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A341413 a(n) = (Sum_{k=1..7} k^n) mod n.

%I A341413 #19 Feb 10 2023 12:53:40
%S A341413 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 A341413 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 A341413 28,43,24,28,56,28,16,28,4,18,20,28,52,25,0,28,68,28,66,19,40
%N A341413 a(n) = (Sum_{k=1..7} k^n) mod n.
%H A341413 Robert Israel, <a href="/A341413/b341413.txt">Table of n, a(n) for n = 1..10000</a>
%F A341413 a(n) = A001554(n) mod n.
%F A341413 a(A056750(n)) = 0.
%F A341413 From _Robert Israel_, Feb 09 2023: (Start)
%F A341413 Given positive integer k, let m = A001554(k).
%F A341413 If p is a prime > m/k and A001554(p*k) == m (mod k), then a(p*k) = m.
%F A341413 This is true for all primes p > m/k for k = 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 14, ...
%F A341413 For k = 5 or 15 it is true for primes p > m/k with p == 1 (mod 4).
%F A341413 For k = 11 it is true for primes p > m/k with p == 1 or 7 (mod 10).
%F A341413 For k = 13 it is true for primes p > m/k with p == 1 (mod 12).
%F A341413  (End)
%p A341413 a:= n-> add(i&^n, i=1..7) mod n:
%p A341413 seq(a(n), n=1..100);  # _Alois P. Heinz_, Feb 11 2021
%t A341413 a[n_] := Mod[Sum[k^n, {k, 1, 7}], n]; Array[a, 100] (* _Amiram Eldar_, Feb 11 2021 *)
%o A341413 (PARI) a(n) = sum(k=1, 7, k^n)%n;
%Y A341413 (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 A341413 Cf. A001554, A056750.
%K A341413 nonn
%O A341413 1,5
%A A341413 _Seiichi Manyama_, Feb 11 2021