A206827 Number of solutions (n,k) of s(k) == s(n) (mod n), where 1 <= k < n and s(k) = k*(k+1)*(2*k+1)/6.

1, 1, 1, 2, 2, 2, 1, 1, 3, 2, 2, 2, 3, 3, 1, 2, 1, 2, 3, 3, 3, 2, 1, 2, 3, 1, 3, 2, 3, 2, 1, 3, 3, 8, 1, 2, 3, 3, 3, 2, 4, 2, 3, 3, 3, 2, 2, 2, 3, 3, 3, 2, 2, 8, 3, 3, 3, 2, 2, 2, 3, 3, 1, 8, 3, 2, 3, 3, 9, 2, 1, 2, 3, 3, 3, 8, 2, 2, 3, 1, 3, 2, 4, 8, 3, 3, 3, 2, 5, 8, 3, 3, 3, 8, 2, 2, 3, 3, 3

Offset: 2

Author

Clark Kimberling, Feb 15 2012

Keywords

• nonn

Comments

For a guide to related sequences, see A206588.

If n is a prime > 3, a(n) = 2. - Robert Israel, Jun 04 2023

Examples

			5 divides exactly two of the numbers s(n)-s(k) for k=1,2,3,4, so a(5)=2.
		

Links

• Robert Israel, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A206588.

Programs

Maple

f:= n -> numboccur(S[n] mod n, S[1..n-1] mod n):
S:= [seq(k*(k+1)*(2*k+1)/6, k=1..100)]:
map(f,[$2..100]); # Robert Israel, Jun 04 2023

Mathematica

s[k_] := k (k + 1) (2 k + 1)/6;
f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
a[n_] := Count[Flatten[t[n]], 1]
Table[a[n], {n, 2, 120}]  (* A206827 *)