A046178 Indices of pentagonal numbers that are also hexagonal.

1, 165, 31977, 6203341, 1203416145, 233456528757, 45289363162681, 8785902997031325, 1704419892060914337, 330648673156820350021, 64144138172531086989705, 12443632156797874055652717, 2414000494280615035709637361, 468303652258282519053613995285

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

The reason we obtain the same Diophantine equation with various parameters is the following: the number that is written 361 in base 4*A046179(n)-2 is the square of 6*A046178(n)-1. That is, 361 in base 110770 is 3*110770^2 + 6*110770 + 1 = 36810643321, i.e., the square of 191861 if we consider the third terms of A046179 and A046178, which are 27693 and 31977, respectively. - Richard Choulet, Oct 03 2007

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->oo} a(n)/a(n-1) = (2 + sqrt(3))^4 = 97 + 56*sqrt(3). - Ant King, Dec 14 2011

Links

• Colin Barker, Table of n, a(n) for n = 1..438
• Eric Weisstein's World of Mathematics, Hexagonal Pentagonal Number.
• Index entries for linear recurrences with constant coefficients, signature (195,-195,1)

Crossrefs

Cf. A046179, A046180.

Programs

Mathematica

LinearRecurrence[{195, -195, 1}, {1, 165, 31977}, 11] (* Ant King, Dec 14 2011 *)

PARI

Vec(x*(3*x^2+30*x-1)/((x-1)*(x^2-194*x+1)) + O(x^20)) \\ Colin Barker, Jun 21 2015

Formula

From Warut Roonguthai Jan 08 2001: (Start)

a(n) = 194*a(n-1) - a(n-2) - 32.

G.f.: x*(1-30*x-3*x^2)/((1-x)*(1-194*x+x^2)). (End)

a(n+1) = 97*a(n) - 16 + 28*sqrt(12*a(n)^2 - 4*a(n) + 1). - Richard Choulet, Oct 09 2007

From Ant King, Dec 14 2011: (Start)

a(n) = 195*a(n-1) - 195*a(n-2) + a(n-3).

a(n) = (1/12)*((sqrt(3)-1)*(2+sqrt(3))^(4n-2) - (sqrt(3)+1)* (2-sqrt(3))^(4n-2) + 2).

a(n) = ceiling((1/12)*(sqrt(3)-1)*(2+sqrt(3))^(4n-2)).

(End)