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A242444
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Number of squares k^2 < n with k^2*p == 1 (mod n) for some prime p < n.
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4
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0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 2, 0, 0, 1, 1, 0, 2, 1, 2, 0, 2, 0, 3, 0, 1, 1, 1, 1, 2, 1, 0, 0, 2, 1, 2, 2, 1, 1, 2, 0, 3, 0, 1, 2, 2, 2, 1, 0, 2, 2, 2, 0, 1, 3, 0, 2, 2, 2, 2, 1, 2, 0, 2, 0, 2, 2, 2, 2, 3, 1, 3, 0
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OFFSET
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1,11
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COMMENTS
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According to the conjecture in A242441, a(n) > 0 for all n > 1848.
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LINKS
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EXAMPLE
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a(9) = 1 since 2^2*7 == 1 (mod 9).
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MATHEMATICA
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r[k_, n_]:=PowerMod[k^2, -1, n]
Do[m=0; Do[If[GCD[k, n]==1&&PrimeQ[r[k, n]], m=m+1], {k, 1, Sqrt[n-1]}]; Print[n, " ", m]; Continue, {n, 1, 80}]
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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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