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 A065113 Sum of the squares of the n-th and the (n+1)st triangular numbers (A000217) is a perfect square. 4

%I

%S 6,40,238,1392,8118,47320,275806,1607520,9369318,54608392,318281038,

%T 1855077840,10812186006,63018038200,367296043198,2140758220992,

%U 12477253282758,72722761475560,423859315570606,2470433131948080,14398739476117878,83922003724759192

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

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

%C a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - _Joerg Arndt_, Feb 15 2012

%H Harvey P. Dale, <a href="/A065113/b065113.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).

%F 2 * A001652 = -1 + A002315 = a(n+1) - A003499(n+1).

%F From _Michael Somos_, Apr 07 2003: (Start)

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

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

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

%F 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

%F 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

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

%t CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]

%t LinearRecurrence[{7,-7,1},{6,40,238},41] (* _Harvey P. Dale_, Dec 27 2011 *)

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

%o (PARI) Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ _Colin Barker_, Mar 05 2016

%Y a(n) = 2 A001652(n) = A002315(n)-1. A003499(n)=a(n)-a(n-1). Cf. A065651.

%K nonn,easy

%O 1,1

%A _Robert G. Wilson v_, Nov 12 2001

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Last modified July 22 07:23 EDT 2019. Contains 325216 sequences. (Running on oeis4.)