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%I #40 Mar 19 2022 06:36:47
%S 1,0,2,4,4,0,20,48,80,0,94,344,424,0,1096,4864,3856,0,16444,52432,
%T 65248,0,182362,720928,671104,0,4152320,11156656,9256396,0,34636834,
%U 135397376,130150588,0,533834992,2773200896,1857304312,0,7065319328,27541477824,26817356776
%N Number of solutions to 1 +- 8 +- 27 +- ... +- n^3 == 0 (mod n).
%H Alois P. Heinz, <a href="/A300269/b300269.txt">Table of n, a(n) for n = 1..1000</a>
%e Solutions for n = 7:
%e -----------------------------------
%e 1 +8 +27 +64 +125 +216 +343 = 784.
%e 1 +8 +27 +64 +125 +216 -343 = 98.
%e 1 +8 +27 -64 +125 -216 +343 = 224.
%e 1 +8 +27 -64 +125 -216 -343 = -462.
%e 1 +8 +27 -64 -125 +216 +343 = 406.
%e 1 +8 +27 -64 -125 +216 -343 = -280.
%e 1 +8 -27 -64 +125 +216 +343 = 602.
%e 1 +8 -27 -64 +125 +216 -343 = -84.
%e 1 -8 +27 +64 +125 -216 +343 = 336.
%e 1 -8 +27 +64 +125 -216 -343 = -350.
%e 1 -8 +27 +64 -125 +216 +343 = 518.
%e 1 -8 +27 +64 -125 +216 -343 = -168.
%e 1 -8 +27 -64 -125 -216 +343 = -42.
%e 1 -8 +27 -64 -125 -216 -343 = -728.
%e 1 -8 -27 +64 +125 +216 +343 = 714.
%e 1 -8 -27 +64 +125 +216 -343 = 28.
%e 1 -8 -27 -64 +125 -216 +343 = 154.
%e 1 -8 -27 -64 +125 -216 -343 = -532.
%e 1 -8 -27 -64 -125 +216 +343 = 336.
%e 1 -8 -27 -64 -125 +216 -343 = -350.
%p b:= proc(n, i, m) option remember; `if`(i=0, `if`(n=0, 1, 0),
%p add(b(irem(n+j, m), i-1, m), j=[i^3, m-i^3]))
%p end:
%p a:= n-> b(0, n-1, n):
%p seq(a(n), n=1..60); # _Alois P. Heinz_, Mar 01 2018
%t b[n_, i_, m_] := b[n, i, m] = If[i == 0, If[n == 0, 1, 0],
%t Sum[b[Mod[n + j, m], i - 1, m], {j, {i^3, m - i^3}}]];
%t a[n_] := b[0, n - 1, n];
%t Table[a[n], {n, 1, 60}] (* _Jean-François Alcover_, Mar 19 2022, after _Alois P. Heinz_ *)
%o (Ruby)
%o def A(n)
%o ary = [1] + Array.new(n - 1, 0)
%o (1..n).each{|i|
%o i3 = 2 * i * i * i
%o a = ary.clone
%o (0..n - 1).each{|j| a[(j + i3) % n] += ary[j]}
%o ary = a
%o }
%o ary[((n * (n + 1)) ** 2 / 4) % n] / 2
%o end
%o def A300269(n)
%o (1..n).map{|i| A(i)}
%o end
%o p A300269(100)
%o (PARI) a(n) = my (v=vector(n,k,k==1)); for (i=2, n, v = vector(n, k, v[1 + (k-i^3)%n] + v[1 + (k+i^3)%n])); v[1] \\ _Rémy Sigrist_, Mar 01 2018
%Y Number of solutions to 1 +- 2^k +- 3^k +- ... +- n^k == 0 (mod n): A300190 (k=1), A300268 (k=2), this sequence (k=3).
%Y Cf. A113263, A195938.
%K nonn
%O 1,3
%A _Seiichi Manyama_, Mar 01 2018
%E More terms from _Alois P. Heinz_, Mar 01 2018
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