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A225117
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Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 3}(x) in descending order.
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9
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1, 2, 1, 4, 13, 1, 8, 93, 60, 1, 16, 545, 1131, 251, 1, 32, 2933, 14498, 10678, 1018, 1, 64, 15177, 154113, 262438, 88998, 4089, 1, 128, 77101, 1475736, 4890287, 3870352, 692499, 16376, 1, 256, 388321, 13270807, 77404933, 117758659, 50476003, 5175013, 65527, 1
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OFFSET
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0,2
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COMMENTS
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The row sums equal the triple factorial numbers A032031 and the alternating row sums, i.e., Sum_{k=0..n}(-1)^k*T(n,k), are up to a sign A000810. - Johannes W. Meijer, May 04 2013
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LINKS
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FORMULA
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Generating function of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 3; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+1, n-k-j)*(1+3*j)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..n-k} (-1)^(n-k-m)*binomial(n-m, k)*A284861(n, m), 0 <= k <= n.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are R(n, x) = (x-1)^n*Sum_{m=0} A284861(n, m)*(1/(x-1))^m, n >= 0, i.e. the row polynomials of A284861 in the variable 1/(x-1) multiplied by (x-1)^n.
The row polynomials with falling powers are P(n, x) = (1-x)^n*Sum_{m=0..n} A284861(n, m)*(x/(1-x))^m, n >= 0.
The e.g.f. of the row polynomials in falling powers of x (A_{n, 3}(x) of the name) is exp((1-x)*z)/(1 - (x/(1 - x)) * (exp(3*(1-x)*z) - 1)) = (1-x)*exp((1-x)*z)/(1 - x*exp(3*(1-x)*z)).
The e.g.f. of the row polynomials R(n, x) (rising powers of x) is then (1-x)*exp(2*(1-x)*z)/(1 - x*exp(3*(1-x)*z)).
Three term recurrence: T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = (3*(n-k)+1)*T(n-1, k-1) + (3*k+2)*T(n-1, k) for n >= 1, k=0..n. (End)
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EXAMPLE
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[0] 1
[1] 2*x + 1
[2] 4*x^2 + 13*x + 1
[3] 8*x^3 + 93*x^2 + 60*x + 1
[4] 16*x^4 + 545*x^3 + 1131*x^2 + 251*x + 1
...
The triangle T(n, k) begins:
n \ k 0 1 2 3 4 5 6 7 ...
0: 1
1: 2 1
2: 4 13 1
3: 8 93 60 1
5: 16 545 1131 251 1
6: 32 2933 14498 10678 1018 1
7: 64 15177 154113 262438 88998 4089 1
8: 128 77101 1475736 4890287 3870352 692499 16376 1
Three term recurrence: T(2,1) = (3*(2-1)+1)*2 + (3*1+2)*1 = 13. - Wolfdieter Lang, Apr 10 2017
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MAPLE
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gf := proc(n, k) local f; f := (x, t) -> x*exp(t*x/k)/(1-x*exp(t*x));
series(f(x, t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
collect(simplify(%), x) end:
seq(print(seq(coeff(gf(n, 3), x, n-k), k=0..n)), n=0..6);
# Recurrence
P := proc(n, x) option remember; if n = 0 then 1 else
(n*x+(1/3)*(1-x))*P(n-1, x)+x*(1-x)*diff(P(n-1, x), x);
expand(%) fi end:
A225117 := (n, k) -> 3^n*coeff(P(n, x), x, n-k):
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MATHEMATICA
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gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 3], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
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PROG
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(Sage)
@CachedFunction
def EB(n, k, x): # Modified cardinal B-splines
if n == 1: return 0 if (x < 0) or (x >= 1) else 1
return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
R.<x> = ZZ[]
if x == 0: return 1
return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
[EulerianPolynomial(n, 3).coefficients()[::-1] for n in (0..5)]
(PARI) T(n, k) = sum(j=0, n - k, (-1)^(n - k - j)*binomial(n + 1, n - k - j)*(1 + 3*j)^n);
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 10 2017
(Python)
from sympy import binomial
def T(n, k): return sum((-1)**(n - k - j)* binomial(n + 1, n - k - j)*(1 + 3*j)**n for j in range(n - k + 1))
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017
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CROSSREFS
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Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,4}(x) = A225118.
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KEYWORD
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AUTHOR
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STATUS
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approved
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