A072221 a(n) = 6*a(n-1) - a(n-2) + 2, with a(0)=1, a(1)=4.

1, 4, 25, 148, 865, 5044, 29401, 171364, 998785, 5821348, 33929305, 197754484, 1152597601, 6717831124, 39154389145, 228208503748, 1330096633345, 7752371296324, 45184131144601, 263352415571284, 1534930362283105, 8946229758127348, 52142448186480985

Offset: 0

Author

Lekraj Beedassy, Jul 04 2002

Keywords

• nonn
• easy

Comments

The product of three consecutive triangular numbers with middle term A000217(m) where m is in this sequence is a square.

k is in this sequence iff the triangle with sides 3,k,k+1 has integer area. Equivalently, numbers k such that 2*(k+2)*(k-1) is a square. - James R. Buddenhagen, Oct 19 2008

Triangular numbers that are equal to a square plus one have this sequence as indices. For example, the 25th triangular number is 25*26/2 = 325 = 18^2 + 1. - Tanya Khovanova and Alexey Radul, Aug 08 2009

The triangle with sides 3, a(n), a(n)+1 has area A075848(n) if n>=0. - Michael Somos, Dec 25 2018

Compare with A016064 for integers m with triangles with sides 4, m, m+2 and integer area. - Michael Somos, May 11 2019

Examples

			From _Michael Somos_, Dec 25 2018: (Start)
For n=1, the triangle (3, 4, 5) has area 6 = A075848(1).
For n=2, the triangle (3, 25, 26) has area 36 = A075848(2). (End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Sylvester Robins, Certain Series of Integral, Rational, Scalene Triangles, The American Mathematical Monthly, Vol. 1, No. 1 (Jan., 1894), pp. 13-14. See Example I.
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Cf. A000217, A001108, A001541, A016064, A075848.

Programs

Haskell

a072221 n = a072221_list !! n
a072221_list = 1 : 4 : (map (+ 2) $
   zipWith (-) (map (* 6) $ tail a072221_list) a072221_list)
-- Reinhard Zumkeller, Apr 27 2012

Mathematica

a[n_] := a[n] = 6a[n - 1] - a[n - 2] + 2; a[0] = 1; a[1] = 4; Table[ a[n], {n, 0, 20}]
LinearRecurrence[{7, -7, 1}, {1, 4, 25}, 25] (* T. D. Noe, Dec 09 2013 *)
a[ n_] := (3 ChebyshevT[ n, 3] - 1) / 2; (* Michael Somos, Dec 25 2018 *)

PARI

{a(n) = (3 * polchebyshev( n, 1, 3) - 1) / 2}; /* Michael Somos, Dec 25 2018 */

Formula

a(n) = (3*A001541(n) - 1)/2.

a(n) = 3*A001108(n) + 1. - David Scheers, Dec 25 2006

From Franz Vrabec, Aug 21 2006: (Start)

a(n) = -1/2 + (3/4)*((3+sqrt(8))^n + (3-sqrt(8))^n) for n >= 0.

a(n) = floor((3/4)*(3+sqrt(8))^n) for n > 0. (End)

G.f.: (1-3x+4x^2)/((1-x)(1-6x+x^2)). - R. J. Mathar, Sep 09 2008

a(n) = a(-n) for all n in Z. - Michael Somos, Dec 25 2018

Extensions

Edited by Robert G. Wilson v, Jul 08 2002