A108920 Number of positive integers k>n such that n+k divides n^2+k^2.

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, 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, 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

Offset: 1

Author

John W. Layman, Jul 19 2005

Keywords

• nonn

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?

Examples

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

Crossrefs

Cf. A066761.