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A212274 Minimal k >= 5n such that n^2 + 2nk + k is a perfect square. 0

%I #16 Feb 13 2013 23:58:30

%S 5,12,16,20,64,76,49,100,112,64,136,148,160,172,184,105,120,220,120,

%T 244,256,121,280,292,161,316,144,176,352,364,221,217,400,217,424,436,

%U 232,225,472,484,496,288,273,532,225,288,253,580,352,604,616,276,640

%N Minimal k >= 5n such that n^2 + 2nk + k is a perfect square.

%C Without any restriction, trivially, a(n)=0 and, if to consider a(n) positive, then, again trivially, a(n)=1. Without the restriction k >= 5n, we have a(n)=4*n+1; on the other hand, if to require a(n)>=5*n and in addition a(n+1)>a(n), then we obtain sequence 5,12,16,20 and, beginning with n=5, we have progression 64+12*(n-5).

%t Table[k = 5*n; While[! IntegerQ[Sqrt[n^2 + 2*n*k + k]], k++]; k, {n, 100}] (* _T. D. Noe_, May 18 2012 *)

%K nonn

%O 1,1

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, May 13 2012

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