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Nonnegative integers n such that 2n^2 + 2n - 3 is square.
9

%I #81 Sep 17 2020 06:18:18

%S 1,2,6,13,37,78,218,457,1273,2666,7422,15541,43261,90582,252146,

%T 527953,1469617,3077138,8565558,17934877,49923733,104532126,290976842,

%U 609257881,1695937321,3551015162,9884647086,20696833093,57611945197,120629983398,335787024098

%N Nonnegative integers n such that 2n^2 + 2n - 3 is square.

%C First differences are apparently in A143608. [_R. J. Mathar_, Jul 17 2009]

%C Alternative definition: T_n and (T_n - 1)/2 are triangular numbers. - _Raphie Frank_, Sep 06 2012

%H Michael De Vlieger, <a href="/A124124/b124124.txt">Table of n, a(n) for n = 1..2613</a>

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://arxiv.org/abs/1810.07895">Classes of Gap Balancing Numbers</a>, arXiv:1810.07895 [math.NT], 2018.

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://ajc.maths.uq.edu.au/pdf/77/ajc_v77_p318.pdf">Counting families of generalized balancing numbers</a>, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.

%H Hermann Stamm-Wilbrandt, <a href="/A124124/a124124.svg">4 interlaced bisections</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,6,-6,-1,1).

%F It appears that a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) if n is even, a(n) = 5*a(n-1)-5*a(n-2)+a(n-3) if n is odd. Can anyone confirm this?

%F Corrected and confirmed (using the g.f.) by _Robert Israel_, Aug 27 2014

%F 2*a(n) = sqrt(7+2*A077442(n-1)^2)-1. - _R. J. Mathar_, Dec 03 2006

%F a(n) = a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [_R. J. Mathar_, Jul 17 2009]

%F For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - _Charlie Marion_, Jan 03 2012

%F From _Raphie Frank_, Sep 06 2012: (Start)

%F If y = A006452(n), then a(n) = 2y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n))).

%F Also see A216134 [a(n) = y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n)))].

%F (End)

%F From _Hermann Stamm-Wilbrandt_, Aug 27 2014: (Start)

%F a(2*n+2) = A098586(2*n).

%F a(2*n+1) = A098790(2*n).

%F a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>6. (End)

%F a(2*n+1)^2 + (a(2*n+1)+1)^2 = A038761(n)^2 + 2^2. - _Hermann Stamm-Wilbrandt_, Aug 31 2014

%p A124124 := proc(n)

%p coeftayl(x*(1+x-2*x^2+x^3+x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)), x=0, n);

%p end proc:

%p seq(A124124(n), n=1..20); # _Wesley Ivan Hurt_, Aug 04 2014

%p # Alternative:

%p a[1]:= 1: a[2]:= 2: a[3]:= 6:

%p for n from 4 to 1000 do

%p a[n]:= (3 + 2*(n mod 2))*(a[n-1]-a[n-2])+a[n-3]

%p od:

%p seq(a[n],n=1..100); # _Robert Israel_, Aug 13 2014

%t LinearRecurrence[{1,6,-6,-1,1},{1,2,6,13,37},40] (* _Harvey P. Dale_, Nov 05 2011 *)

%t CoefficientList[Series[(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)*(x^2 - 2*x - 1)*(x^2 + 2*x - 1)), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Aug 04 2014 *)

%o (PARI)

%o for(n=1,10^10,if(issquare(2*n^2+2*n-3),print1(n,", "))) \\ _Derek Orr_, Aug 13 2014

%Y Cf. A001108, A001652, A001653, A006452, A008844, A046090, A046172, A077442, A098790, A216134.

%K nonn,easy

%O 1,2

%A _John W. Layman_, Nov 29 2006

%E More terms from _Harvey P. Dale_, Feb 07 2011

%E More terms from _Wesley Ivan Hurt_, Aug 04 2014