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

0, 1, 4, 9, 26, 55, 154, 323, 900, 1885, 5248, 10989, 30590, 64051, 178294, 373319, 1039176, 2175865, 6056764, 12681873, 35301410, 73915375, 205751698, 430810379, 1199208780, 2510946901, 6989500984, 14634871029, 40737797126, 85298279275, 237437281774

Offset

0,3

Comments

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

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

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.

Wikipedia, Pell numbers

Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Formula

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

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).

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).

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).

a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013

From Raphie Frank, Jan 04 2013: (Start)

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

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

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

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

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

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

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

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

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

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

(End)

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

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

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

Mathematica

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* T. D. Noe, Sep 03 2012 *)

Prog

(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
(PARI) isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ Michel Marcus, Aug 13 2014

Crossrefs

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

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

Sequence in context: A090117 A329125 A020181 * A226908 A328657 A335983

Adjacent sequences: A216131 A216132 A216133 * A216135 A216136 A216137

Keyword

nonn,easy

Author

Raphie Frank, Sep 01 2012

Status

approved