A206827 - OEIS

%I #20 Jun 04 2023 08:40:34

%S 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,

%T 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,

%U 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

%N 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.

%C For a guide to related sequences, see A206588.

%C If n is a prime > 3, a(n) = 2. - _Robert Israel_, Jun 04 2023

%H Robert Israel, <a href="/A206827/b206827.txt">Table of n, a(n) for n = 2..10000</a>

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

%p f:= n -> numboccur(S[n] mod n, S[1..n-1] mod n):

%p S:= [seq(k*(k+1)*(2*k+1)/6, k=1..100)]:

%p map(f,[$2..100]); # _Robert Israel_, Jun 04 2023

%t s[k_] := k (k + 1) (2 k + 1)/6;

%t f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];

%t t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]

%t a[n_] := Count[Flatten[t[n]], 1]

%t Table[a[n], {n, 2, 120}]  (* A206827 *)

%Y Cf. A206588.

%K nonn

%O 2,4

%A _Clark Kimberling_, Feb 15 2012