A016129 Expansion of 1/((1-2x)(1-6x)).

1, 8, 52, 320, 1936, 11648, 69952, 419840, 2519296, 15116288, 90698752, 544194560, 3265171456, 19591036928, 117546237952, 705277460480, 4231664828416, 25389989101568, 152339934871552, 914039609753600

Offset

0,2

Comments

a(n) = sum of n-th row in triangle A100851. - Reinhard Zumkeller, Nov 20 2004

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..900

Index entries for linear recurrences with constant coefficients, signature (8,-12)

Formula

a(n)= A071951(n+2, 2) = 9*(2*3)^(n-1) - (2*1)^(n-1) = (2^(n-1))*(3^(n+1)-1), n>=0. - Wolfdieter Lang, Nov 07 2003

From Lambert Klasen (lambert.klasen(AT)gmx.net), Feb 05 2005: (Start)

G.f.: 1/((1-2*x)*(1-6*x)).

E.g.f.: (-exp(2*x) + 3*exp(6*x))/2.

a(n) = (6^(n+1) - 2^(n+1))/4. (End)

a(n)^2 = A144843(n+1). - Philippe Deléham, Nov 26 2008

a(n-1) = ((4+sqrt4)^n - (4-sqrt4)^n)/4 in Fibonacci form. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008

a(n) = 8*a(n-1) - 12*a(n-2). - Philippe Deléham, Jan 01 2009

a(n) = det(|ps(i+2,j+1)|, 1 <= i,j <= n), where ps(n,k) are Legendre-Stirling numbers of the first kind (A129467). - Mircea Merca, Apr 06 2013

Mathematica

Join[{a=1, b=8}, Table[c=8*b-12*a; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *)
CoefficientList[Series[1/((1-2x)(1-6x)), {x, 0, 30}], x] (* or *) LinearRecurrence[{8, -12}, {1, 8}, 30] (* Harvey P. Dale, Jan 15 2015 *)

Prog

(Sage) [lucas_number1(n, 8, 12) for n in range(1, 21)] # Zerinvary Lajos, Apr 23 2009
(Sage) [(6^n - 2^n)/4 for n in range(1, 21)] # Zerinvary Lajos, Jun 04 2009
(Magma) [(6^(n+1)-2^(n+1))/4 : n in [0..20]]; // Vincenzo Librandi, Oct 09 2011
(PARI) Vec(1/(1-2*x)/(1-6*x)+O(x^99)) \\ Charles R Greathouse IV, Apr 17 2012

Crossrefs

Cf. A089278, A089500.

Sequence in context: A257285 A125345 A111996 * A006631 A205218 A287813

Adjacent sequences: A016126 A016127 A016128 * A016130 A016131 A016132

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved