A224496 Smallest k such that k*2*p(n)^2+1=q is prime, k*2*q^2+1=r, k*2*r^2+1=s, k*2*r^2+1=t, r, s, and t are also prime.

386, 2769, 96656, 5366, 420, 34454, 65039, 192215, 458367, 24735, 27155, 777, 736254, 80297, 279927, 113429, 650474, 238919, 8229, 1284345, 642789, 333141, 11510, 1009271, 932, 395126, 1202174, 25811, 204534, 16286, 22094, 2661131, 22530, 128225, 56225, 900

Offset: 1

Author

Pierre CAMI, Apr 08 2013

Keywords

• nonn

Comments

Conjecture: a(n) exist for all n

t=k*2*(k*2*(k*2*(k*2*p(n)^2+1)^2+1)^2+1)^2+1

s=k*2*(k*2*(k*2*p(n)^2+1)^2+1)^2+1

r=k*2*(k*2*p(n)^2+1)^2+1

q=k*2*p(n)^2+1

Links

• Pierre CAMI, Table of n, a(n) for n = 1..80

Crossrefs

Cf. A224489, A224490, A224491, A224492, A224193, A224494, A224495.

Programs

Mathematica

a[n_] := For[k = 1, True, k++, p = Prime[n]; If[PrimeQ[q = k*2*p^2 + 1] && PrimeQ[r = k*2*q^2 + 1] && PrimeQ[s = k*2*r^2 + 1] && PrimeQ[k*2*s^2 + 1], Return[k]]]; Table[ Print[an = a[n]]; an , {n, 1, 36}] (* Jean-François Alcover, Apr 12 2013 *)