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%I #8 Aug 01 2015 12:10:08
%S 25,99,609,650,1189,3479,4901,5429,11659,16275,29469,38479,62525,
%T 73814,78089,117739,142449,201116,203319,240199,328769,381275,406900,
%U 504889,576909,743775,839629,1005731,1058819,1183259,1464709,1622919,1960244
%N Values of n such that the expression sqrt(4!*(n+1) + 1) yields a perfect power.
%e 25 is in the sequence since sqrt(4!*(25+1) + 1) = 25 = 5^2;
%e 99 is in the sequence since sqrt(4!*(99+1) + 1) = 49 = 7^2. - _Jon E. Schoenfield_, Aug 01 2015
%t lst = {}; Do[a = Sqrt[4! (n + 1) + 1]; If[IntegerQ@ a && GCD @@ Last /@ FactorInteger@a > 1, AppendTo[lst, n]], {n, 0, 1977428}]; lst (* _Robert G. Wilson v_, Sep 24 2008 *)
%Y Subset of A144065.
%K nonn
%O 1,1
%A _Alexander R. Povolotsky_, Sep 22 2008
%E More terms from _Robert G. Wilson v_, Sep 24 2008
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