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Numbers n such that s! + n^2 and (s + 2)! + n^2 are squares for some s, ordered by s.
3

%I #21 Oct 12 2012 14:00:57

%S 1,179,204,108,996,2934,81720,2152080,851040,271106640,7935621120,

%T 1143137318400,52250931532800

%N Numbers n such that s! + n^2 and (s + 2)! + n^2 are squares for some s, ordered by s.

%C The corresponding numbers s are: 5, 6, 7, 8, 9, 11, 12, 14, 16, 17, 19, 25, 27.

%e 5! + 1 = 11^2 and 7! + 1 = 71^2.

%e 6! + 179^2 = 181^2 and 8! + 179^2 = 269^2.

%e 7! + 204^2 = 216^2 and 9! + 204^2 = 636^2.

%e 8! + 108^2 = 228^2 and 10! + 108^2 = 1908^2.

%o (PARI) for(n=4,32,a=n!;b=((n+2)*(n+1)-1)*a;c=divisors(b);for(i=2,#c-1,s=c[i];r=b\s;if(r<s,next(2),d=abs(s-r)/2;t=d^2-a;if(issquare(t),print([n,d,sqrtint(t)])))))

%Y Cf. A217277, A217541, A217550, A217551, A217553.

%K nonn

%O 1,2

%A _Robin Garcia_, Oct 06 2012