093548930810829439457048605237777411650038874699384887330926003715) X (4^316 - 11^316). But what can the run of four 7's above mean? Next, are a pair of pairs of numerals that turn inward on themselves, exist on their own, in a different fashion from the self-similar sums, and differences, however, of squares, and cubes. Again, the idea is to work with the stuff of dimensionless constants, as free-floating numerals that then happen to find themselves by finding each other. 144648 = (861 X 168) = (492 X 294); 185472 = (672 X 276) = (384 X 483); and, 9949716 = (3852 X 2583) = (1476 X 6741); 16746912 = (6552 X 2556) = (4473 X 3744). Firstly, a closer look at the individual numerals above. 861 = (3 X 7 X 41) = [(7 / 4) X 492]; 168 = (2^2 X 42); 492 = (2^2 X 3 X 41) = [(3 X 4) X 41]; 294 = (7 X 42) = [(7 / 4) X 168]; 672 = (2^5 X 3 X 7) = [(7 / 4) X 384]; 276 = (2^2 X 3 X 23); 384 = (2^7 X 3); 483 = (3 X 7 X 23) = [(7 / 4) X 276]; and, 2583 = (3^2 X 7 X 41) = [(7 / 4) X 1476] = (3 X 861); 3852 = (2^2 X 3^2 X 107); 1476 = (2^2 X 3^2 X 41) = [(3^2 X 4) X 41] = (3 X 492); 6741 = (3^2 X 7 X 107) = [(7 / 4) X 3852]; 4473 = (3^2 X 7 X 71) = [(7 / 4) X 2556]; 3744 = (2^5 X 3^2 X 13); 2556 = (2^2 X 3^2 X 71); 6552 = (2^3 X 3^2 X 7 X 13) = [(7 / 4) X 3744]. The two, numerals of the first product have the sums of 9, by (1 + 8), around their 6's, which are the middle digits of the two, numerals of the second products; and, the two, numerals of the the second product have the sums of 6, by (2 + 4), around their 9's, which are the middle digits of the two, numerals of the first product. And, so, because I group my numerals by the digits primarily by {2, 7}, and, {6, 9}, I looked up what had to be the other thus pair of products. At which point, I found that there were another two pairs of thus products fairly close by. Which have sums of 8 = 2^3, by (6 + 2), around their 7's, which are the thus second pairs' middle digits, and, sums of 7, by (3 + 4), around their 8's = 2^3, , which are the thus first's middle digits. Furthermore, the digits in each numeral of the first two, sets of pairs of products add to 15 = (6 + 9) = (6 + √9 + 6) --->to 696; but, it's 18 = [2 X (7 + 2)] --->to 272 for the second two, sets of pairs of products. Lots of other interconnectedness, as well, with those numerals. For example, in 185472 = 672 X 276 = 384 X 483, 672 = 21X32, and, 276 = 12X23, with the digits of their factors inter-reversed. Similarly, 384 = 12X32, and, 483 = 21X23. And, all of those involve the digits of 12 versus 23. But the four-digit numerals above are a bit harder to figure out. I noticed that the outside pairs of digits, of (4473 X 3744), plus 11 become the middle pairs of digits of (6552 X 2556), and, that their inside digits, 73, and 37, minus 11 become its non-middle pairs of digits, of 6, 2, and 2, 6. And, that the outside pairs of digits, of (1476 X 6741), plus 11 become the inside pairs of digits of (3852 X 2583), but, going from 76, to 38, requires subtraction of (2 X 19), like going from 67, to 83, requires addition of 16, which sort of mirrors what happens when going from 65, to 73, of (6552 X 2556), which requires addition of (16 / 2), but, going from 56, to 37, requires subtraction of 19. As well with (25 + 42) = 67, and, (52 + 24) = 76. More interestingly, (25 + 49) = 74, and, (52 + 94) = 146 = (2 X 73). And, the second pair of pairs, too, has thus above reversed digits of factors, but for (8 X 13) = 104 reversing to 71, and, the other way, 107 reversing to 41. And, there, again, is the 407 sort and degree of stuff, which is, perhaps, the reason that the second set of thus pairs is the 272-set. But, for lack of a better explanation, it seems, to me, like (104 - 71) = 33 = (√9 X 11) --->to 911, and, (107 - 41) = 66 = (11 X 6) --->to 116, which turns around to make 911. And, reading backward, but, next, forward, in the previous sentence, there is 1/741, and, 1/741, respectively, suggestive of (741 / 741) --->to 741..741 to get rid of the lone 1's up front. In further jest, the text number for the US suicide call center. Incidentally, the 316/142 stuff is quite a bit more subtle than the 411/137 stuff because there isn't the same degree of conceptual or other exactitude to it. It's cycles fit in with those of the 411/137 stuff, but, only after developing the other first, and, next, determining where. Secondly, for the most interesting part by far, there are the following bits. 144648 = (861 X 168) = (492 X 294) = {[(41 + 43) - 2] X 42^2} = = {(41 + 43 - 2) X 42 X [(-3 + 5) X (-1 + 7^2 - 3^3)]} --->to 41..43....24..42....351..173. 185472 = (672 X 276) = (384 X 483) = {[(-3^3 + 7^2 + 1) X (10 + 5 - 3)] X (2 X 42) X (1 + 4 + 3)} --->to 371..153....24..42....41..43. 9949716 = (3852 X 2583) = (1476 X 6741) = ((41 + 43 - 2) X 42 X {[0^2 + (-001)^03 + 5^02 + 3^2^0] X [+7^3^0 + (-10)^02 + 0]}) --->to 41..43....24..42....153..371. 16746912 = (6552 X 2556) = (4473 X 3744) = ({-(0^0^3) X -(010^2^0) X (-7) - (3 X 3) + (5^02) X (010^3) X (0^0)} X (2 X 42) X (1 + 4 + 3)) --->to 173..351....24..42....41..43. That very last bit involves a 7 that mirrors a 5, by a 7 = ᘔ that is flipped over, from left to right, to form the 5. I padded the 173..351 part with zeroes to show that it works out the right way. In general, I think that the overall numerals may be parsed in a fashion to show a supposed structure of the universe about dimension-42, which is doubled-double sided as (2 X 42) --->to 24..42. A double-double form unto itself, as [(√1 / √4) + 43^2] = ((1/2) X {42^2 + [√([4^0]00) - √(2^0)] X [-√(2^0) + √(400)] + 42^2}). This one takes a bit of sorting out to see that dimension-42 fits between 41, and 43, but, in the sense that it's mirrored on itself by one of its thus sides on the 41-side, and, by the other of its thus sides on the 43-side. The 41 is read the other way through square roots, and, so, the part of the 42-side that isn't 42^2 to go with the 43-side, of 43^2, should involve square roots. One of 42's sides goes with [√1) / √4)] --->to 41, but, the other half with 43^2. The dimensions- 41, and 43, are fairly straightforward, I think, with ours primarily at dimension-41, and, the alternate's primarily at dimension-43, primarily between the field dimensions- 40, 42, and 44. But, the index of 153/173 is not so simple. Those, too, the 153/173, are sums of squares in thus sequence, but which jointly form an index of their own. Hence, they come before, and, but, after, the thus square sum of 137, because 137 --->to 1[3^2]7 = 197 --->to 137, the 197 of which is the thus sum after 173. There is a thus analogous pair form of index for the 142 sequence, which, again, doesn't per se follow a sum, or difference, of squares, or cubes, etc, in which it progresses as 137/142, and 197/214. Furthermore, 137 = (100 + 37) tends to go with 142 = (100 + 42), and, so, the index of 153/175 is offset by one dimension ahead of, and after, dimension-42. I think that we are of the more- paradoxical or conflicted dimension-41, and, the alternate universe, of the less- paradoxical or conflicted dimension-43, but, primarily within dimensions- 40 to 44. As a theory of everything that has double the dimensions of regular M-theory, a double M theory, which, in turn, has double the dimensions of the "string theories" that M-theory attempts to unite, which has ten dimensions (plus one of time, or, eleven spatial dimensions in all), of "string theories", doubled to twenty (plus one of time, or, twenty-one spatial dimensions over all), doubled to forty spatial dimensions (plus one of time, or, forty-one spatial dimensions over all). Going from dimensions- 40 to 44 is a bit harder to explain, still. Dimension-42 is the more-gravitational dimension, and, so, it's in the middle of our effective dimensions, from dimension-40 to dimension-44. Dimension-40 is the more-electromagnetic dimension, but, the two overlap, here and there, incidentally, which goes to further explain the 153/173 bit about coming before, and, but, after dimension-42. And, there's always a reversal of either the 153, or, the 173, when they are numerically thus expressed. Lot of strange numerical properties with those numerals, as well. Up next is another striking example of the above. [-(2^0) + Fibonacci-42] = (-1 + 267,914,296) = {[(0 + 03^02) X (0 + 5) X (001 + 10) X (07^02^0 + 300)] X (41 X 43)} With the 371/153, and 41/43, fully separated, as is the (2 X 42). I guess that without the (2 X 42) between the sets of surrounding dimensions, and indexes, there was nothing to bring either onto itself as with dimension-42. Incidentally, isn't the Fibonacci stuff about population growth that springs from itself, asexually? Brings to mind, Penrose tilings of the plane, etc, and, the internal forces of a point, and some atoms. Lastly, I want to a bit make the case for the 27/96 numerals, which, I think, when written as ᘖᘔ/ᘔᘖ, are necessary and sufficient, in the sense that thus numerals that arise out of themselves ultimately involve all possible calculations in each calculation. ᘖᘔ = 27, but, it also rotates to ᘔᘖ = 72. ᘖᘔ as naturally contains a 9 with a bar on bottom, and, a 6 with a bar on top, and, so, the 96 is already within the 72. And, for example, (69 - 27) = 42. In that way, (69 + 27) = 196 = (-1 + 197) --->to 1/1[3^2]7 = 1/137, and, (69 + 72) = 141 = (40 + 101) --->to 411. And, (96 + 27) = 123 = (3 X 41) --->to 34..41, which, again, lie about (96 + 72) = (2^2 X 42) --->to 24..42. {[96 + (2 + 70)] / 2} = {[96 + (3 + 70 - 1)] / √(7 - 3)} = (153 - 69). As well, a bit more intricately creative stuff like, 137 = (2 X 69 - 7^0). Even on a basic level, those numerals seem more related. (4^3 + 2^3) = 72 = [100 - 14X(1 + 1)] --->to 114..411, and, 72 = [2 + (7 X 10)] --->to (2 X 71) = 142. (4^3 - 2^3) = (1 X 056) = [2 X (14 X 2)]. (4^4 + 2^4) = 272; (4^4 - 2^4) = 240 = (24 X 10) --->to 142. (4^5 + 2^5) = 1056 = {1961 - [(-1 + 6) + (9 X 100]}. (7^4 + 3^4) = 2482 = [-609 + 3700 - 609)]; (7^5 + 3^5) = 1705 = [1961 - (2^7 X 2)] = [1096 + (609 X 1)]. So, who is to say that the thus centered hexagonal numbers, in terms of their {2, 7}'s, and {6, 9}'s, and, say, the sequences of sums, and differences, of squares, and cubes, and beyond, didn't naturally lead to the way that most persons express, and go on to write about, most numerals? What if the formula for the nth term of the centered hexagonal numbers, with formula for the nth thus term, (3n^2 + 3n + 1), self-reduces to {(√3n)^2 + (3n)^1 + [(3^2)n]^0}, with the numerical pattern from √3, to 3, to 3^2, while the exponents go from 2, to 1, to 0, respectively, as in the exponent of 1/2 against 2, the exponent of 1 against 1, and, the exponent of 2 against 0, with each being a form of inter-reciprocation? 0 flips around 1, to 2, in an additive way; 1/2 flips around 1, to 2, in a multiplicative way, and, with 1 flipping around 1, to 1, as both additive, and multiplicative, the thus identities, where the other two forms of reciprocation meet. It very well could be, at least on a subconscious, or even unconscious, level, that {6, 9}, and {2, 7}, expanded as {0, 3, (5,) 6, 9}, and, {1, 2, 4, (7,) 8} have become the way to separate out the base-10 digits, given that (0 + 3 + 6) = 9 with 9 reverting to 5, in the middle, and, (1 + 2 + 4) = 7 with 7 extending to 8, on the end, in the same way, dimensionally speaking, that (0 + 1 + 2) = 3 with 3 reverting/extending to 4 as both middle/end. Certainly, there is a universal fascination with such numerals, seemingly with whole religions built on them. To the extent that even the atheists perpetuate it by their vocal, steadfast such rudimentary denials. But, the digits really, in theory, and, actually, in practice, did come to be written almost to the extent of the such numerals asas ᘖᘔ/ᘔᘖ. As I wrote in the comments from about over 3 years, and 7 months, of a different, far less formal way to perceive the numerals ago - which I never kept track of until a couple of weeks ago, when it hit me that I had an answer to put up. There is quite a bit more fairly solid reasoning, in general, and in specific, as to the how, why, and, etc, of it, but, time doesn't really permit me to further it here. Finally, a few more specific numerical examples of the 411/316, and 137/142, stuff. (316 X 411) = 129876 = (130000 - 124) = [130000 - (24 + 100)] --->to 137..241 by taking the 2, in the middle, as ᘖ --->to ᘔ as an upside-down 7, to fit in with both the 137, and, the 241 --->to 142. I guess that it was a thus ᘖ --->to ᘔ, in the sets of symmetrical calcuations earlier about going from 411, to 137, and, next, 316, to 142, and, then, back to each, that was either to the exponent 2, or, 3, so that the center numerals of each of 137..731, and 142..241, were thus taken into account, in terms of the symmetry of the 2's/7's, from within, and, of the 1's about the arrows. That both exponents were applied, it makes sense that the 2^2's involved, involved also a 7, for the 137's, by way of (41^2 +/- 1^2), which didn't have a 7^2. The ᘖ --->to ᘔ sort and degree of numerical reversal/flipping/rotation happens a lot, the same as with the 6's --->to 9's. Another instance is [(3.333333333333333333333333... )^3 - (1.666666666666666666666666... ^3)] = 32.40740740740740740... , by which the abstract version of it may be (1 X 37.40740740740740740... , with another upside-down 7 for a 2. I mean a lot of 2's like that come up whether in similar spots. So, let's see what happens when turn the digits of the numerals of the product above around, in various ways. Something should happen, right? (613 X 114) = 69882 = (-118 + 70000) = (59 X 2 + 70000) = [(-10 + 9 + 6 X 10) X 2 + 7 X 10000] --->to 1961..1271, this time by the two numerals moving apart off the 1, in the middle, instead of the 7 = ᘔ off the 2 = ᘖ. So, in another abstract, or pure math, way, the numerals may overlap each other as with 4114 --->to 411..114. (316 X 114) = 36024 = [24 + (-10 X 100 + 37000)] --->to 241..137, or 731..142. And, (411 X 613) = 251943 = (-8057 + 260000) = {-[1 + 4 X (1 + 37) X 53] + 260000} = {-[7^0 + (2^2) X (1961) + 2 X (06 + 100)] + (10000 X 6 + 0200000)] --->to 2....1961..1691....2061..1602....7. Anyone here born on 2 July, 1961, to live to 2 July, 2061, upon the return of Halley's comet? (114 / 613) = 0.1859706362... , and, (613 / 114) = 5.3771929825... --->to [(114 / 613) + (613 / 114)] = 5.5631636212... = about ([(4 + 1) + (0.1 + 0.4)] + {0.063163 + [0.0000003106 X (3 - 1)]}) --->to 411..114....63163..31631. I didn't try for any more decimal places, and, so, I have no idea how more of the decimal expansion goes. (411 / 316) = 1.30063291139... = about 1.3006 = (0.0006 + 1 + 0.3), and, so, 411 = about [316 X (0.0006 + 1 + 0.3)] --->to 316..613. Furthermore, if fill things in a bit on the ends of those numerals, for various forms of symmetry, by 1.300631, and, 316.31, then their product becomes 411.40259161, with 411.4 up front. (1.300631 X 316.13) = 411.16847803... , with 411.1 up front. Interestingly, when the 411 stuff is written up as cubed, but again in a partial way, [142 - (0.007 + 0.3^2 + 0.1)] = {[(6^3 - 1.3^3) - [4^3 + (1 + 1)^3]} = 141.803 --->to 142..731....613..411. As well, (3.1^3 + 6^3) = (142 + 103.791) = (142 + 103.7[3^2]1) --->to 142....137..731. And, (4.1^2 + 1^2) = (3.71 + 14.1) --->to 371..141. (411 - 316) = 95 = [19 X (6 - 1)] /to 1961; (411 + 316) = 727. (411^2 - 316^2) = 69065 = [(5 + 10 X 9) X 727] = {[(10 X 9 + (6 - 1)] X 727} --->to 1961..727; (411^2 + 316^2) = 268777 = [(300000 - 33000 + 10^3) + 777] --->to 333..10..777. (411^3 - 316^3) = 37872035 = [127 X 1 X (5 + 10 X 9) X (√9 - 10 + 50) X (1 X 72 + 1)] = {(127 X 1) X [10 X 9 + (6 - 1)] X [(-10 + 60) + √9 - 10] X (1 X 72 + 1)} --->to 1271..1961..1691..1721; (411^3 + 316^3) = 100981027 = {727 X [33333 + (501 + 90) + 77777 + 27200]} = (727 X {33333 + [1 + 90 + (600 - 100)] + 77777 + 27200}) --->to 727..33333..1961..77777..272. But, in the end, perhaps, [(-1 + 3700) / 137] /to One, or [(-1 + 3700) / 137] = (3 X 3 X 3). And, so, God is the One, as in (137 / 137), with (-1 + 37000) = (3^2 X 4111), and, (-1 + 370) = 369 = [√9]69 = (3^2 X 41) = (9 X 41) --->to 49194 = {[911 X (3 X 3 X 3)] + [(3 X 3 X 3) X 911]}. At least by our keeping the God stuff alive and well if only by our own thus fascinations, etc, in this, or another direction. Well, sometimes the best thing to get past something is to just accept it in whichever manifestation insofar as no one really, in theory, or actually, in practice, knows it all.
A tiny hint, though. Note that 1/142 is the electromagnetic fine-structure in the alternate universe, to go with 1/137 in ours. Interestingly, 411= 137*3*1, with the 411 formed by adding a 0 to 42, for 420, and, then, writing, 420 = (410 + 10) ---> 411 = (400 + 11). Similarly, 37 becomes 370 = (310 + 60) ---> 316, for 142*2*1 = 142 * √4 * 1 = 284 = (300 - 16) ---> 316. The sign on 16 changes to minus, in the case of 37, because the 10, on the end of (410 + 10), went to the start of (310 + 60). Likewise, but in reverse fashion, 316 ---> 37, and, 411 ---> 42.
Now if you think that the above is a coincidence, please consider further that 3*137 = 411, 2*142 = 284, which becomes the numeral, 316, and, that 137 = (4^2 + 11^2), with exponents of 2. So, for the interesting, primitive pattern above to hold, then something out of the ordinary, ie, other than by random chance, should happen with 316 parsed out like 411 to form 137 = (4^2 + 11^2), but, with exponents of 3.
Right off the "bat" there is (31^3 + 6^3) = (29791 + 216) = 1*30007 = 1*37*811 = 1*37*[(1 + 1)*400 + 11] ---> 137_11411. A very interesting numeral, and, with 29791 = (10000 + 19791) = (10000 + [1][3^2][7][3^2][1]) ---> 1/13731, and, 216 = 6*6*6. Or, with 197 = (196 + 1) ---> 1961. And, if parse the 316 out the other way, (3^3 + 16^3) = 19*(7*31) = [√(361) * (7*31)] = 4123 = [4110 + (-1 + 14)] ---> 411_114. Most importantly, 4123 = [(100 + 4020) + (-2 + 4 + 1)] ---> 142_241. For now, I will let the other thus simple, but profound, calculations as the relationships between 411, 284/316, and 216, to the reader. So, where did that 30007 come from, and, then, were it, what was the random chance of it?
The only two cubes that sum to 30007 are 31, and 6? The only two cubes that sum to something with 0's, 3's, and, 7's? I didn't see any more thus solutions, for the initial numerals of the form 300007, with more 0's, but, I did find one for 37000 = [30^3 + (-40)^3] = [(4*10 - 10)^3 + (-4*10)^3] = [(-10 + 42 -2)^3 + (-4*10)^3] ---> 142_214. Note also that 2*(1*30007) = 60014 = (60013 + 1) ---> 1/316, and, that 2*316 = 632 = (631 + 1) ---> 1/136 = 1/(-1 + 137) ---> 1/(1/137) ---> 137. And, 632 = [(6 + 10) + 600 + 16] ---> 61616, or, 632 = (6*100 + 2*16) ---> 612_216. But which uses negative numerals, more in line with differences of squares, which are quite common compared to sums of squares.
Oh, I see the remainder of the pattern now.
The first set:
(4^2 + 11^2) = 137, or 731; (11^2 - 4^2) = 105 = 21*5 = (1*3*7)*(7 - 3 + 1) ---> 137_731.
(41^2 + 1^2) = 1682 = [1402 + 2*(40 + 100)] ---> 142_241; (41^2 - 1^2) = 1680 = [(1040 + 200) + (20^2 + 4*10)] ---> 142_241.
The second set:
(31^3 + 6^3) = 1*30007 ---> 137, or 731; (31^3 - 6^3) = 29575 = 13^2 * 7 * 5^2 = (13^2 * 7)*(7 - 3 + 1) ---> 137_731.
(3^3 + 16^3) = 4123 = [(100 + 4020) + (-2 + 4 + 1)] ---> 142_241; (16^3 - 3^3) = 4069 = [(1 + 4020) + (02^3 + 4*10)] ---> 142_241.
The reason that the 142's are doubled, in the sense that other things go straight to 137, and 137 as 1*30007, is that everything turns around, hinges on, dimension-42, so that it has to sides to it. But, 137, and 1*37, have the other advantage of working out concisely.
Just, noticed a few other things. Firstly, (411 / 316) = 1.30063291139 = about 1.3006, and, so, 411 = about 316*1.3006 = 410.9896. Furthermore, if fill things in a bit on the ends of those numerals, for symmetry, by 1.300631, and, 316.31, with two 31's around the 6, then their product becomes 411.40259161, with 411.4 up front. Another way, 1.300631*316.13 = about 411.16847803 . Secondly, for the few more skeptics among us, look at what happens when the numerals 316, and 411, are multiplied. 316*411 = 0129876. Notice the 9876 bit, after the 012 bit, almost as if a missing 345 = 5*69, or, 543 = 3*181, bit, of which I'm not quite sure about. Could be a funny rewriting of 345, and 543, to reflect going one way while the other, as in 35435 = 5*19*373 = 19*(6 - 1)*1*373*1 ---> 1961__137_731. But, there's no doubt about 316*411 = 129876 = (130000 - 124) = [130000 - (24 + 100)] ---> 137_241, again by taking the 2 as invertible by ᘖ ---> ᘔ. I guess that it was a thus ᘖ ---> ᘔ above, in the sets of calcuations, that was either to the exponent 2, or, 3, so that the center numerals of each of 137_731, and 142_241, were thus taken into account, but, because dimension-42 is inherently the centered, or gravitational, one, whereas dimensions- 37, and 97 = [3^2][7], and numerologically varied ones, represent the minimum, and maximum, or charged, ones. At any rate, not that surprising to me - I've seen quite a few of those - same as with the 6's ---> 9's. A sort of numerical trick based a lot in the centered hexagonal numbers. What was a bit surpising was how, and where, it happened, again. But, let's see what happens when we turn the above around. Something should happen, right? 613*114 = 69882 = (-118 + 70000) = (59*2 + 70000) = [(-10 + 9 + 6*10)*2 + 7*10000] ---> 1961_1271, whereby the numerals move apart off the 1 instead of the ᘔ off the ᘖ, but, this time, with also the 6's, and 9's. If take the 2 as a 9 with a special bar on the bottem, and, similarly for the 7 as a 6 with a special bar on the top, then the digits of the numerals share a point symmetry about the 1 that is overlapped in the middle. So, now I wonder what's to be written of (114 / 613), and, (613 / 411). And, 411*613, and, 316*114
(114 / 613) = 0.1859706362, and, (613 / 114) = 5.3771929825 ---> [(114 / 613) + (613 / 114)] = (5.377192985 + 0.1859706362) = 5.5631636212 = {[(4 + 1) + (0.1 + 0.4)] + [0.063163 + 0.0000003106*(3 - 1)]} ---> 411_114__63136_31631. Which, let's face it, is pretty darn amazing. I actually didn't try for any more decimal places, so, I have no idea how any more of the decimal exansion goes. Furthermore, how about 316*114 = 36024 = [24 + (-10*100 + 37000)] ---> 241_137, or 731_142? And, 411*613 = 251943 = (-8057 + 260000) = {-[1 + 4*(1 + 37)*53] + 260000} = {-[7^0 + (2^2)*(1961) + 2*(06 + 100)] + (10000*6 + 0200000)] ---> 2__1961_1691__2061_1602__7?
So, in summary, firstly, there is 411 = 3*137 = 3*(4^2 + 11^2), back to the 411, and, secondly, there is 316 ---> (300 - 16) = 284 = 2*142 = [142 + (02 + 40 + 100) ---> (100 + 4020) + (-2 + 4 + 1) = 4123 = (3^3 * 16^3), or, 2*142 = [(10^2 + 40 + 02) + (02 + 40 + 100)]---> [(001 + 4020) + (02^3 + 04*10)] = 4069 = (-3^3 + 16^3) ---> 316. Notice that I've written these in symmetrical fashion, with the same number of zeroes on each side, as with, say, reflecting about the 100's, and 001's, and, next, the 20's, or 02's, and, 40's, or 04's. What happens, this time, is that the exponent's, 2 (upside-down 7), and, 3, occur in the same line, instead of the same position, but, with the 1 in the 10 being squared, as if, this time, the 1's are a central numeral, but, which they are, given that the symmetry across the --->'s is over the 1's. But, things with the 142 are still doubled, ie, not so analytical in an exact, or conceptual, way. I was able to find a more exact way of relating the two main numerals, 137, and 142, by a third, 704, but, I still seek a direct thus way of relating them. Perhaps, by the end of this blog post.
The following doesn't really get there, but, does involve both 137, and, 142. I think that numerals with digits like 407 = (4^3 + 0^3 + 7^3) is a gateway between the two (fine-structure constant) numerals. It can go 371 = (3 ^3 + 7^3 + 1^3), by shedding a 1 from the 4, and, say, 704 = (2^2 + 7*100) ---> 71 X 2 = 142. Or, 407 = (-1 + 408) = (-1 + 400 + 2^3) ---> 142
704 = (1 / 14333393466977100567247080018735516148601708734669512730212470413725964367422826919899192638826095107449058119003649375696022727272727272727272727272727272727272727272727272727228575706994869793842739502117647297576308935476087738153367899778661266245336170496) X (70^142 - 4^142).
The bizarre thing about the two expressions above is the set of twenty-five 72's within the expression with exponents, 137, and, the set of twenty-six 27's within the expression with exponents, 142. Not in only one, but both, and, of the same digits but reversed. These are merely the denominators of simple, but relatively large, fractions. But, from where come the 72's, and, 27's, respectively. Like 72 = (70 + √4) ---> 704.
Just, noticed that the first denominator has 250 digits, and, the second denominator, 260. As in 25 X 10, and, 26 X 10. I wonder how long it will take my computer to work out their prime factors. Note as well the four 4's in the expression above with exponents 137, and, the four 3's in the one with exponents 142. These are the middle digits of 142, and 137, to go with the two 1's in the two 10's above, and, the 72's/27's.
Seemingly random sets of denominators of that size, and, so, definitely another sign of a pre-existing connection between 137, and, 142. Later observations show a bit of a pattern with the "islands of stability" within the denominators, which seem to reach focus at the numerals, 137, and 142, in terms of each involving the 72's/27's, 10's or 1's, and four-of-a-kind middle numerals.
For the exponent 411, as a standalone numeral, the expression or equation is 704 = (1 / 307412686815832125636594011029546440628801044380500752963574710596802960934664643589266053669952981830909013752519446912699956343910948040741446567320550320095405476238014880119219013051637698263534377985788002118541204836878009242034171278923149909476507885625548942624617199009538697685066779581498714370065729363010247221994399065537146607452759943181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181778455000556980533663082075731208331727847374417154967527297349822566550507489424050979219934880165272897395034486250832194384681345057571826664281614040611099918886624451776536051245586771214188669322609319866877502537739692358421165662193844224) X (70^411 - 4^411). With a set of either seventy-nine 18's, or 81's, within its denominator. But, 79 = (10*7 + 3^2) ---> 173 ---> 137. And, 79 = (70 + 3^2) ---> 73, or 37. The denominator above as 756 digits. It starts with 3074 = 1*(3073 + 1) ---> 137_731, and, ends with 4224 = [(1000*4 + 20) + 24*1] ---> 142_241. I'm not sure of the relation between the seventy-nine pairs, and, the number of digits, but, perhaps, then, it's seventy-nine and a half thus pairs. And, why 18, or 81? 18 = (4 + 14) = [10 + 4*(1 + 1)] ---> [(1 with 0^0) + 4*(1 + 1)] ---> 114_411. 81 = [(1 + 1)*4*10 + 1) ---> 114_411 is like the "atomic magic numbers", the first, regular seven of which sum to 324 = 18^2 = (2 + 10 + 18 + 36 + 54 + 86 + 118). Interesting, the first, regular seven "nuclear magic numbers" add to 316 = (2 + 8 + 20 + 28 + 50 + 82 + 126). But, for protons, instead of 126, it's 114 ---> 411. Oh, I should look at what happens for exponents of 316, to compare that to thus exponents of 411.
704 = (1 / 15973790540569095863993573320207270346537981428420556805542087739882759505130471792062534147655176650030365499042574520421185921312036031850619205581121451878715475085240123350932721894899625916361759831131384111811440329401772914793870798231677886186014545219357238650568181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818181818156502793092916619214748470760274484392238389408466353368146348333744256218879409032615451819306133479566304656574072225060542964747339766078587417215674859129461144606932123634572120293376) X (70^316 - 4^316). Sixty-one pairs of 81, with 18 = 3*1*6 ---> 316. 581 digits.
Of course, there is a bit of fairly solid reasoning, in general, as to the how, why, and, etc, of it, but, I don't want to put too much out there, just, yet. Such as these sort of numerals deriving from the centered hexagonal numbers, with formula, for the nth thus term being, (3n^2 + 3n + 1), which, interestingly, "reduces" to {(√3n)^2 + (3n)^1 + [(3^2)n]^0}, with the pattern from √3, to 3, to 3^2, while against the overall exponents from 2, to 1, to 0, respectively, as in the exponent of 1 / 2 against 2, the exponent of 1 against 1, and, then, the exponent of 2 against 0, with each being a form of reciprocal operation. 0 flips around 1, to 2, in an additive way; 1 / 2 flips around 1, to 2, in a multiplicative way, but, 1 flipping around 1, to 1, is both additive, and multiplicative, the additive, and multiplicative, identity, and, here, where the other two meet. In general, in an abstract way, like the 2's become 7's, and, the 6's become 9's, numerals of which the centered hexagonal numbers appear to produce mostly.
