A230852 Smallest k<3*2^n such that 3*2^n+k is the smallest of four consecutive primes in arithmetic progression or 0 if no solution.

0, 0, 0, 0, 0, 59, 0, 0, 205, 229, 167, 353, 1595, 4459, 6407, 6215, 14995, 4559, 4697, 11399, 365, 10199, 19327, 39103, 3185, 13649, 15787, 2693, 21455, 24929, 32209, 30509, 13421, 5389, 36947, 12869, 27277, 38389, 973, 69199, 58835, 165629, 52597, 25463, 17923, 38629, 90263, 17153, 48143, 2171, 1255

Offset: 1

Author

Pierre CAMI, Oct 31 2013

Keywords

• nonn

Comments

Conjecture: if n>8 there is always a solution.

Examples

			3*2^6+59=251, and 251, 257, 263, 269 are four consecutive primes in arithmetic progression 6 so a(6)=59.
3*2^9+205=1741, and 1741, 1747, 1753, 1759 are four consecutive primes in arithmetic progression 6 so a(9)=205.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..110

Crossrefs

Cf. A054800, A230699.

Programs

PARI

cpap4(p)=my(q=nextprime(p+1),g=q-p);nextprime(q+1)-q==g&&nextprime(p+2*g+1)==p+3*g
a(n)=forprime(p=3<Charles R Greathouse IV, Oct 31 2013