A347908 Even numbers k such that 2^(2*k) == 2 (mod k).

2, 14, 1022, 20066, 485918, 2531678, 3677198, 8277458, 8893262, 21122318, 26358638, 39852014, 42448478, 76712318, 131492498, 144322478, 164360606, 175126478, 176647378, 196705598, 249126626, 306789074, 317051378, 438023138, 497041538, 696970718, 996520658

Offset: 1

Author

Jianing Song, Sep 18 2021

Keywords

• nonn

Comments

Numbers of the form 2*t where 2^(4*t-1) == 1 (mod t).

Even terms in A130421. Complement of A347906 in A130421.

If k > 14 is a term, then k/2 must be composite, since for odd primes p we have 2^(4*p) == 16 (mod 2*p). If k = 2*t > 2 is a term, then 2*k-1 = 4*t-1 must also be composite, since ord(2,t) | (4*k-1) and ord(2,t) <= eulerphi(t) <= t-1 < 4*t-1.

If k = 2*t > 2 is a term, then (2^(2*k) - 2)/k = (2^(4*k-1) - 1)/t is composite. See A347907 for a proof.

2*(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

			14 is a term since 14 divides 2^28 - 2.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..1298 (contains all terms below 10^15; based on Max Alekseyev's b-file for A130421)

Crossrefs

Cf. A130421, A347906, A347907.

Programs

PARI

isA347908(k) = if(k%4==2, k=k>>1; if(isprime(k) && k!=7, 0, Mod(2, k)^(4*k-1)==1), 0)

Formula

a(n) = A347907(n)*2.