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A225118
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Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 4}(x) in descending order.
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8
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1, 3, 1, 9, 22, 1, 27, 235, 121, 1, 81, 1996, 3446, 620, 1, 243, 15349, 63854, 40314, 3119, 1, 729, 112546, 963327, 1434812, 422087, 15618, 1, 2187, 806047, 12960063, 37898739, 26672209, 4157997, 78117, 1, 6561, 5705752, 162711868, 840642408, 1151050534
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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 quadruple factorial numbers A047053 and the alternating row sums, i.e., sum((-1)^k*T(n,k),k=0..n), are up to a sign A079858. - Johannes W. Meijer, May 04 2013
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LINKS
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FORMULA
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G.f. 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 = 4; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
E.g.f. of row polynomials (rising powers of x): (1-x)*exp(3*(1-x)*z)/(1-y*exp(4*(1-x)*z)), i.e. e.g.f. of the triangle.
E.g.f. for the row polynomials with falling powers of x (A_{n, 4}(x) of the name): (1-x)*exp((1-x)*z)/(1 - x*exp(4*(1-x)*z)).
T(n, k) = Sum_{j=0..k} (-1)^(k-j) * binomial(n+1,k-j) * (3+4*j)^n, 0 <= k <= n.
Recurrence: T(n, k) = (4*(n-k) + 1)*T(n-1, k-1) + (3 + 4*k)*T(n-1, k), n >= 1, with T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k. (End)
In terms of Euler's triangle = A123125: T(n, k) = Sum_{m=0..n} (binomial(n, m)*3^(n-m)*4^m*Sum_{p=0..k} (-1)^(k-p)*binomial(n-m, k-p)*A123125(m, p)), 0 <= k <= n. - Wolfdieter Lang, Apr 13 2017
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EXAMPLE
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[0] 1
[1] 3*x + 1
[2] 9*x^2 + 22*x + 1
[3] 27*x^3 + 235*x^2 + 121*x + 1
[4] 81*x^4 + 1996*x^3 + 3446*x^2 + 620*x + 1
...
The triangle T(n, k) begins:
n\k
0: 1
1: 3 1
2: 9 22 1
3: 27 235 121 1
4: 81 1996 3446 620 1
5: 243 15349 63854 40314 3119 1
6: 729 112546 963327 1434812 422087 15618 1
7: 2187 806047 12960063 37898739 26672209 4157997 78117 1
...
row n=8: 6561 5705752 162711868 840642408 1151050534 442372648 39531132 390616 1,
row n=9: 19683 40156777 1955297356 16677432820 39523450714 29742429982 6818184988 367889284 1953115 1.
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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, 4), x, n-k), k=0..n)), n=0..6);
# Recurrence:
P := proc(n, x) option remember; if n = 0 then 1 else
(n*x+(1/4)*(1-x))*P(n-1, x)+x*(1-x)*diff(P(n-1, x), x);
expand(%) fi end:
A225117 := (n, k) -> 4^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, 4], {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, 4).coefficients()[::-1] for n in (0..5)]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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