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A060392
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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.
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5
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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
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OFFSET
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1,3
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COMMENTS
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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
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REFERENCES
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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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nice,nonn
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AUTHOR
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Luis Rodriguez-Torres (ludovicusmagister(AT)yahoo.com), Apr 03 2001
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EXTENSIONS
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STATUS
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approved
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