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A249782
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a(1)=2; thereafter, a(n) is the smallest prime not yet used which is compatible with the condition that a(n) is a quadratic residue modulo a(k) for the next n indices k = n+1, n+2, ..., 2n.
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1
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2, 7, 3, 37, 11, 83, 107, 139, 43, 211, 193, 881, 751, 2777, 2633, 5981, 6563, 10531, 3407, 9871, 12421, 23873, 6449, 164789, 3547, 39877, 248909, 370081, 528883, 1144453, 574813, 1201153, 1428929, 2225053, 1397719
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OFFSET
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1,1
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COMMENTS
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L(a(n)/a(k)) = 1 for the next n indices k = n+1, n+2, ..., 2n where L(a/p) is the Legendre symbol.
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LINKS
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EXAMPLE
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a(1) = 2 because the next term is 7 and L(2/7) = 1;
a(2) = 7 because the next two terms are (3,37) => L(7/3) = 1 and L(7/37) = 1;
a(3) = 3 because the next three terms are (37,11,83) => L(3/37) = 1, L(3/11) = 1 and L(3/83) = 1.
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PROG
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(PARI) m=35; v=vector(m); u=vectorsmall(10000*m); for(n=1, m, for(i=1, 10^9, if(!u[i], for(j=(n+1)\2, n-1, if(kronecker(v[j], prime(i))==-1 | | kronecker(v[j], prime(i))==0, next(2))); v[n]=prime(i); u[i]=1; break))); v
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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