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Primes p of the form 4k+1 for which s=13 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.
7

%I #5 Jan 12 2020 23:45:54

%S 2749,2897,3049,3529,3557,3929,4073,4253,4657,4817,5081,5281,5417,

%T 5449,5657,5693,5869,6053,6121,6529,6793,6833,7109,7393,7541,7829,

%U 7877,7993,8209,8329,8377,8429,8501,8741,8761,8893,9001,9109,9157,9209,9257,9293

%N Primes p of the form 4k+1 for which s=13 is the least positive integer such that sp-(floor(sqrt(sp)))^2 is a square.

%e a(1)=2749 since p=2749 is the least prime of the form 4k+1 for which sp-(floor(sqrt(sp)))^2 is not a square for s=1..12, but 13p-(floor(sqrt(13p)))^2 is a square (for p=2749 it is 16).

%Y Cf. A145016, A145022, A145023, A145047.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Sep 30 2008