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A216134 Numbers k such that 2 * A000217(k) + 1 is triangular. 12

%I #97 Nov 27 2022 10:38:47

%S 0,1,4,9,26,55,154,323,900,1885,5248,10989,30590,64051,178294,373319,

%T 1039176,2175865,6056764,12681873,35301410,73915375,205751698,

%U 430810379,1199208780,2510946901,6989500984,14634871029,40737797126,85298279275,237437281774

%N Numbers k such that 2 * A000217(k) + 1 is triangular.

%C Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - _Alex Ratushnyak_, Apr 18 2013

%C For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - _Stefano Spezia_, Nov 25 2022

%H Colin Barker, <a href="/A216134/b216134.txt">Table of n, a(n) for n = 0..1000</a>

%H Ahmet Tekcan and Alper Erdem, <a href="https://arxiv.org/abs/2211.08907">General Terms of All Almost Balancing Numbers of First and Second Type</a>, arXiv:2211.08907 [math.NT], 2022.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Pell_number">Pell numbers</a>

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

%F G.f.: x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)). - _R. J. Mathar_, Sep 08 2012

%F sqrt(2) = lim_{k->infinity} ((a(2k+1) + a(2k) + 1)/2)/(a(2k+1) - a(2k)) = lim_{k->infinity} A001333(2k + 1)/A000129(2k + 1).

%F 1 + (sqrt 2) = lim_{k->infinity} (a(2k + 1) - a(2k))/(a(2k + 1) - 2*a(2k) + a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A000129(2k).

%F 1 + 1/(sqrt 2) = lim_{k->infinity} (a(2k+1) - a(2k))/(a(2k) - a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A001333(2k).

%F a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - _Raphie Frank_, Jan 04 2013

%F From _Raphie Frank_, Jan 04 2013: (Start)

%F A124174(n) = a(n)*(a(n) + 1)/2.

%F A079496(n) = a(n + 1) - a(n).

%F A000129(2n) = a(2n) - 2*a(2n - 1) + a(2n - 2).

%F A000129(2n) = a(2n + 1) - 2*a(2n) + a(2n - 1).

%F A000129(2n + 1) = a(2n + 1) - a(2n).

%F A001333(2n) = a(2n) - a(2n - 1).

%F A001333(2n + 1) = (a(2n + 1) + a(2n) + 1)/2.

%F A006451(n + 1) = (a(n + 2) + a(n))/2.

%F A006452(n + 2) = (a(n + 2) - a(n))/2.

%F A124124(n + 2) = (a(n + 2) + a(n))/2 + (a(n + 2) - a(n)).

%F (End)

%F a(n + 2) = sqrt(8*a(n)^2 + 8*a(n) + 9) + 3*a(n) + 1; a(0) = 0, a(1) = 1. - _Raphie Frank_, Feb 02 2013

%F a(n) = (3/8 + sqrt(2)/4)*(1 + sqrt(2))^n + (-1/8 - sqrt(2)/8)*(-1 + sqrt(2))^n + (3/8 - sqrt(2)/4)*(1 - sqrt(2))^n + (-1/8 + sqrt(2)/8)*(-1 - sqrt(2))^n - 1/2. - _Robert Israel_, Aug 13 2014

%F E.g.f.: (1/4)*(-2*cosh(x) - 2*sinh(x) + 2*cosh(sqrt(2)*x)*(cosh(x) + 2*sinh(x)) + sqrt(2)*(cosh(x) + 3*sinh(x))*sinh(sqrt(2)*x)). - _Stefano Spezia_, Dec 10 2019

%t LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* _T. D. Noe_, Sep 03 2012 *)

%o (PARI) Vec( x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)) + O(x^66) ) \\ _Joerg Arndt_, Aug 13 2014

%o (PARI) isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ _Michel Marcus_, Aug 13 2014

%Y Cf. A124174, A000129, A001333, A006451, A006452, A124124, A079496.

%Y Cf. A000217, A069017 (triangular numbers of the form k^2 + k + 1).

%K nonn,easy

%O 0,3

%A _Raphie Frank_, Sep 01 2012

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Last modified March 28 07:20 EDT 2024. Contains 371235 sequences. (Running on oeis4.)