A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192

Offset: 1

Author

Robert G. Wilson v, Nov 12 2001

Keywords

• nonn
• easy

Comments

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

Examples

			T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008)
• Index entries for two-way infinite sequences
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Cf. A001652, A002315, A003499 (first differences), A065651.

Programs

Mathematica

CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]
LinearRecurrence[{7,-7,1},{6,40,238},41] (* Harvey P. Dale, Dec 27 2011 *)

PARI

a(n)=-1+subst(poltchebi(abs(n+1))-poltchebi(abs(n)),x,3)/2

PARI

Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

Formula

a(n) = 2*A001652(n) = -1 + A002315(n).

a(n) - a(n-1) = A003499(n).

From Michael Somos, Apr 07 2003: (Start)

G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).

a(n) = 6*a(n-1) - a(n-2) + 4.

a(-1-n) = -a(n) - 2. (End)

a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011

a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012

a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016

From Klaus Purath, Sep 05 2021: (Start)

(a(n+1) - a(n) - a(n-1) + a(n-2))/8 = A005319(n), for n >= 3.

((a(n) - a(n-1))^2)/2 - 2 = A005319(n)^2 = 2*A132592(n), for n>= 2.

a(n) = A265278(2*n+1).

a(n) = A293004(2*n+1).

a(n) = A213667(2*n).

a(n) = Sum_{k=1..n} A003499(k). (End)