A220235 -id:A220235 - cp's OEIS frontend

A341409 a(n) = (Sum_{k=1..3} k^n) mod n.

0, 0, 0, 2, 1, 2, 6, 2, 0, 4, 6, 2, 6, 0, 6, 2, 6, 2, 6, 18, 15, 14, 6, 2, 1, 14, 0, 14, 6, 14, 6, 2, 3, 14, 31, 2, 6, 14, 36, 18, 6, 38, 6, 10, 36, 14, 6, 2, 13, 24, 36, 46, 6, 2, 1, 42, 36, 14, 6, 38, 6, 14, 36, 2, 16, 2, 6, 30, 36, 14, 6, 2, 6, 14, 51, 22, 17, 14, 6, 18, 0, 14, 6, 38, 21

Offset: 1

Seiichi Manyama, Feb 11 2021

Seiichi Manyama, Table of n, a(n) for n = 1..10000

(Sum_{k=1..m} k^n) mod n: A096196 (m=2), this sequence (m=3), A341410 (m=4), A341411 (m=5), A341412 (m=6), A341413 (m=7).

Cf. A001550, A045576, A056645, A220235.

a:= n-> add(i&^n, i=1..3) mod n:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 11 2021

a[n_] := Mod[Sum[k^n, {k, 1, 3}], n]; Array[a, 100] (* Amiram Eldar, Feb 11 2021 *)

PARI
```
a(n) = sum(k=1, 3, k^n)%n;
```

a(n) = A001550(n) mod n.

a(A056645(n)) = 0.

A289259 Numbers k such that k^2 divides 2^k + 3^k.

Original entry on oeis.org

1, 5, 55, 1971145, 3061355, 109715901845, 340799222665

Offset: 1

Robert Israel, Jun 29 2017

If k is in the sequence and p is a prime factor, coprime to k, of 2^k + 3^k, then k*p is in the sequence.
55 = 5 * 11
1971145 = 5 * 11 * 35839
3061355 = 5 * 11 * 55661
109715901845 = 5 * 11 * 35839 * 55661
340799222665 = 5 * 11 * 55661 * 111323
See Known Terms link for additional terms.
From Felix Fröhlich, Jun 29 2017: (Start)
For k in the sequence, A220235(k) = 0.
Subsequence of A045576. (End)

			2^5 + 3^5 = 275 is divisible by 5^2, so 5 is in the sequence.

Robert Israel and Ray Chandler, Known Terms
A. Velampalli et al., Mathematics StackExchange, Can you prove or disprove that there exist infinitely many integers n such that n^2 divides 2^n+3^n?

Cf. A007689, A045576, A220235.

select(t -> 2&^t + 3&^t mod t^2 = 0, [$1..10^6]);

is(n) = Mod(2, n^2)^n==-3^n \\ Felix Fröhlich, Jun 29 2017

is(n) = Mod(2,n^2)^n+Mod(3,n^2)^n==0 \\ Charles R Greathouse IV, Jun 29 2017

a(6)-a(7) confirmed as next terms by Ray Chandler, Jul 02 2017

Known terms updated and moved to a-file by Ray Chandler, Jul 03 2017

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A341409 a(n) = (Sum_{k=1..3} k^n) mod n.

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A289259 Numbers k such that k^2 divides 2^k + 3^k.

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