A046193 Indices of heptagonal numbers (A000566) which are also triangular numbers (A000217).

1, 5, 221, 1513, 71065, 487085, 22882613, 156839761, 7368130225, 50501915861, 2372515049741, 16261460067385, 763942477886281, 5236139639782013, 245987105364332645, 1686020702549740705, 79207083984837225313, 542893430081376724901, 25504435056012222218045

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

From Ant King, Oct 19 2011: (Start)

lim_{n->infinity} a(2n+1)/a(2n) = (1/2)*(47+21*sqrt(5)).

lim_{n->infinity} a(2n)/a(2n-1) = (1/2)*(7+3*sqrt(5)).

(End)

From Raphie Frank, Nov 30 2012: (Start)

Where L_n is a Lucas number and F_n is Fibonacci number:

lim_{n->infinity} a(2n+1)/a(2n) = (1/2)*(L_8 + F_8*sqrt(5)),

lim_{n->infinity} a(2n)/a(2n-1) = (1/2)*(L_4 + F_4*sqrt(5)),

a(n) = L_1*a(n-1) + L_12*a(n-2) - L_12*a(n-3)- L_1*a(n-4) + L_1*a(n-5).

(End)

Values of n such that 2*n-1 and 10*n-1 are both perfect squares. - Colin Barker, Dec 03 2016

Links

• Colin Barker, Table of n, a(n) for n = 1..798
• Eric Weisstein's World of Mathematics, Heptagonal Triangular Number.
• Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).

Crossrefs

Cf. A039835, A046194.

Programs

Magma

[n: n in [1..2*10^8] |IsSquare(2*n-1) and IsSquare(10*n-1)]; // Vincenzo Librandi, Dec 04 2016

Mathematica

LinearRecurrence[{1,322,-322,-1,1},{1,5,221,1513,71065},17] (* Ant King, Oct 19 2011 *)
Select[Range@240000000, IntegerQ@Sqrt[2 # - 1] && IntegerQ@Sqrt[10 # - 1] &] (* Vincenzo Librandi, Dec 04 2016 *)

PARI

Vec(-x*(x^4+4*x^3-106*x^2+4*x+1)/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^50)) \\ Colin Barker, Jun 23 2015

Formula

For n odd, a(n+2) = 322*a(n+1) - a(n) - 96; for n even, a(n+1) = 161*a(n) - 48 + 36*sqrt(20*a(n)^2 - 12*a(n)+1). - Richard Choulet, Sep 29 2007, Oct 09 2007

From Ant King, Oct 19 2011: (Start)

a(n) = 322*a(n-2) - a(n-4) - 96.

a(n) = a(n-1) + 322*a(n-2) - 322*a(n-3) - a(n-4) + a(n-5).

a(n) = (1/20)*((sqrt(5)-(-1)^n)*(sqrt(5)+2)^(2n-1) + (sqrt(5)+(-1)^n)*(sqrt(5)-2)^(2n-1)+6).

a(n) = ceiling((1/20)*(sqrt(5)-(-1)^n)*(2+sqrt(5))^(2n-1)).

G.f.: x*(1 + 4*x - 106*x^2 + 4*x^3 + x^4)/((1-x)*(1-18*x+x^2)*(1+18*x+x^2)).

(End)