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

1,1

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, ...).

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)

Formula

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

Example

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

Prog

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

Crossrefs

Cf. A130421, A347906, A347907.

Sequence in context: A319540 A190634 A130421 * A227403 A156736 A334247

Adjacent sequences: A347905 A347906 A347907 * A347909 A347910 A347911

Keyword

nonn

Author

Jianing Song, Sep 18 2021

Status

approved