A339122 Number of elements of the Rubik's Cube group of order A338883(n).

1, 170911549183, 33894540622394, 4346957030144256, 133528172514624, 140621059298755526, 153245517148800, 294998638981939200, 55333752398428896, 34178553690432192, 44590694400, 2330232827455554048, 23298374383021440, 14385471333209856, 150731886270873600

Offset: 1

Author

Ben Whitmore, Nov 24 2020

Keywords

• nonn
• fini
• full

Comments

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

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.

Examples

			a(1) = 1 because the only element of order A338883(1) = 1 is the identity element.
a(73) = 51490480088678400 is the number of elements of order A338883(73) = 1260.
		

Links

• Ben Whitmore, Table of n, a(n) for n = 1..73
• Tomas Rokicki, SpeedSolving Puzzles Community.

Crossrefs

Cf. A057731, A075152, A338883.

Programs

Mathematica

pN[p_] := Total[p]!/Times@@p/Times@@Factorial[Flatten[Tally[p]][[2 ;; ;; 2]]]
oddQ[p_] := OddQ[Total[p - 1]]
ord[p_] := LCM @@ p
oriN[p_, o_] := Module[{i, t, a = 0, ns = 0, s = 0, r}, t = ord[p]/p;
  For[i = 1, i <= Length[p], i++,
   If[Mod[t[[i]], o] == 0, a += p[[i]], ns += 1; s += p[[i]]]];
     {If[a == 0, r = o^(s - ns), r = o^a o^(s - ns - 1)], o^(a + s - 1) - r}]
val[p1_, o1_, p2_, o2_] :=
Module[{z}, z = pN[p1] pN[p2];
     {{LCM[ord[p1], ord[p2]],z oriN[p1, o1][[1]] oriN[p2, o2][[1]]},
     {{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]]}},
  {{LCM[ord[p1] o1, ord[p2] o2], z oriN[p1, o1][[2]] oriN[p2, o2][[2]] }}}]
p8 = IntegerPartitions[8]; p12 = IntegerPartitions[12];
ce = Select[p8, ! oddQ[#] &]; co = Select[p8, oddQ[#] &];
ee = Select[p12, ! oddQ[#] &]; eo = Select[p12, oddQ[#] &];
res = {}; max = 0;
For[i = 1, i <= Length[ce], i++,
For[j = 1, j <= Length[ee], j++,
  AppendTo[res, val[ce[[i]], 3, ee[[j]], 2]]]]
For[i = 1, i <= Length[co], i++,
For[j = 1, j <= Length[eo], j++,
  AppendTo[res, val[co[[i]], 3, eo[[j]], 2]]]]
p = Partition[res // Flatten, 2]; c // Clear;
For[i = 1, i <= Length[p], i++,
  If [IntegerQ[c[p[[i, 1]]]], c[p[[i, 1]]] += p[[i, 2]],
   c[p[[i, 1]]] = p[[i, 2]]]; If[p[[i, 1]] > max, max = p[[i, 1]]]];
Select[Table[c[i], {i, 1, max}], IntegerQ[#] &] (* Herbert Kociemba, Jun 30 2022 *)

Python

# See post #11 in SpeedSolving Puzzles Community link.

Formula

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

Extensions

a(10) corrected by Ben Whitmore, Jun 27 2022