cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A244626 Composite numbers k congruent to 5 (mod 8) such that 2^((k-1)/2) mod k = k-1.

Original entry on oeis.org

3277, 29341, 49141, 80581, 88357, 104653, 196093, 314821, 458989, 489997, 800605, 838861, 873181, 1004653, 1251949, 1373653, 1509709, 1678541, 1811573, 1987021, 2269093, 2284453, 2387797, 2746477, 2909197, 3400013, 3429037, 3539101, 3605429, 4360621, 4502485, 5590621, 5599765
Offset: 1

Views

Author

Gary Detlefs, Jul 02 2014

Keywords

Comments

This sequence contains the n mod 8 = 5 pseudoprimes to the following modified Fermat primality criterion:
Conjecture 1: if p is an odd prime congruent to {3,5} (mod 8) then 2^((p-1)/2) mod p = p-1.
This conjecture has been tested to 10^8.
This criterion produces far fewer pseudoprimes than the 2^(n-1) mod n = 1 test and thus has a higher probability of success. The number of pseudoprimes for the two tests up to 10^k are:
10^5 5 26 19.23%
10^6 13 78 16.66%
10^7 40 228 17.54%
There are 40 terms < 10^7. If an additional constraint 3^(n-1) mod n = 1 and 5^(n-1) mod n = 1 is added, only 4 terms remain: (29341, 314821, 873181, 9863461).
This sequence appears to be a subset of A175865, A001262, A047713, A020230.
Number of terms below 10^k for k = 5..15: 5, 13, 40, 132, 369, 975, 2534, 6592, 17403, 45801, 122473. The corresponding numbers for 2^(n-1) mod n = 1: 26, 78, 228, 637, 1718, 4505, 11645, 29902, 76587, 197455, 513601. - Jens Kruse Andersen, Jul 13 2014
Also composite numbers 2n+1 with n even such that 2n+1 | 2^n+1. - Hilko Koning, Jan 27 2022
Conjecture 1 is true. With p = 2k+1 then 2^k mod (2k+1) == 2k. So 2k+1 | 2k-2^k. Prime numbers 2k+1 == +-3 (mod 8) are the prime numbers such that 2k+1 | 2^k+1 (Comments A007520). A reflection across the x-axis and +1 translation across the y-axis of the graph (2k-2^k) / (2k+1) gives the graph (2^k+1) / (2k+1). So the k values of both 2k+1 | 2k-2^k and 2k+1 | 2^k+1 are identical. - Hilko Koning, Feb 04 2022

Crossrefs

Programs

  • Maple
    for n from 5 to 10^7 by 8 do if 2^((n-1)/2) mod n = n-1 and not isprime(n) then print(n) fi od;

Extensions

a(18) corrected by Jens Kruse Andersen, Jul 13 2014