A090040 (3*6^n + 2^n)/4.

1, 5, 28, 164, 976, 5840, 35008, 209984, 1259776, 7558400, 45349888, 272098304, 1632587776, 9795522560, 58773127168, 352638746624, 2115832446976, 12694994616320, 76169967566848, 457019805138944, 2742118830309376

Offset: 0

Author

Paul Barry, Nov 20 2003

Keywords

• easy
• nonn

Comments

A090040 is the Q-residue of the triangle A175840, where Q is the triangular array (t(i,j)) given by t(i,j)=1; see A193649 for the definition of Q-residue. - Clark Kimberling, Aug 07 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..300
• Index entries for linear recurrences with constant coefficients, signature (8,-12).

Crossrefs

Cf. A081335.

Programs

Magma

[(3*6^n+2^n)/4: n in [0..30]]; // Vincenzo Librandi, Jun 10 2011

Mathematica

LinearRecurrence[{8,-12},{1,5},30] (* Harvey P. Dale, Nov 23 2014 *)

PARI

a(n)=(3*6^n+2^n)/4 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: (1-3*x)/((1-2*x)*(1-6*x)).

E.g.f.: (3*exp(6*x)+exp(2*x))/4 = exp(4*x)*(cosh(2*x)+sinh(2*x)/2).

a(n) = 8*a(n-1) -12*a(n-2), a(0)=1, a(1)=5.

a(n) = (3*6^n+2^n)/4.

a(n)=6*a(n-1)-2^(n-1). - Paul Curtz, Jan 09 2009

Fourth binomial transform of (1, 1, 4, 4, 16, 16, ...). a(n)=sum{k=1..floor(n/2), C(n, 2k)4^(n-k-1)}. - Paul Barry, Nov 22 2003

a(n) = A019590 (mod 4), proof via a(n)=8*a(n-1)-12*a(n-2). - R. J. Mathar, Feb 25 2009

a(n) = Sum_{k, 0<=k<=n} A117317(n,k)*3^k. - Philippe Deléham, Jan 28 2012