A385326 The number of positive k <= 2*n + 1 such that 2*n + 1 divides (2^k + 2*n + 1)^2 - 1.

1, 3, 2, 2, 3, 2, 2, 7, 4, 2, 7, 2, 2, 3, 2, 6, 6, 5, 2, 6, 4, 6, 7, 2, 2, 12, 2, 5, 6, 2, 2, 21, 10, 2, 6, 2, 8, 7, 5, 2, 3, 2, 21, 6, 8, 15, 18, 5, 4, 6, 2, 2, 17, 2, 6, 6, 8, 5, 19, 9, 2, 12, 2, 18, 18, 2, 14, 7, 4, 2, 6, 4, 10, 7, 2, 10, 12, 15, 6, 6, 4, 2, 16, 2, 2, 19, 2, 5, 6, 2, 2, 6, 10, 9, 21, 2, 4, 32, 2, 2, 6

Offset: 0

Author

Juri-Stepan Gerasimov, Jun 25 2025

Keywords

• nonn

Examples

			1 is the term because 2*0 + 1 = 1 is divisor of (2^1 + 2*0 + 1)^2 - 1 = 3^2 - 1 = 8.
		

Crossrefs

Cf. A003462 (numbers m > 0 such that a(m) = 3), A005384 (primes p such that a(p) = 2), A005408 (odd numbers), A076481 (primes q such that a(q) = 3), A081858 (numbers k numbers k >= 0 such that 2k + 1 divides 2^k - 1), A102781 (numbers k such that 2k + 1 divides (2^k + 2*k + 1)^2 - 1), A224486 (numbers k such that 2k + 1 divides 2^k + 1).

Programs

Magma

[#[k: k in [1..2*n+1] | ((2^k+2*n+1)^2 - 1) mod (2*n + 1) eq 0]: n in [0..100]];

Mathematica

a[n_]:=Length[Select[Range[2n+1],Divisible[(2^#+2n+1)^2-1,2n+1] &]]; Array[a,101,0] (* Stefano Spezia, Jun 25 2025 *)

PARI

a(n) = sum(k=1, 2*n+1, !Mod((2^k + 2*n + 1)^2 - 1, 2*n + 1)); \\ Michel Marcus, Jun 25 2025