OFFSET
1,1
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).
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
LINKS
Jens Kruse Andersen, Table of n, a(n) for n = 1..10000
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;
CROSSREFS
KEYWORD
nonn
AUTHOR
Gary Detlefs, Jul 02 2014
EXTENSIONS
a(18) corrected by Jens Kruse Andersen, Jul 13 2014
STATUS
approved