A374913 Numbers k such that k^(k + 1) == k + 1 (mod 2*k + 1).

2, 3, 6, 11, 14, 15, 18, 23, 26, 30, 35, 39, 50, 51, 54, 63, 74, 75, 78, 83, 86, 90, 95, 98, 99, 111, 114, 119, 131, 134, 135, 138, 146, 155, 158, 174, 179, 183, 186, 191, 194, 198, 210, 215, 219, 230, 231, 239, 243, 251, 254, 270, 278, 299, 303, 306, 315, 323, 326, 330, 338, 350

Offset

1,1

Links

Michel Marcus, Table of n, a(n) for n = 1..10000

Formula

Conjecture (Superseeker): a(n) = A263458(n)/2. - R. J. Mathar, Aug 02 2024

The conjectured formula is false. There exist numbers k such that 2*k + 1 is composite and k^(k + 1) == k + 1 (mod 2*k + 1). For example, when k = 1023: 1023^1024 == 1024 (mod 2047) and 2047 = 23*89 is composite. - Jedrzej Miarecki, Jan 16 2025

Mathematica

Select[Range[350], Mod[#^(#+1), 2#+1]==#+1 &] (* Stefano Spezia, Jul 23 2024 *)

Prog

(Magma) [n: n in [0..350] | n^(n+1) mod (2*n+1) eq n+1];
(PARI) isok(k) = Mod(k, 2*k+1)^(k+1) == k+1; \\ Michel Marcus, Feb 05 2025

Crossrefs

Supersequence of A002515 and A374914.

Cf. A374912.

Sequence in context: A294510 A218155 A042944 * A074737 A361124 A163781

Adjacent sequences: A374910 A374911 A374912 * A374914 A374915 A374916

Keyword

nonn,changed

Author

Juri-Stepan Gerasimov, Jul 23 2024

Status

approved