A273870 Numbers m such that 4^(m-1) == 1 (mod (m-1)^2 + 1).

1, 3, 5, 17, 217, 257, 387, 8209, 20137, 37025, 59141, 65537, 283801, 649801, 1382401, 373164545, 535019101, 2453039425, 4294967297

Offset

1,2

Comments

Also, numbers m such that (4^k)^(m-1) == 1 (mod (m-1)^2 + 1) for all k >= 0.

a(20) > 2*10^12, if it exists. - Giovanni Resta, Feb 26 2020

Links

Table of n, a(n) for n=1..19.

Formula

a(n) = sqrt(A273999(n)-1) + 1. - Jinyuan Wang, Feb 24 2020

Example

5 is a term because 4^(5-1) == 1 (mod (5-1)^2+1), i.e., 255 == 0 (mod 17).

Prog

(Magma) [n: n in [1..100000] | (4^(n-1)-1) mod ((n-1)^2+1) eq 0]
(PARI) isok(n) = Mod(4, (n-1)^2+1)^(n-1) == 1; \\ Michel Marcus, Jun 02 2016

Crossrefs

Prime terms are in A273871.

Contains A000215 (Fermat numbers) as subsequence.

Contains 1 + A247220 as subsequence.

Cf. A019434, A273999.

Sequence in context: A056826 A370879 A278138 * A272060 A333873 A058910

Adjacent sequences: A273867 A273868 A273869 * A273871 A273872 A273873

Keyword

nonn,more

Author

Jaroslav Krizek, Jun 01 2016

Extensions

a(14)-a(15) from Michel Marcus, Jun 02 2016

Edited by Max Alekseyev, Apr 30 2018

a(16)-a(19) from Jinyuan Wang, Feb 24 2020

Status

approved