A329168 Numbers m that divide 4^(m + 1) + 1.

1, 17, 4097, 7361, 85073, 658529, 3999137, 72281281, 285143057, 628944689, 854112113, 1423081169, 2561019281, 3111576929, 4298117633, 5921265041, 14224884929, 21336998129, 34317377233, 50723421713, 63797137889, 144269032049, 163834314353, 187397322209, 212565453281

Offset

1,2

Comments

Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.

Conjecture: For k > 1, if f(1) = p(1) equals one of the prime factors of k^2 + 1, p(i+1) equals one of the prime factors of k^(f(i)+1) + 1 greater than p(i), f(i+1) = f(i)*p(i+1), then k^(f(i) + 1) == -1 (mod f(i)) for all integers i. (Especially in this sequence, k = 4, so {f(i)} can be 17, 4097, 4298117633, ...) - Jinyuan Wang, Nov 16 2019

Links

Table of n, a(n) for n=1..25.

Prog

(Magma) [n + 1: n in [0..5000000] | Modexp(4, n + 2, n + 1) eq n ];
(PARI) isok(m) = Mod(4, m)^(m+1) == -1; \\ Jinyuan Wang, Nov 16 2019

Crossrefs

Cf. A055685.

Solutions to k^(m + 1) + 1 == -1 (mod m): A296369 (k = 2), A328230 (k = 3).

Sequence in context: A032909 A367536 A239165 * A194015 A015058 A015034

Adjacent sequences: A329165 A329166 A329167 * A329169 A329170 A329171

Keyword

nonn

Author

Juri-Stepan Gerasimov, Nov 06 2019

Extensions

a(13)-a(25) from Giovanni Resta, Nov 08 2019

Status

approved