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

Author

Eric W. Weisstein

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A028230, A036428, A251963.

Cf. A006253.

Programs

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

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 *)

PARI

Vec(x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)) + O(x^100)) \\ Colin Barker, Dec 11 2014

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 = 2, 26, 362, 5042

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

a(t+2) = 14*a(t+1) - a(t) - 4 with = 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