A046184 Indices of octagonal numbers which are also squares.

1, 9, 121, 1681, 23409, 326041, 4541161, 63250209, 880961761, 12270214441, 170902040409, 2380358351281, 33154114877521, 461777249934009, 6431727384198601, 89582406128846401, 1247721958419651009, 17378525011746267721, 242051628206028097081

Offset

1,2

Comments

The equation a(t)*(3*a(t)-2) = m*m is equivalent to the Pell equation (3*a(t)-1)*(3*a(t)-1) - 3*m*m = 1. - Paul Weisenhorn, May 12 2009

As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 16 2011

Also numbers n such that the octagonal number N(n) is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014

Also nonnegative integers y in the solutions to 2*x^2 - 6*y^2 + 4*x + 4*y + 2 + 2 = 0, the corresponding values of x being A251963. - Colin Barker, Dec 11 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Eric Weisstein's World of Mathematics, Octagonal Square Number.

Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Formula

{n: A000567(n) in A000290}.

Nearest integer to (1/6) * (2+sqrt(3))^(2n-1). - Ralf Stephan, Feb 24 2004

a(n) = A045899(n-1) + 1 = A051047(n+1) + 1 = A003697(2n-2). - N. J. A. Sloane, Jun 12 2004

a(n) = A001835(n)^2. - Lekraj Beedassy, Jul 21 2006

From Paul Weisenhorn, May 12 2009: (Start)

With A=(2+sqrt(3))^2=7+4*sqrt(3) the equation x*x-3*m*m=1 has solutions

x(t) + sqrt(3)*m(t) = (2+sqrt(3))*A^t and the recurrences

x(t+2) = 14*x(t+1) - x(t) with <x(t)> = 2, 26, 362, 5042

m(t+2) = 14*m(t+1) - m(t) with <m(t)> = 1, 15, 209, 2911

a(t+2) = 14*a(t+1) - a(t) - 4 with <a(t)> = 1, 9, 121, as above. (End)

From Ant King, Nov 15 2011: (Start)

a(n) = 14*a(n-1) - a(n-2) - 4.

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

a(n) = (1/6)*( (2+sqrt(3))^(2n-1) + (2-sqrt(3))^(2n-1) + 2 ).

a(n) = ceiling( (1/6)*(2 + sqrt(3))^(2n-1) ).

a(n) = (1/6)*( (tan(5*Pi/12))^(2n-1) + (tan(Pi/12))^(2n-1) + 2 ).

a(n) = ceiling ( (1/6)*(tan(5*Pi/12))^(2n-1) ).

G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)

a(n) = A006253(2n-2). - Andrey Goder, Oct 17 2021

Mathematica

LinearRecurrence[ {15, -15, 1}, {1, 9, 121}, 17 ] (* Ant King, Nov 16 2011 *)
CoefficientList[Series[x (1-6x+x^2)/((1-x)(1-14x+x^2)), {x, 0, 30}], x] (* Harvey P. Dale, Sep 01 2021 *)

Prog

(Magma) I:=[1, 9, 121]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Nov 17 2011
(PARI) Vec(x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)) + O(x^100)) \\ Colin Barker, Dec 11 2014

Crossrefs

Cf. A028230, A036428, A251963.

Cf. A006253.

Sequence in context: A302941 A183514 A138978 * A084769 A246467 A202835

Adjacent sequences: A046181 A046182 A046183 * A046185 A046186 A046187

Keyword

nonn,easy

Author

Eric W. Weisstein

Status

approved