OFFSET
1,4
EXAMPLE
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.
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
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
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
STATUS
approved