A239666 a(n) = least number k such that n*k^n+1 is prime, or 0 if no such number exists.

1, 1, 4, 1, 8, 1, 4, 3, 10, 1, 42, 1, 60, 15, 22, 1, 8, 1, 198, 42, 10, 1, 8, 115, 34, 21, 0, 1, 54, 1, 130, 3, 4, 7, 72, 1, 778, 204, 30, 1, 108, 1, 178, 15, 14, 1, 924, 28, 234, 63, 1376, 1, 44, 3, 16, 27, 256, 1, 180, 1, 706, 51, 98, 0, 546, 1, 4, 153, 150, 1, 170

Offset

1,3

Comments

a(n) = 1 iff n+1 is prime.

If a(n) = 0, then n is in A097792. Note that the converse is not true: a(4) = 1, not 0.

If n is in A097792 and n > 4, then a(n) = 0. For a sketch of this proof, either n = 4b^4 for some positive integer b > 2 or n = (bp)^p for some prime p > 2 and some positive integer b. In the first case, n*k^n+1 can be factored by Sophie Germain's identity into two trinomials where neither can equal 1 since b > 2, so n*k^n+1 must be composite. In the second case, (bpk^{b^p p^(p-1)}+1) is a factor of n*k^n+1 since p is odd. - William Dean, Oct 23 2024

Links

William Dean, Table of n, a(n) for n = 1..1000

Example

3*1^3+1 = 4 is not prime. 3*2^3+1 = 25 is not prime. 3*3^3+1 = 82 is not prime. 3*4^3+1 = 193 is prime. Thus, a(3) = 4.

Prog

(PARI)
is_A097792(n)={my(p, t); n%4==0 && ispower(n\4, 4) || ((2 < p = ispower(n, , &t)) && if(isprime(p), t%p==0, foreach(factor(p)[, 1], q, q%2 && n%q==0 && return(1))))}
a(n) = if(n!=4 && is_A097792(n), 0, for(k=1, oo, if(ispseudoprime(n*k^n+1), return(k)))); \\ [corrected by Andrew Howroyd, Oct 25 2024]

Crossrefs

Cf. A097792, A116954, A089001.

Sequence in context: A384999 A089146 A297402 * A101512 A085994 A179836

Adjacent sequences: A239663 A239664 A239665 * A239667 A239668 A239669

Keyword

nonn

Author

Derek Orr, Mar 29 2014

Status

approved