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A273870
Numbers m such that 4^(m-1) == 1 (mod (m-1)^2 + 1).
2
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
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.
Sequence in context: A056826 A370879 A278138 * A272060 A333873 A058910
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