A008781 For any circular arrangement of 0..n-1, let S be the sum of cubes of every sum of two contiguous numbers; then a(n) is the number of distinct values of S.
1, 1, 1, 3, 12, 46, 163, 405, 770, 1252, 1921, 2816, 3977, 5464, 7313
Offset: 1
Keywords
Examples
Consider n = 5: and the circular arrangements of {0,1,2,3,4}. Here are the values of [ A, B, C, D, E ] (A+B)^3 + (B+C)^3 +(C+D)^3 +(D+E)^3 +(E+A)^3: [0,1,2,3,4], (0+1)^3 + (1+2)^3 +(2+3)^3 +(3+4)^3 +(4+0)^3 = 560; [0,1,2,4,3], (0+1)^3 + (1+2)^3 +(2+4)^3 +(4+3)^3 +(3+0)^3 = 614; [0,1,3,2,4], (0+1)^3 + (1+3)^3 +(3+2)^3 +(2+4)^3 +(4+0)^3 = 470; [0,1,4,2,3], (0+1)^3 + (1+4)^3 +(4+2)^3 +(2+3)^3 +(3+0)^3 = 494; [0,1,3,4,2], (0+1)^3 + (1+3)^3 +(3+4)^3 +(4+2)^3 +(2+0)^3 = 632; [0,1,4,3,2], (0+1)^3 + (1+4)^3 +(4+3)^3 +(3+2)^3 +(2+0)^3 = 602; [0,2,1,3,4], (0+2)^3 + (2+1)^3 +(1+3)^3 +(3+4)^3 +(4+0)^3 = 506; [0,2,1,4,3], (0+2)^3 + (2+1)^3 +(1+4)^3 +(4+3)^3 +(3+0)^3 = 530; [0,3,1,2,4], (0+3)^3 + (3+1)^3 +(1+2)^3 +(2+4)^3 +(4+0)^3 = 398; [0,4,1,2,3], (0+4)^3 + (4+1)^3 +(1+2)^3 +(2+3)^3 +(3+0)^3 = 368; [0,3,1,4,2], (0+3)^3 + (3+1)^3 +(1+4)^3 +(4+2)^3 +(2+0)^3 = 440; [0,4,1,3,2], (0+4)^3 + (4+1)^3 +(1+3)^3 +(3+2)^3 +(2+0)^3 = 386; There are 12 different values, so a(5) = 12.
Programs
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Maple
A008781 := proc(n) local msu,p,c,i ; msu := {} ; for p in combinat[permute](n-1) do c := [0,op(p)] ; s := 0 ; for i from 0 to n-1 do s := s+(c[i+1]+c[1+modp(i+1,n)])^3 ; end do: msu := msu union {s} ; end do: nops(msu) ; end proc: # R. J. Mathar, Jul 18 2017
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Mathematica
f[perm_] := Total[#]^3& /@ Partition[Join[{0}, perm, {0}], 2, 1] // Total; a[n_] := a[n] = f /@ Permutations[Range[n - 1]] // Union // Length; Reap[Do[Print[n, " ", a[n]]; Sow[a[n]], {n, 1, 12}]][[2, 1]] (* Jean-François Alcover, Feb 24 2020 *)
Extensions
Corrected by Naohiro Nomoto, Sep 10 2001
More terms from Vit Planocka (planocka(AT)mistral.cz), Sep 29 2002
a(12) from Alois P. Heinz, May 26 2013
a(13)-a(15) from Sean A. Irvine, Apr 04 2018