OFFSET

1,3

EXAMPLE

Let b(p) be the smallest m such that p*2^m + 1 is prime. We have a(1) = b(2) = 0.

The least prime p such that b(p) > 0 is p = 3 with b(3) = 1, so a(2) = 1.

The least prime p such that b(p) > 1 is p = 7 with b(7) = 2, so a(3) = 2.

The least prime p such that b(p) > 2 is p = 17 with b(17) = 3, so a(4) = 3.

The least prime p such that b(p) > 3 is p = 19 with b(19) = 6, so a(5) = 6.

The least prime p such that b(p) > 6 is p = 31 with b(31) = 8, so a(6) = 8.

The least prime p such that b(p) > 8 is p = 47 with b(47) = 583, so a(7) = 583.

PROG

(PARI) b(p) = for(k=0, oo, if(isprime(p*2^k+1), return(k)))

list(lim) = if(lim>=2, my(v=[0], r=0); forprime(p=2, lim, if(b(p)>r, r=b(p); v=concat(v, r))); v) \\ gives a(n) for A350118(n) <= lim

CROSSREFS

KEYWORD

sign,fini,hard,more

AUTHOR

Jianing Song, Dec 14 2021

STATUS

approved