A046179 -id:A046179 - cp's OEIS frontend

A046180 Hexagonal pentagonal numbers.

1, 40755, 1533776805, 57722156241751, 2172315626468283465, 81752926228785223683195, 3076689623521787481625080301, 115788137209866023854693048367775, 4357570752679408318225730700647767185, 163992817590548715438241125333485021875651

Offset: 1

Eric W. Weisstein

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1) = (2+sqrt(3))^8 = 18817 + 10864*sqrt(3). - Ant King, Dec 13 2011

Dickson calls the terms "triangular, pentagonal and hexagonal" (all hexagonal numbers are also triangular). - Jonathan Sondow, May 06 2014

Colin Barker, Table of n, a(n) for n = 1..219
L. E. Dickson, History of the Theory of Numbers, vol. II, pp. 19-20.
Eric Weisstein's World of Mathematics, Hexagonal Pentagonal Number.
Index entries for linear recurrences with constant coefficients, signature (37635,-37635,1).

Cf. A046178, A046179.

LinearRecurrence[{37635, -37635, 1}, {1, 40755, 1533776805}, 8] (* Ant King, Dec 13 2011 *)

Vec(x*(1+3120*x+15*x^2)/((1-x)*(1-37634*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 21 2015

a(n) = 37634*a(n-1) - a(n-2) + 3136; g.f.: x*(1+3120*x+15*x^2)/((1-x)*(1-37634*x+x^2)). - Warut Roonguthai Jan 08 2001
a(n+1) = 18817*a(n)+1568+1358*(192*a(n)^2+32*a(n)+1)^0.5 - Richard Choulet, Sep 19 2007
From Ant King, Dec 13 2011: (Start)
a(n) = 37635*a(n-1) - 37635*a(n-2) + a(n-3).
a(n) = (1/48)*((2+sqrt(3))^(8n-5)+(2-sqrt(3))^(8n-5)-4).
a(n) = floor((1/48)*(2+sqrt(3))^(8n-5)).
a(n) = (1/48)*((tan(5*Pi/12))^(8n-5)+(tan(Pi/12))^(8n-5)-4).
a(n) = floor((1/48)*(tan(5*pi/12))^(8n-5)).
(End)

A046178 Indices of pentagonal numbers that are also hexagonal.

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein

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

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)

Cf. A046179, A046180.

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

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

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)

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A046180 Hexagonal pentagonal numbers.

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