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Least k such that n! + k^2 + k is a perfect square.
1

%I #22 Nov 27 2013 06:48:13

%S 0,0,1,5,3,119,719,260,180,495,3628799,39916799,6561647,9899,722799,

%T 15308432,735092,779904000,1193692724,77841862127,1947879910469375,

%U 7981643124,231453720692,37427202522827,1793894810624,251579908188224,286527829489835787,33866880878997899

%N Least k such that n! + k^2 + k is a perfect square.

%C a(n) <= n!-1, because with k=n!-1 we have n!+k^2+k = n! + (n!)^2 - 2*n! + 1 + n! - 1 = (n!)^2, a perfect square.

%H Jon E. Schoenfield, <a href="/A230389/b230389.txt">Table of n, a(n) for n = 0..42</a>

%e 4! + 3^2+3 = 24+12 = 36, a perfect square, so a(4)=3.

%Y Cf. A000290, A000142, A002378.

%K nonn

%O 0,4

%A _Alex Ratushnyak_, Oct 18 2013

%E a(18)-a(21) from _Jon E. Schoenfield_, Oct 24 2013

%E a(22) from _Jon E. Schoenfield_, Oct 26 2013

%E a(23) from _Jon E. Schoenfield_, Oct 29 2013

%E More terms from _Jon E. Schoenfield_, Nov 26 2013

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