A127671 Cumulant expansion numbers: Coefficients in expansion of log(1 + Sum_{k>=1} x[k]*(t^k)/k!).
1, 1, -1, 1, -3, 2, 1, -4, -3, 12, -6, 1, -5, -10, 20, 30, -60, 24, 1, -6, -15, -10, 30, 120, 30, -120, -270, 360, -120, 1, -7, -21, -35, 42, 210, 140, 210, -210, -1260, -630, 840, 2520, -2520, 720, 1, -8, -28, -56, -35, 56, 336, 560, 420, 560, -336, -2520, -1680, -5040, -630, 1680, 13440, 10080, -6720
Offset: 1
Examples
Row n=3: [1,-3,2] stands for the polynomial 1*x[3] - 3*x[1]*x[2] + 2*x[1]^3 (the Abramowitz-Stegun order of the p(3)=3 partitions of n=3 is [3],[1,2],[1^3]).
References
- C. Itzykson and J.-M. Drouffe, Statistical field theory, vol. 2, p. 413, eq.(13), Cambridge University Press, (1989).
Links
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 831-2.
- Tom Copeland, The creation / raising operators for Appell sequences.
- Wolfdieter Lang, First 10 rows of cumulant numbers and polynomials
- E. Schröder, Ueber unendlich viele Algorithmen zur Auflösung der Gleichungen, Mathematische Annalen vol. 2, 317-365, 1870.
- G. Stewart, On infinitely many algorithms for solving equations, 1993, (translation into English of Schröder's paper above).
Formula
E.g.f. for multivariate row polynomials A(t) := log(1 + Sum_{k>=1} x[k]*(t^k)/k!).
Row n polynomial p_n(x[1],...,x[n]) = [(t^n)/n!]A(t).
a(n,m) = A264753(n, m) * M_3(n,m) with M_3(n,m) = A036040(n,m) (Abramowitz-Stegun M_3 numbers). - corrected by Johannes W. Meijer, Jul 12 2016
p_n(x[1],...,x[n]) = -(n-1)!*F(n,x[1],x[2]/2!,..,x[n]/n!) in terms of the Faber polynomials F(n,b1,..,bn) of A263916. - Tom Copeland, Nov 17 2015
With D = d/dz and M(0) = 1, the differential operator R = z + d(log(M(D))/dD = z + d(log(1 + x[1] D + x[2] D^2/2! + ...))/dD = z + p.*exp(p.D) = z + Sum_{n>=0} p_(n+1)(x[1],..,x[n]) D^n/n! is the raising operator for the Appell sequence A_n(z) = (z + x[.])^n = Sum_{k=0..n} binomial(n,k) x[n-k] z^k with the e.g.f. M(t) e^(zt), i.e., R A_n(z) = A_(n+1)(z) and dA_n(z)/dz = n A_(n-1)(z). The operator Q = z - p.*exp(p.D) generates the Appell sequence with e.g.f. e^(zt) / M(t). - Tom Copeland, Nov 19 2015
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