A347907 - OEIS

%I #33 Apr 16 2023 16:16:32

%S 1,7,511,10033,242959,1265839,1838599,4138729,4446631,10561159,

%T 13179319,19926007,21224239,38356159,65746249,72161239,82180303,

%U 87563239,88323689,98352799,124563313,153394537,158525689,219011569,248520769,348485359,498260329,636381799,638395369,685333399,689463889

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

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

%C Numbers k such that 2*k is in A130421.

%C 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.

%C 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.

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

%H Jianing Song, <a href="/A347907/b347907.txt">Table of n, a(n) for n = 1..1298</a> (contains all terms below 5*10^14; based on Max Alekseyev's b-file for A130421)

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

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

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

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

%Y Cf. A130421, A347908.

%Y Cf. A347906 (a similar sequence).

%K nonn

%O 1,2

%A _Jianing Song_, Sep 18 2021