A215968 Smallest k > 0 such that 240*k*p+1 , 6*k*p*(240*k*p+1)+1 , and 40*(6*k*p*(240*k*p+1)+1)+1 are prime or 0 if no solution, where p = prime(n).

11, 21, 36, 8, 2, 140, 389, 45, 56, 145, 235, 71, 0, 121, 155, 56, 280, 80, 109, 37, 187, 217, 21, 97, 89, 7, 66, 28, 2, 166, 26, 101, 129, 93, 148, 51, 39, 71, 28, 139, 65, 20, 78, 14, 149, 3, 411, 516

Offset: 1

Author

Pierre CAMI, Aug 29 2012

Keywords

• nonn

Comments

Let 240*k*p(n)+1 = prime R, 6*k*p(n)*R+1 = prime Q, and 40*Q+1 = prime P.

P=(240*k*p(n))^2+240*k*p(n)+41=x^2+x+41 with x=240*k*p(n)

As R and Q are provable primes so is P a provable prime of the Euler polynomial x^2+x+41.

The only 0 is for p(13)=41 as R is always composite except for k=0 then P and Q are unity.

Examples

			240*11*2+1=5281 prime, 6*11*2*5281+1=697093 prime, (240*11*2)^2+(240*11*2)+41=27883721 prime. p(1)=2 so k(1)=11.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..3500
• David Broadhurst, A 132738-digit prime of the form x^2+x+41, Aug 14 2012 (archive of the nmbrthry mailing list).

Crossrefs

Cf. A215697.

Programs

Mathematica

a[n_] := (Clear[k]; p = Prime[n]; R = 240*k*p + 1; Q = 6*k*p*R + 1; P = 40*Q + 1; If[FactorList[P][[1, 1]] > 1, Return[0], For[k = 1, True, k++, If[PrimeQ[P] && PrimeQ[Q] && PrimeQ[R], Return[k]]]]); Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Sep 10 2012 *)