Victor Miller:

So where did this calculation come from?

I'm trying to extend Sloane's A176762 (a(0) to a(19) are 1, 10, 13, 23, 19, 7, 356, 4, 2, 11, 15, 5, 3, 14, 45, 36, 6, 112, 269, 15999. Smallest number that takes n steps to reach a cycle under iteration of sum-of-squares-of-digits map.). I'm not particularly capable of symbolic mathematics, so forgive the following... If it is true that, from a(18) on, a(n) is the sum-of-squares-of- digits of a(n+1), I have an empirically derived table that would allow one to derive a(n+1) from a(n) mod 81... mod x y 0 0 0 1 0 1 2 2 688 3 2 1688 4 0 2 5 0 12 6 2 2688 7 3 37888 8 1 58 9 0 3 10 0 13 11 2 3688 12 1 258 13 3 8888 14 3 18888 15 2 788 16 0 4 17 1 77 18 1 177 19 1 68 20 1 168 21 1 277 22 3 38888 23 1 268 24 2 3788 25 0 5 26 0 15 27 2 5688 28 1 368 29 0 25 30 2 888 31 2 1888 32 1 78 33 1 178 34 2 2888 35 1 468 36 0 6 37 0 16 38 3 58888 39 2 3888 40 0 26 41 1 378 42 1 577 43 5 8888888 44 1 568 45 0 36 46 2 4888 47 1 88 48 1 188 49 0 7 50 0 17 51 1 288 52 0 46 53 0 27 54 0 127 55 2 5888 56 1 388 57 1 578 58 0 37 59 0 137 60 4 888888 61 0 56 62 3 78888 63 1 488 64 0 8 65 0 18 66 2 6888 67 2 16888 68 0 28 69 0 128 70 2 26888 71 3 378888 72 1 588 73 0 38 74 0 57 75 0 157 76 1 2588 77 3 88888 78 3 188888 79 2 7888 80 0 48 More specifically, I am suggesting that if a(n) mod 81 = z -> {x,y}, then a(n+1) = (y+1)*10^((a(n)-z-81*x)/81)-1. So... a(19) mod 81 = 42 -> a(20) = 578*10^196-1 a(20) mod 81 = 55 -> a(21) = 5889*10^((578*10^196-218)/81)-1 a(21) mod 81 = 2 -> a(22) = 689*10^((5889*10^((578*10^196-218)/ 81)-165)/81)-1 a(22) mod 81 = 31 -> a(23) = 1889*10^(689*10^((5889*10^((578*10^196-218)/81)-165)/81)-194)-1 Care to continue? ;)