A164055 Triangular numbers that are one plus a perfect square.

1, 10, 325, 11026, 374545, 12723490, 432224101, 14682895930, 498786237505, 16944049179226, 575598885856165, 19553418069930370, 664240615491776401, 22564627508650467250, 766533094678624110085, 26039560591564569275626

Offset: 1

Author

Tanya Khovanova & Alexey Radul, Aug 08 2009

Keywords

• nonn
• easy

Comments

a(n+1) is the second element of the n-th set of three consecutive triangular numbers whose product is a perfect square. - Arkadiusz Wesolowski, Apr 27 2012

The triangular numbers of this sequence satisfy the Diophantine equation T(k) = k*(k+1)/2 = m^2 + 1, which is equivalent to (2k+1)^2 - 2*(2m)^2 = 9. Now, with x=2k+1 and y=2m, the Pell-Fermat equation x^2 - 2*y^2 = 9 appears. The solutions x and y of this equation are respectively in A106329 and A075848. The indices k=(x-1)/2 of the triangular numbers of this sequence are in A072221, while the indices m=y/2 of the corresponding square numbers are in A106328. - Bernard Schott, Mar 09 2019

Examples

			10 is in this sequence because it is a triangular number A000217(4) and is equal to a square plus 1: 10 = 3^2 + 1.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..500
• K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A001110, A106329, A075848 (solutions of x^2 - 2 * y^2 = 9).

Cf. A072221 (indices of the triangular terms of this sequence), A106328 (indices of the corresponding square numbers).

Programs

Haskell

a164055 n = a164055_list !! (n-1)
a164055_list = 1 : 10 : 325 : zipWith (+) a164055_list
   (map (* 35) $ tail $ zipWith (-) (tail a164055_list) a164055_list)
-- Reinhard Zumkeller, Apr 29 2012

Mathematica

LinearRecurrence[{35,-35,1}, {1,10,325}, 50] (* G. C. Greubel, Sep 09 2017 *)

PARI

my(x='x+O('x^50)); Vec(x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1))) \\ G. C. Greubel, Sep 09 2017

Formula

A000217 INTERSECT A002522.

a(n) = A000217(A072221(n-1)).

From R. J. Mathar, Sep 22 2009: (Start)

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).

G.f.: x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1)). (End)

A010054(a(n)) * A010052(a(n) - 1) = 1. - Reinhard Zumkeller, Apr 29 2012

a(n) = (A106329(n) - 1)*(A106329(n) + 1)/8 = (A106328(n))^2 + 1. - Bernard Schott, Mar 10 2019

a(n) = (14 + 9*(17+12*sqrt(2))^(1-n) - 9*(-17+12*sqrt(2))*(17+12*sqrt(2))^n) / 32. - Colin Barker, Mar 23 2019

a(n) = 9 * A001110(n) + 1 (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Extensions

Comment molded into formula by R. J. Mathar, Sep 22 2009