A272175 Least number k such that (k^2+1) mod s = prime(n) where s is the sum of the distinct primes dividing k^2+1, or 0 if no such k exists.

13, 3, 68, 182, 5, 32, 191, 333, 73, 70, 1068, 132, 507, 173, 774, 50, 11, 30, 1553, 3990, 338, 2307, 246, 2917, 1228, 80, 14369, 76, 114, 1590, 2529, 100, 28, 4952, 82, 659, 948, 7083, 2190, 8938, 19, 489, 11393, 1968, 2941, 21124, 3549, 1725, 64, 1382, 2540

Offset: 1

Author

Michel Lagneau, Apr 28 2016

Keywords

• nonn

Comments

Conjecture: a(n)> 0 for all n > 0.

The primes in the sequence are 3, 5, 11, 13, 19, 29, 73, 173, 191,...

The squares in the sequence are 25, 64, 100,...

Examples

			a(1)=13 because 13^2+1 = 170 = 2*5*17 => 170 mod(2+5+17) = 170 mod 24 = 2 = prime(1).
		

Crossrefs

Cf. A005574, A193462, A262965.

Programs

Mathematica

Table[k=0;While[Mod[k^2+1,Plus@@First[Transpose[FactorInteger[k^2+1]]]]!=Prime[n],k++];k, {n,50}]

PARI

a(n) = {k = 1; while ((m=k^2+1) && (lift(Mod(m, vecsum(factor(m)[,1]))) != prime(n)) , k++); k;} \\ Michel Marcus, Apr 29 2016