A301643 Strong pseudo safe-primes: numbers n = 2m+1 with 2^m == +-1 (mod n) and m a strong pseudoprime A001262.

715523, 2651687, 2882183, 10032383, 14924003, 15640403, 30278399, 32140859, 45698963, 86727203, 129210083, 202553159, 257330639, 271938803, 274831643, 294056003, 307856267, 332164619, 413008067, 437894243, 447564527, 494832203, 654796019, 689552603, 735119003

Offset: 1

Author

Francois R. Grieu, Mar 25 2018

Keywords

• nonn

Comments

Equivalently, numbers n = 2m+1 that are not safe primes A005385 even though n and m are strong probable primes (that is, prime or strong pseudoprime A001262). That follows from a result by Fedor Petrov.

All known terms are prime, including the 542622 less than 2^65 (obtained by post-processing Jan Feitsma and William Galway's table).

Examples

			n = 715523 is in the sequence because n = 2m+1 with m = 357761, and 2^m mod n = 715522 which is among 1 or n-1 (the latter), and m is a strong pseudoprime A001262. The latter holds because m = 131*2731 is composite, and m passes the strong probable prime test. The latter holds because when writing m-1 as d*(2^s) with d odd, it holds that 2^d mod m = 1 or there exists an r with 0 <= r < s and 2^(d*(2^r)) == -1 (mod m); specifically, d = 2795, s = 7, 2^2795 mod 357761 = 357760 = m-1, thus 2^(d*(2^r)) == -1 (mod m) for r = 0.
		

Links

• Jan Feitsma, The pseudoprimes below 2^64
• William Galway, Tables of pseudoprimes and related data [Includes a file with pseudoprimes up to 2^64.]
• Fedor Petrov, The question's congruence and m prime imply n prime, MathOverflow.

Crossrefs

Subsequence of A300193.

Cf. A001262, A005385.

Programs

Mathematica

For[m=3,(n=2m+1)<13^8,m+=2,If[MemberQ[{1,n-1},PowerMod[2,m,n]]&&(d=m-1;t=1;While[EvenQ[d],d/=2;++t];If[(x=PowerMod[2,d,m])!=1,While[--t>0&&x!=m-1,x=Mod[x^2,m]]];t>0)&&!PrimeQ[m],Print[n]]]

PARI

is_A001262(n, a=2)={ (bittest(n, 0) && !isprime(n) && n>8) || return; my(s=valuation(n-1, 2)); if(1==a=Mod(a, n)^(n>>s), return(1)); while(a!=-1 && s--, a=a^2); a==-1; } \\ after A001262
isok(n) = if (n%2, my(m = (n-1)/2, r = Mod(2, n)^m); ((r==1) || (r==-1)) && is_A001262(m)); \\ derived from Michel Marcus, May 07 2018