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%I #7 Oct 19 2017 03:14:29
%S 1,5,1,13,1,5,25,3805,125,53,569,1,851525,73,149,9305,385645,85,
%T 82596761,126985,1,113,1517,4574225,1,5,535979945,63445,145,7170685,
%U 19805,55335641,493,3793,265,65,1027776565,1
%N 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.
%t 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
%Y Cf. A002144, A094048.
%K nonn
%O 1,2
%A _Matthijs Coster_, Apr 29 2004
%E Edited by _Don Reble_, Apr 30 2004
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