A140945 Triangle read by rows: counts series-parallel networks by the number of series connections.
1, 1, 1, 1, 6, 1, 1, 25, 25, 1, 1, 90, 290, 90, 1, 1, 301, 2450, 2450, 301, 1, 1, 966, 17451, 41580, 17451, 966, 1, 1, 3025, 112035, 544971, 544971, 112035, 3025, 1, 1, 9330, 671980, 6076350, 12122502, 6076350, 671980, 9330, 1, 1, 28501, 3846700, 60738700, 217523922, 217523922, 60738700, 3846700, 28501, 1
Offset: 1
Examples
Triangle begins: 1; 1, 1; 1, 6, 1; 1, 25, 25, 1; 1, 90, 290, 90, 1; 1, 301, 2450, 2450, 301, 1; 1, 966, 17451, 41580, 17451, 966, 1; ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50; first 17 rows from Brian Drake, Jul 24 2008)
- Brian Drake, An inversion theorem for labeled trees and some limits of areas under lattice paths (Example 1.5.1), A dissertation presented to the Faculty of the Graduate School of Arts and Sciences of Brandeis University.
- Luis Ferroni and Matt Larson, Kazhdan-Lusztig polynomials of braid matroids, arXiv:2303.02253 [math.CO], 2023.
- Nicholas Proudfoot, Yuan Xu, and Ben Young, On the enumeration of series-parallel matroids, arXiv:2406.04502 [math.CO], 2024.
Programs
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Maple
N:=6: 1/a*log(1+a*y)+1*log(1+b*y)/b-y=x: solve(%, y):series(%, x, N): simplify(%, symbolic): convert(%, polynom): subs(b=1, %): R:= [seq(i!*coeff(%, x, i), i=1..N-1)]: seq( seq(coeff(R[i], a, j), j=0..i-1), i=1..N-1);
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PARI
T(n) = {[Vecrev(p) | p<-Vec(serlaplace(intformal(serreverse(log(1 + x*y + O(x*x^n))/y + log(1 + x + O(x*x^n)) - x))))]} { my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Mar 08 2023
Formula
E.g.f. is reversion of log(1+ax)/a+log(1+bx)/b-x.
Let f(x,t) = (1+x)*(1+x*t)/(1-x^2*t) and let D be the operator f(x,t)*d/dx. Then the n-th row polynomial equals (D^n)(f(x,t)) evaluated at x = 0. - Peter Bala, Sep 29 2011
Comments