OFFSET
0,2
COMMENTS
With interpolated zeros this has a(n) = (6*0^n + 4^n + (-4)^n + 4*2^n + 4*(-2)^n)/16 and counts closed walks of length n at a vertex of the 4-cube. [Typo corrected by Alexander R. Povolotsky, May 26 2008]
Also, cogrowth sequence of the 16-element group C2^4. - Sean A. Irvine, Nov 10 2024
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
G. R. Franssens, On a number pyramid related to the binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
Katarzyna Grygiel, Pawel M. Idziak and Marek Zaionc, How big is BCI fragment of BCK logic, arXiv preprint arXiv:1112.0643 [cs.LO], 2011. [From N. J. A. Sloane, Feb 21 2012]
L. Reyzin, Mathoverflow, Number of closed walks on an n-cube.
Index entries for linear recurrences with constant coefficients, signature (20,-64).
FORMULA
G.f.: (1-16*x+24*x^2)/((1-4*x)*(1-16*x)).
a(n) = 3*0^n/8 + 16^n/8 + 4^n/2.
From Peter Bala, Nov 13 2006: (Start)
E.g.f.: cosh^4(x).
O.g.f.: 1/(1-4*1*x/(1-3*2*x/(1-2*3*x/(1-1*4*x)))) (continued fraction). (End)
(-1)^n*a(n) = Sum_{k=0..n} A086872(n,k)*(-5)^(n-k). - Philippe Deléham, Aug 17 2007
a(n) = 20*a(n-1) - 64*a(n-2); a(0) = 1, a(1) = 4, a(2) = 40. - Harvey P. Dale, Aug 23 2011
a(n) = 4*A026244(n-1), n > 0. - R. J. Mathar, Oct 24 2014
a(n) = (1/2^4)*Sum_{j = 0..4} binomial(4, j)*(4 - 2*j)^(2*n). See Reyzin link. - Peter Bala, Jun 03 2019
MATHEMATICA
CoefficientList[Series[(1-16x+24x^2)/((1-4x)(1-16x)), {x, 0, 30}], x] (* or *) Join[{1}, LinearRecurrence[{20, -64}, {4, 40}, 30]] (* Harvey P. Dale, Aug 23 2011 *)
PROG
(Magma) [3*0^n/8+16^n/8+4^n/2: n in [0..30]]; // Vincenzo Librandi, May 31 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 11 2004
EXTENSIONS
Title improved by Sean A. Irvine at the suggestion of Peter Bala, Jun 04 2019
STATUS
approved