A175035 Offsets i such that i + n*(n+1)/2 is a perfect square for some positive integer n.

1, 3, 4, 6, 8, 9, 10, 13, 15, 16, 19, 21, 22, 24, 25, 26, 28, 30, 33, 34, 35, 36, 39, 43, 45, 46, 48, 49, 53, 54, 55, 58, 60, 61, 63, 64, 66, 71, 72, 75, 76, 78, 79, 80, 81, 85, 89, 90, 91, 93, 94, 97, 99, 100, 103, 105, 106, 108, 111, 114, 115, 116, 118, 120

Offset: 1

Author

Ctibor O. Zizka, Nov 10 2009

Keywords

• easy
• nonn

Comments

The ansatz n*(n+1)/2+i=s^2 can be transformed into (2*n+1)^2-2*(2*s)^2 =1-8*i.

A necessary condition for solutions to this Diophantine equation is that D=2 is a quadratic residue of the squarefree part of 8*i-1 (see A057126).

A sufficient condition is then available by a sequence of tests on the continued fractions of a quadratic surd that originates from a solution of this congruence.

See Mollin and Matthews for details. - R. J. Mathar, Nov 16 2009

Links

• R. A. Mollin, Simple continued fraction solutions for Diophantine Equations, Exposit. Mathem. 19 (2001) 55-73.
• Keith Matthews, The Diophantine equation x^2-Dy^2=N,D >0, Exposit. Mathem. 18 (4) (2000) 323-331 [MR1788328].

Crossrefs

Cf. A001108, A006451, A154138 to A154154.

Cf. A230044.

Programs

Mathematica

Take[Rest[Ceiling[Sqrt[#]]^2-#&/@Accumulate[Range[1000]]//Union],70] (* Harvey P. Dale, Sep 07 2019 *)

PARI

is(n)=#bnfisintnorm(bnfinit(z^2-8),-8*n+1) /* Ralf Stephan, Oct 14 2013 */

Extensions

Extended by R. J. Mathar, Nov 26 2009