A083522 Smallest k such that k*(k+1)*(k+2)*...*(k+n-1) + 1 is prime, or 0 if no such number exists.

1, 1, 1, 0, 3, 3, 4, 4, 6, 2, 1, 10, 5, 3, 9, 6, 6, 4, 5, 8, 6, 7, 19, 25, 11, 2, 1, 3, 9, 23, 7, 7, 39, 5, 7, 2, 1, 5, 78, 2, 1, 15, 19, 12, 17, 6, 3, 14, 8, 21, 23, 17, 14, 40, 16, 6, 8, 13, 15, 5, 15, 82, 46, 51, 39, 43, 6, 11, 61, 57, 16, 2, 1, 26, 54, 2, 1, 13, 4, 62, 31, 69, 27, 155, 21

Offset: 1

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 05 2003

Keywords

• nonn

Comments

The product of four consecutive integers + 1 is always composite (a square), so a(4) = 0. Are there any more zeros in the sequence?

Since rather large numbers (up to 193 digits) are encountered in the computation, the Pocklington-Lehmer "P-1" primality test is used, as implemented in PARI 2.1.3.

Examples

			1*2*3*4*5 + 1 = 121 = 11*11 and 2*3*4*5*6 + 1 = 721 = 7*103 are composite, but 3*4*5*6*7 + 1 = 2521 is prime, so a(5) = 3.
		

Crossrefs

Cf. A083520, A083521.

Programs

PARI

m=1000; for(n=1,85,b=0; k=1; while(b<1&&k
				

Extensions

Edited and extended by Klaus Brockhaus and Don Reble, May 06 2003