A347907 Numbers k such that 2^(4*k-1) == 1 (mod k).

1, 7, 511, 10033, 242959, 1265839, 1838599, 4138729, 4446631, 10561159, 13179319, 19926007, 21224239, 38356159, 65746249, 72161239, 82180303, 87563239, 88323689, 98352799, 124563313, 153394537, 158525689, 219011569, 248520769, 348485359, 498260329, 636381799, 638395369, 685333399, 689463889

Offset: 1

Author

Jianing Song, Sep 18 2021

Keywords

• nonn

Comments

Odd numbers k such that ord(2,k) divides 4*k-1, where ord(2,k) is the multiplicative order of 2 modulo k.

Numbers k such that 2*k is in A130421.

Terms > 7 must be composite, since for odd primes p we have 2^(4*p-1) == 8 (mod p). If k > 1 is a term, then 4*k-1 must also be composite, since ord(2,k) | (4*k-1) and ord(2,k) <= eulerphi(k) <= k-1 < 4*k-1.

If k > 1 is a term, then (2^(4*k-1) - 1)/k is composite. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1, then (2^(4*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(4*k-1) - 1, there exist positive integers a,b such that a*b = k and that a | 2^u - 1 and b | 2^(u*(v-1)) + ... + 2^u + 1. Note that (2^u - 1)/a, (2^(u*(v-1)) + ... + 2^u + 1)/b >= (2^u - 1)/k >= (2^sqrt(4*k-1) - 1)/k > 1, so (2^(4*k-1) - 1)/k is the product of two integers > 1, so it is composite.

2^t - 1 is a term if and only if 2^(t+2) == 5 (mod t) (t = 1, 3, 9, 871, 2043, 2119, 8769, ...).

Examples

			7 is a term since 7 divides 2^27 - 1.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..1298 (contains all terms below 5*10^14; based on Max Alekseyev's b-file for A130421)

Crossrefs

Cf. A130421, A347908.

Cf. A347906 (a similar sequence).

Programs

Mathematica

Join[{1},Parallelize[Select[Range[69*10^7],PowerMod[2,4#-1,#]==1&]]] (* Harvey P. Dale, Apr 16 2023 *)

PARI

isA347907(k) = if(k%2 && (!isprime(k) || k==7), Mod(2, k)^(4*k-1)==1, 0)

Formula

a(n) = A347908(n)/2.