A094049 Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least k for which m^2+1=p(n)*k^2 has a solution.
1, 5, 1, 13, 1, 5, 25, 3805, 125, 53, 569, 1, 851525, 73, 149, 9305, 385645, 85, 82596761, 126985, 1, 113, 1517, 4574225, 1, 5, 535979945, 63445, 145, 7170685, 19805, 55335641, 493, 3793, 265, 65, 1027776565, 1
Offset: 1
Keywords
Programs
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Mathematica
f[n_] := Block[{y = 1}, While[ !IntegerQ[ Sqrt[n*y^2 - 1]], y++]; y]; lst = {}; Do[p = Prime@n; If[Mod[p, 4] == 1, AppendTo[lst, f@p]; Print[{n, f@p}]], {n, 66}]; lst
Extensions
Edited by Don Reble, Apr 30 2004