A066761 -id:A066761 - cp's OEIS frontend

A082604 a(n) is the total number of positive integral solutions (x,y), order being taken into account, to the equation 1/x + 1/y + 1/z = 1/n.

10, 46, 106, 148, 196, 310, 232, 394, 442, 538, 352, 910, 334, 778, 1126, 736, 418, 1210, 460, 1702, 1414, 904, 538, 2128, 904, 1006, 1336, 1990, 616, 2884, 520, 1576, 1876, 1228, 2668, 3142, 490, 1258, 2212, 3556, 706, 3640, 604, 2782, 3808, 1186, 790

Offset: 1

Yuval Dekel (dekelyuval(AT)hotmail.com), May 23 2003

nonn

Cf. A082986.

a(n) = 6*(A004194(n)-A066761(n))-8. - Vladeta Jovovic, May 26 2003

More terms from Vladeta Jovovic, May 26 2003

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

Original entry on oeis.org

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

John W. Layman, Jul 19 2005

nonn

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?

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

Cf. A066761.

cp's OEIS Frontend

A082604 a(n) is the total number of positive integral solutions (x,y), order being taken into account, to the equation 1/x + 1/y + 1/z = 1/n.

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

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