A077999 Expansion of (1-x)/(1-2*x-2*x^3).

1, 1, 2, 6, 14, 32, 76, 180, 424, 1000, 2360, 5568, 13136, 30992, 73120, 172512, 407008, 960256, 2265536, 5345088, 12610688, 29752448, 70195072, 165611520, 390727936, 921846016, 2174915072, 5131286016, 12106264064, 28562358272, 67387288576, 158987105280

Offset: 0

Author

N. J. A. Sloane, Nov 17 2002

Keywords

• nonn
• easy

Comments

a(n) = number of permutations on [n] that avoid nonconsecutive instances of the patterns 321 and 312. For example, a(4) does not count pi=4231 because 431 forms a 321 pattern in pi but 431 is not a consecutive (that is, contiguous) string in pi; also, the first 3 letters form a 312 pattern but that's not disqualifying because they do occur consecutively. Counting these permutations by various statistics yields the listed formulas/recurrences. - David Callan, Oct 26 2006

a(n) = term (1,1) of M^n, M = the 4 X 4 matrix [1,0,1,1; 1,1,0,0; 0,1,0,1; 1,0,0,1]. a(n)/a(n-1) tends to 2.3593040859..., an eigenvalue of the matrix and a root to the characteristic polynomial x^4 - 3x^3 + 2x^2 - 2x + 2. - Gary W. Adamson, Oct 01 2008

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• S. R. Finch, Several Constants Arising in Statistical Mechanics, arXiv:math/9810155 [math.CO], 1998-1999. see p. 8.
• Index entries for linear recurrences with constant coefficients, signature (2,0,2)

Crossrefs

Cf. A052912, A110524.

Programs

GAP

a:=[1,1,2];; for n in [4..40] do a[n]:=2*(a[n-1]+a[n-3]); od; a; # G. C. Greubel, Jun 27 2019

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-x)/( 1-2*x-2*x^3) )); // G. C. Greubel, Jun 27 2019

Mathematica

CoefficientList[Series[(1-x)/(1-2x-2x^3),{x,0,40}],x] (* or *) LinearRecurrence[{2,0,2},{1,1,2},40] (* Harvey P. Dale, Sep 10 2016 *)

PARI

my(x='x+O('x^40)); Vec((1-x)/(1-2*x-2*x^3)) \\ G. C. Greubel, Jun 27 2019

Sage

((1-x)/(1-2*x-2*x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jun 27 2019

Formula

a(n) = 2*(a(n-1) + a(n-3)) counts the above permutations by first entry. a(n) = a(n-1) + a(n-2) + 3*Sum_{k=0..n-3} a(k) counts by last entry. a(n) = 2^(n-1) + Sum_{k=0..n-3} 2^(n-2-k)*a(k) counts by location of first 3xx pattern. a(n) = Sum_{k=0..floor(n/3)} ((n-k)/(n-2k))* binomial(n-2*k,k) * 2^(n-2*k-1) counts by number of 3xx patterns. - David Callan, Oct 26 2006

a(n) = A052912(n) - A052912(n-1). - R. J. Mathar, May 30 2014

a(n) = (-1)^n * A110524(n). - G. C. Greubel, Jun 27 2019