A379983 Numbers k such that there exists a number 1 <= m <= k-1 and at least two different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2.

385, 425, 432, 450, 504, 585, 616, 630, 665, 693, 728, 770, 792, 800, 810, 850, 864, 900, 910, 935, 952, 1008, 1015, 1040, 1155, 1170, 1197, 1232, 1260, 1275, 1287, 1296, 1320, 1330, 1350, 1360, 1365, 1386, 1456, 1512, 1540, 1547, 1584, 1600, 1620, 1672, 1680

Offset: 1

Author

Jianing Song, Jan 07 2025

Keywords

• nonn

Comments

Numbers k = A355812(r) such that A379895(r) < A355813(r).

The smallest k such that there exists a number 1 <= m <= k-1 and at least three different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2 is k = 1872: we have 1/300^2 - 1/325^2 = 1/468^2 - 1/585^2 = 1/624^2 - 1/1040^2 = 1/720^2 - 1/1872^2. See the Mathematics Stack Exchange link for more examples, and A380150.

Examples

			See a-file for examples.
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..2000
• Mathematics Stack Exchange, Finding multiple ways of representing a number by a difference of inverse squares
• Jianing Song, All examples with k <= 1500

Crossrefs

Cf. A355812, A355813, A379895, A380150.

Programs

PARI

is(n) = my(v=[], m2); for(y=1, n-1, for(x=1, y-1, m2=1/(1/x^2-1/y^2+1/n^2); if(m2==m2\1 && issquare(m2), v=concat(v, [m2])); if(#Set(v)<#v, return(1)))); return(0) \\ See also A379895 for its program

Extensions

More terms from Jinyuan Wang, Jan 08 2025