A289314 Number of n X n Fishburn matrices with entries in the set {0,1,2}.
1, 2, 12, 264, 19632, 4606752, 3311447232, 7202118117504, 47151987852663552, 927337336972381327872, 54741643544083873448266752, 9696222929066933463021344262144, 5152757080697434799933013959862300672, 8215035458438940398186389046297459974152192
Offset: 0
Examples
a(2) = 12: The twelve 2 X 2 Fishburn matrices with entries 0, 1 or 2 are /1 0\ /1 0\ /2 0\ /2 0\ \0 1/ \0 2/ \0 1/ \0 2/ /1 1\ /1 2\ /1 1\ /1 2\ /2 1\ /2 2\ /2 1\ /2 2\. \0 1/ \0 1/ \0 2/ \0 2/ \0 1/ \0 1/ \0 2/ \0 2/ Alternatively, the twelve 2-colored primitive Fishburn matrices of dimension 2 (using +1 and -1 for the two different colored versions of 1) are /+-1 0\ (4 possibilities) \0 +-1/ and /+-1 +-1\ (8 possibilities). \ 0 +-1/
Links
- Robert Israel, Table of n, a(n) for n = 0..64
- Hsien-Kuei Hwang, Emma Yu Jin, and Michael J. Schlosser, Asymptotics and statistics on Fishburn Matrices: dimension distribution and a conjecture of Stoimenow, arXiv:2012.13570 [math.CO], 2020.
- Vít Jelínek, Counting general and self-dual interval orders, Journal of Combinatorial Theory, Series A, Volume 119, Issue 3, April 2012, pp. 599-614; arXiv preprint, arXiv:1106.2261 [math.CO], 2011.
Programs
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Maple
N:= 20: # to get a(0)..a(N) g:= add(x^n*mul((3^i-1)/(1+x*(3^i-1)),i=1..n),n=0..N): S:= series(g,x,N+1): seq(coeff(S,x,j),j=0..N); # Robert Israel, Jul 11 2017
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Mathematica
QP = QPochhammer; nmax = 14; Sum[(-1)^n (1-x)^(-n-1) x^n QP[3, 3, n]/QP[x/(x-1), 3, n+1], {n, 0, nmax}] + O[x]^nmax // CoefficientList[#, x]& (* Jean-François Alcover, Sep 19 2018 *)
Formula
O.g.f.: A(x) = Sum_{n >=0} x^n Product_{i = 1..n} (3^i - 1)/(1 + x*(3^i - 1)) = 1 + 2*x + 12*x^2 + 264*x^3 + ... (use Jelínek, Theorem 2.1 with v = w = x = y = 2).
Two conjectural continued fractions for the o.g.f.:
A(x) = 1/(1 - 2*x/(1 - 4*x/(1 - 24*x/(1 - 64*x/(1 - 234*x/(1 - 676*x/(1 - ... - 3^(n-1)*(3^n - 1)*x/(1 - (3^n - 1)^2*x/(1 - ...))))))))) and
A(x) = 1 + 2*x/(1 - 6*x/(1 - 16*x/(1 - 72*x/(1 - 208*x/(1 - ... - 3^n*(3^n - 1)*x/(1 - (3^(n+1) - 1)*(3^n - 1)*x/(1 - ...))))))).
a(n) ~ c * 3^(n*(n+1)/2), where c = QPochhammer(1/3)^2 = 0.313741223174946734265526469975707962872482170305592991802056615373429729... - Vaclav Kotesovec, Aug 31 2023, updated Mar 17 2024
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