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Number of positive integers k>n such that n+k divides n^2+k^2.
0

%I #3 Mar 30 2012 17:37:53

%S 0,1,2,2,2,4,2,3,4,5,2,7,2,5,7,4,2,8,2,7,8,5,2,10,4,5,6,7,2,15,2,5,8,

%T 5,7,13,2,5,8,10,2,15,2,8,12,5,2,13,4,9,8,8,2,12,8,10,8,5,2,23,2,5,13,

%U 6,8,15,2,8,8,16,2,17,2,5,13,8,7,16,2,13,8,5,2,23,8,5,8,10,2,26,7,8,8,5,8

%N Number of positive integers k>n such that n+k divides n^2+k^2.

%C If n+k divides n^2+k^2 then k<=n(2n+1). If n>2 then there are at least two values of k>n such that n+k divides n^2+k^2; they are k=n(n-1) and k=n(2n-1). Further, if n is prime, these are the only two values. If n=2^j, then there are exactly j values of k>x such that n+k divides n^2+k^2; they are k=3n, k=7n, k=15n,..., k=(2x-1)n. Is this sequence the same as A066761 except for the prepended a(1)=0?

%e 6+k divides 36+k^2 only for k=12,18,30 and 66, so a(6)=4.

%Y Cf. A066761.

%K nonn

%O 1,3

%A _John W. Layman_, Jul 19 2005