cp's OEIS Frontend

Some terms above 10^15: 26435035805519327, 158975398896178078, 64044049390757098943, 1063423126446412987081943, 1091220919655042176978844783, 81074850280355100090334498663, 6317483763950169936179578094903, 5672799393875320397186007124651847, 170923900137537174138295268515194974, 195746953975871672436191077726091399305155458, 325665752547333314939363628501536024940097079718953, 953533053776414279913696071891872697927468471633033, 85791212788381063775490416118630897060666265030605503, 334519297382630382793758729321508383611586565722054114034741260213364710519401967713. - Max Alekseyev, Jun 18 2014

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

Numbers k such that 2*k is in A006935. For k > 1, k is a term if and only if 2*k is an even pseudoprime to base 2.

Odd terms in A130421. Complement of A347908 in A130421.

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

If k > 1 is a term, then (2^(2*k-1) - 1)/k is composite. Proof: since 2*k-1 is composite, write 2*k-1 = u*v, u >= v > 1, then (2^(2*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(2*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(2*k-1) - 1)/k > 1, so (2^(2*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+1) == 3 (mod t) (t = 1, 111481, 465793, ... in A296370).

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, ...).