A115139 Array of coefficients of polynomials related to integer powers of the generating function of Catalan numbers A000108.
1, 1, 1, -1, 1, -2, 1, -3, 1, 1, -4, 3, 1, -5, 6, -1, 1, -6, 10, -4, 1, -7, 15, -10, 1, 1, -8, 21, -20, 5, 1, -9, 28, -35, 15, -1, 1, -10, 36, -56, 35, -6, 1, -11, 45, -84, 70, -21, 1, 1, -12, 55, -120, 126, -56, 7, 1, -13, 66, -165, 210, -126, 28, -1, 1, -14, 78, -220, 330, -252, 84, -8, 1, -15, 91, -286, 495, -462, 210, -36, 1
Offset: 1
Examples
The irregular triangle a(n, m) begins: n\m 0 1 2 3 4 5 6 7 8 1: 1 2: 1 3: 1 -1 4: 1 -2 5: 1 -3 1 6: 1 -4 3 7: 1 -5 6 -1 8: 1 -6 10 -4 9: 1 -7 15 -10 1 10: 1 -8 21 -20 5 11: 1 -9 28 -35 15 -1 12: 1 -10 36 -56 35 -6 13: 1 -11 45 -84 70 -21 1 14: 1 -12 55 -120 126 -56 7 15: 1 -13 66 -165 210 -126 28 -1 16: 1 -14 78 -220 330 -252 84 -8 17: 1 -15 91 -286 495 -462 210 -36 1 ... Reformatted and extended. - _Wolfdieter Lang_, Jan 27 2016 1/c(x) = P(2,x) - x*P(1,x)*c(x) = 1 - x*c(x), with the o.g.f. of A000108 (Catalan). 1/c(x)^2 = P(3,x) - x*P(2,x)*c(x) = (1-x) - x*c(x). c(x)^2 = (-P(1,x) + P(2,x)*c(x))/x^1 = (-1 + 1*c(x))/x. c(x)^3 = (-P(2,x) + P(3,x)*c(x))/x^2 = (-1 + (1-x)*c(x))/x^2. P(3,x) = 1-x = x*S(2,1/sqrt(x)) with Chebyshev's S(2,y) = U(2,y/2) = y^2 - 1.
References
- J. A. Clapperton, P. J. Larcombe and E. J. Fennessey, On iterated generating functions for integer sequences and Catalan polynomials, Utilitas Mathematica, 77 (2008), 3-33.
- J. A. Clapperton, P. J. Larcombe, E. J. Fennessey and P. Levrie, A class of auto-identities for Catalan polynomials and Padé approximation, Congressus Numerantium, 189 (2008), 77-95.
- N. G. de Bruijn, D. E. Knuth and S. O. Rice, The average height of planted plane trees, in: Graph Theory and Computing (ed. T. C. Read), Academic Press, New York, 1972, pp. 15-22.
- R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
Links
- Alois P. Heinz, Rows n = 1..200, flattened
- Hong Duc Bui, Counting Number of Triangulations of Point Sets: Reinterpreting and Generalizing the Triangulation Polynomials, arXiv:2504.09615 [cs.CG], 2025. See p. 3.
- Giulio Cerbai, Anders Claesson, and Luca Ferrari, Stack sorting with restricted stacks, arXiv:1907.08142 [cs.DS], 2019.
- Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009.
- R. Florez and J. C. Saunders, Irreducibility of generalized Fibonacci polynomials , Integers 22 (2022).
- Atsushi Komaba, Hisashi Johno, and Kazunori Nakamoto, A novel statistical approach for two-sample testing based on the overlap coefficient, arXiv:2206.03166 [math.ST], 2022.
- G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationelle, Cahier no. 15, Paris, 1970, pp. 3-41.
- Wolfdieter Lang, First 16 rows
- Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419.
- Peter J. Larcombe and Eric J. Fennessey, A Non-Linear Identity for A Particular Class of Polynomial Families, Fibonacci Quart. 52 (2014), no. 1, 75-79. Mentions this sequence.
Programs
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Maple
seq(seq((-1)^k*binomial(n-k,k),k=0..floor(n/2)),n=0..16); # Peter Luschny, May 10 2016
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Mathematica
p[x_, n_] := p[x, n] = p[x, n - 1] + x*p[x, n - 2]; p[x_, -1] = p[x_, 0] = 1; p[x_, 1] = 1 + x; Flatten[ Table[ CoefficientList[p[-x, n - 1], x], {n, 0, 16}]] (* Jean-François Alcover, Jun 20 2011 *) Flatten[Map[CoefficientList[#,x]&, Table[Sum[Binomial[t - i, i] x^(i) (-1)^i, {i, 0, t}], {t, 1,15}]]] (* Rigoberto Florez, Aug 28 2022 *)
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Python
import math L1 = [math.comb(t - i, i)*(-1)**i for t in range(16) for i in range(t)] L1 = list(filter((0)._ne_, L1)) print(L1) # Rigoberto Florez, Sep 03 2022
Formula
a(n, m) = ((-1)^(m))*binomial(n-1-m, m), n>=1, m=0..ceiling(n/2)-1.
a(n, m) = [x^m]P(n,x), n>=1, m=0..ceiling(n/2)-1, with P(n,x) given above in terms of Chebyshev's S-polynomials.
P(n,x) = (u^(2*n) - v^(2*n))/(u^2 - v^2), where u and v are defined by u^2 + v^2 =1 and u*v = sqrt(x). Example: P(3,x) = (u^6 - v^6)/(u^2 - v^2) = u^4 + u^2*v^2 + v^4 = 1 - x. - Emeric Deutsch, Jun 16 2011
G.f.: 1/(1- x + y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1- x*y)*x/((2*k+2- x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
T(n, k) = GegenbauerC(k, (n+1)/2-k, -1) assuming the triangle (0,0) based. - Peter Luschny, May 10 2016
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