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A216503
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a(n) = number of positive integers k such that n = x^2 + k*y^2 has a solution with x>0, y>0.
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7
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0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 5, 3, 3, 4, 6, 5, 5, 5, 5, 5, 4, 6, 7, 5, 5, 6, 8, 5, 6, 7, 7, 7, 5, 9, 10, 6, 6, 9, 11, 6, 8, 9, 10, 7, 6, 10, 11, 7, 7, 11, 11, 8, 7, 10, 11, 9, 8, 9, 14, 7, 9, 10, 11, 8, 10, 13, 11, 9, 8, 15, 17, 9, 9, 14, 11, 8, 11, 12, 14
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OFFSET
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1,5
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LINKS
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EXAMPLE
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a(17) = 6 because
17 = 4^2 + 1*1^2.
17 = 3^2 + 2*2^2.
17 = 1^2 + 4*2^2.
17 = 3^2 + 8*1^2.
17 = 2^2 + 13*1^2.
17 = 1^2 + 16*1^2.
Therefore there are 6 different values of k for which 17 can be written in the form a^2 + k*b^2.
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MATHEMATICA
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Table[cnt = 0; Do[b = 1; found = False; While[q = n - k*b^2; ! found && q > 0, If[IntegerQ[Sqrt[q]], cnt++; found = True]; b++], {k, n}]; cnt, {n, 100}] (* T. D. Noe, Sep 11 2012 *)
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PROG
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(PARI) for(n=1, 100, sol=0; for(k=1, n-1, for(x=1, n, if(issquare(n-k*x*x)&&n-k*x*x>0, sol++; break))); print1(sol", ")) /* V. Raman, Oct 16 2012 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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