%I #19 Feb 10 2023 12:53:40

%S 0,0,1,0,3,2,0,4,1,0,6,8,2,0,4,4,11,14,9,16,7,8,5,20,8,10,1,0,28,20,

%T 28,4,25,4,14,32,28,26,4,36,28,20,28,12,28,2,28,20,0,0,19,48,28,32,34,

%U 28,43,24,28,56,28,16,28,4,18,20,28,52,25,0,28,68,28,66,19,40

%N a(n) = (Sum_{k=1..7} k^n) mod n.

%H Robert Israel, <a href="/A341413/b341413.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A001554(n) mod n.

%F a(A056750(n)) = 0.

%F From _Robert Israel_, Feb 09 2023: (Start)

%F Given positive integer k, let m = A001554(k).

%F If p is a prime > m/k and A001554(p*k) == m (mod k), then a(p*k) = m.

%F This is true for all primes p > m/k for k = 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 14, ...

%F For k = 5 or 15 it is true for primes p > m/k with p == 1 (mod 4).

%F For k = 11 it is true for primes p > m/k with p == 1 or 7 (mod 10).

%F For k = 13 it is true for primes p > m/k with p == 1 (mod 12).

%F (End)

%p a:= n-> add(i&^n, i=1..7) mod n:

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Feb 11 2021

%t a[n_] := Mod[Sum[k^n, {k, 1, 7}], n]; Array[a, 100] (* _Amiram Eldar_, Feb 11 2021 *)

%o (PARI) a(n) = sum(k=1, 7, k^n)%n;

%Y (Sum_{k=1..m} k^n) mod n: A096196 (m=2), A341409 (m=3), A341410 (m=4), A341411 (m=5), A341412 (m=6), this sequence (m=7).

%Y Cf. A001554, A056750.

%K nonn

%O 1,5

%A _Seiichi Manyama_, Feb 11 2021