OFFSET
1,3
COMMENTS
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?
EXAMPLE
6+k divides 36+k^2 only for k=12,18,30 and 66, so a(6)=4.
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Layman, Jul 19 2005
STATUS
approved