A046177 Squares (A000290) which are also hexagonal numbers (A000384).

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007

Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009

As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011

Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015

Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

References

• M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Links

• Colin Barker, Table of n, a(n) for n = 1..327
• Eric Weisstein's World of Mathematics, Hexagonal Square Number.
• Eric Weisstein's World of Mathematics, Square Triangular Number.
• Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Crossrefs

Cf. A008844, A046176, A253826.

Cf. A001110 (Numbers that are both triangular and square).

Cf. A000290, A000384, A016754, A253826.

Programs

Mathematica

LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)

PARI

Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

Formula

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007

a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009

a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009

From Ant King, Nov 07 2011: (Start)

a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).

a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).

a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).

a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).

a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).

G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).

(End)

a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016

a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016