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 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 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. LINKS Table of n, a(n) for n=1..85. EXAMPLE 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. PROG (PARI) m=1000; for(n=1, 85, b=0; k=1; while(b<1&&k

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