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

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

Offset: 1

Author

Robert Israel, Jun 29 2017

Keywords

• nonn
• more

Comments

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)

Examples

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

Links

• 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?

Crossrefs

Cf. A007689, A045576, A220235.

Programs

Maple

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

PARI

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

PARI

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

Extensions

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