A137514 A triangular sequence from umbral calculus expansion of Simon Plouffe's rational polynomial for A002890: p(x,t) = exp(x*t)*(1 - 6*t + 9*t^2 - 4*t^3 + t^4)/(4*t - 1)/(2*t - 1).
1, 0, 1, 2, 0, 1, 12, 6, 0, 1, 120, 48, 12, 0, 1, 1680, 600, 120, 20, 0, 1, 31680, 10080, 1800, 240, 30, 0, 1, 766080, 221760, 35280, 4200, 420, 42, 0, 1, 22579200, 6128640, 887040, 94080, 8400, 672, 56, 0, 1, 778014720, 203212800, 27578880, 2661120, 211680, 15120, 1008, 72, 0, 1
Offset: 1
Examples
Triangle begins: {1}, {0, 1}, {2, 0, 1}, {12, 6, 0, 1}, {120, 48, 12, 0, 1}, {1680, 600, 120, 20, 0, 1}, {31680, 10080, 1800, 240, 30, 0, 1}, {766080, 221760, 35280, 4200, 420, 42, 0, 1}, {22579200, 6128640, 887040, 94080, 8400, 672, 56, 0, 1}, {778014720, 203212800, 27578880, 2661120, 211680, 15120, 1008, 72, 0, 1}, ...
References
- Terrel L. Hill, Statistical Mechanics: Principles and Selected Applications, Dover, New York, 1956, page 336 ff
Programs
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Mathematica
Clear[p, f, g] p[t_] = Exp[x*t]*(1 - 6*t + 9*t^2 - 4*t^3 + t^4)/(4*t - 1)/(2*t - 1); Table[ ExpandAll[n!*SeriesCoefficient[Series[p[t], {t, 0, 30}], n]], {n, 0, 10}] a = Table[ CoefficientList[n!*SeriesCoefficient[; FullSimplify[Series[p[t], {t, 0, 30}]], n], x], {n, 0, 10}]; Flatten[a]
Formula
p(x,t) = exp(x*t)*(1 - 6*t + 9*t^2 - 4*t^3 + t^4)/(4*t - 1)/(2*t - 1) = Sum_{n>=0} P(x,n)*t^n/n!; out_n,m=n!*Coefficients(P(x,n)).
Comments