A060392 Let f(m) = smallest prime that divides k^2 + k + m for k = 0,1,2,...; sequence gives smallest m >= 0 such that f(m) is the n-th prime.

0, 1, 5, 47, 11, 221, 17, 1217, 941, 2747, 8081, 9281, 41, 55661, 19421, 333491, 1262201, 601037, 5237651, 9063641, 12899891, 26149427, 24073871, 28537121, 352031501, 398878547, 160834691, 67374467, 146452961, 24169417397

Offset: 1

Author

Luis Rodriguez-Torres (ludovicusmagister(AT)yahoo.com), Apr 03 2001

Keywords

• nice
• nonn

Comments

Chris Nash (see the Prime Puzzles link) has shown that such an m always exists.

For n>1, least odd number d such that the Legendre symbol (1-4d/prime(k)) = -1 for k = 2,...,n, but not for n+1. - T. D. Noe, Apr 19 2004

Examples

			k^2 + k takes the values 0, 2, 6, 12, ... for k = 0,1,2,...; the smallest prime divisor of these numbers is 2, so f(0) = 2.
		

References

• R. F. Lukes, C. D. Patterson and H. C. Williams, Numerical sieving devices: their history and some applications. Nieuw Arch. Wisk. (4) 13 (1995), no. 1, 113-139. Math. Rev. 96m:11082

Links

• G. W. Fung and H. C. Williams, Quadratic polynomials with high density of primes, Mathematics of Computation, Vol. 55, 1990.
• C. Rivera, www.primepuzzles.net, Conjecture 17
• Eric Weisstein's World of Mathematics, Prime-Generating Polynomial

Crossrefs

Cf. A060380, A060393-A060398. A060394 gives associated values of k.

Programs

Mathematica

nn=20; a=Table[0, {nn}]; d=-1; While[Length[Select[a, # == 0&]] != 1, d=d+2; i=2; While[JacobiSymbol[1-4d, Prime[i]]==-1, i++ ]; If[i<=nn && a[[i]]==0, a[[i]]=d]]; a (* corrected by Jean-François Alcover, Feb 06 2019 *)

PARI

lista(nn) = {va = vector(nn); d = -1; while (#select(x->(x==0), va) != 1, d += 2; i = 2; while(kronecker(1-4*d, prime(i)) == -1, i++); if ((i <= nn) && (va[i] == 0), va[i] = d);); va;} \\ Michel Marcus, Feb 05 2019

Extensions

Corrected by T. D. Noe, Apr 19 2004