A339122 - OEIS

%I #23 Aug 30 2022 14:17:46

%S 1,170911549183,33894540622394,4346957030144256,133528172514624,

%T 140621059298755526,153245517148800,294998638981939200,

%U 55333752398428896,34178553690432192,44590694400,2330232827455554048,23298374383021440,14385471333209856,150731886270873600

%N Number of elements of the Rubik's Cube group of order A338883(n).

%C The most common order is 60, with a(33) = 4199961633799421952 elements, or about 9.71% of the group.

%C The least common order (excluding 1) is 11, with a(11) = 44590694400 elements, or about 0.0000001% of the group. Elements of order 11 are rare because they cannot affect the corner pieces of the cube.

%H Ben Whitmore, <a href="/A339122/b339122.txt">Table of n, a(n) for n = 1..73</a>

%H Tomas Rokicki, <a href="https://www.speedsolving.com/threads/average-scramble-loop-length.85241/#post-1451996">SpeedSolving Puzzles Community</a>.

%F Sum_{n=1..73} a(n) = 43252003274489856000 = A075152(3).

%e a(1) = 1 because the only element of order A338883(1) = 1 is the identity element.

%e a(73) = 51490480088678400 is the number of elements of order A338883(73) = 1260.

%t pN[p_] := Total[p]!/Times@@p/Times@@Factorial[Flatten[Tally[p]][[2 ;; ;; 2]]]

%t oddQ[p_] := OddQ[Total[p - 1]]

%t ord[p_] := LCM @@ p

%t oriN[p_, o_] := Module[{i, t, a = 0, ns = 0, s = 0, r}, t = ord[p]/p;

%t   For[i = 1, i <= Length[p], i++,

%t    If[Mod[t[[i]], o] == 0, a += p[[i]], ns += 1; s += p[[i]]]];

%t      {If[a == 0, r = o^(s - ns), r = o^a o^(s - ns - 1)], o^(a + s - 1) - r}]

%t val[p1_, o1_, p2_, o2_] :=

%t Module[{z}, z = pN[p1] pN[p2];

%t      {{LCM[ord[p1], ord[p2]],z oriN[p1, o1][[1]] oriN[p2, o2][[1]]},

%t      {{LCM[ord[p1] o1,ord[p2]],z oriN[p1, o1][[2]] oriN[p2, o2][[1]]}},   {{LCM[ord[p1],ord[p2] o2],z oriN[p1, o1][[1]] oriN[p2, o2][[2]]}},

%t   {{LCM[ord[p1] o1, ord[p2] o2], z oriN[p1, o1][[2]] oriN[p2, o2][[2]] }}}]

%t p8 = IntegerPartitions[8]; p12 = IntegerPartitions[12];

%t ce = Select[p8, ! oddQ[#] &]; co = Select[p8, oddQ[#] &];

%t ee = Select[p12, ! oddQ[#] &]; eo = Select[p12, oddQ[#] &];

%t res = {}; max = 0;

%t For[i = 1, i <= Length[ce], i++,

%t For[j = 1, j <= Length[ee], j++,

%t   AppendTo[res, val[ce[[i]], 3, ee[[j]], 2]]]]

%t For[i = 1, i <= Length[co], i++,

%t For[j = 1, j <= Length[eo], j++,

%t   AppendTo[res, val[co[[i]], 3, eo[[j]], 2]]]]

%t p = Partition[res // Flatten, 2]; c // Clear;

%t For[i = 1, i <= Length[p], i++,

%t   If [IntegerQ[c[p[[i, 1]]]], c[p[[i, 1]]] += p[[i, 2]],

%t    c[p[[i, 1]]] = p[[i, 2]]]; If[p[[i, 1]] > max, max = p[[i, 1]]]];

%t Select[Table[c[i], {i, 1, max}], IntegerQ[#] &] (* _Herbert Kociemba_, Jun 30 2022 *)

%o (Python) # See post #11 in SpeedSolving Puzzles Community link.

%Y Cf. A057731, A075152, A338883.

%K nonn,fini,full

%O 1,2

%A _Ben Whitmore_, Nov 24 2020

%E a(10) corrected by _Ben Whitmore_, Jun 27 2022