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Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 3}(x) in descending order.
+10
9
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
OFFSET
0,2
COMMENTS
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
LINKS
Zhe Wang and Zhi-Yong Zhu, The spiral property of q-Eulerian numbers of type B, The Australasian Journal of Combinatorics, Volume 87(1) (2023), Pages 198-202. See p. 199.
FORMULA
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).
From Wolfdieter Lang, Apr 10 2017: (Start)
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)
EXAMPLE
[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
... - Wolfdieter Lang, Apr 08 2017
Three term recurrence: T(2,1) = (3*(2-1)+1)*2 + (3*1+2)*1 = 13. - Wolfdieter Lang, Apr 10 2017
MAPLE
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):
seq(print(seq(A225117(n, k), k=0..n)), n=0..5); # Peter Luschny, Mar 08 2014
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,4}(x) = A225118.
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, May 02 2013
STATUS
approved
a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.
+10
4
-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406
OFFSET
0,2
LINKS
FORMULA
a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).
For a recurrence see the Maple program.
G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014
G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014
E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014
a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014
From Peter Bala, Nov 13 2016: (Start)
a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.
a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)
MAPLE
B := proc(n, u, k) option remember;
if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);
MATHEMATICA
CoefficientList[Series[-E^(-2*x)*Sech[3*x], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
Table[-6^n EulerE[n, 1/6], {n, 0, 19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)
PROG
(Sage)
from mpmath import mp, polylog, im
mp.dps = 32; mp.pretty = True
def A225147(n): return im(-2*I*(1+add(binomial(n, j)*polylog(-j, I)*3^j for j in (0..n))))
[int(A225147(n)) for n in (0..19)]
CROSSREFS
Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.
Bisections are A002438 and A000191.
KEYWORD
sign,easy
AUTHOR
Peter Luschny, Apr 30 2013
STATUS
approved
Array read by ascending antidiagonals: A(n, k) = (-2*n)^k * Euler(k, (n - 1)/(2*n)) for n >= 1 and A(0, k) = 1.
+10
3
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -3, -2, 1, 1, 1, -8, -11, 0, 1, 1, 1, -15, -26, 57, 16, 1, 1, 1, -24, -47, 352, 361, 0, 1, 1, 1, -35, -74, 1185, 1936, -2763, -272, 1, 1, 1, -48, -107, 2976, 6241, -38528, -24611, 0, 1
OFFSET
0,13
COMMENTS
This is the counterpart of A377666, where A(1, n) are the secant numbers A122045(n). Here A(1, n) are the tangent numbers A155585(n).
FORMULA
A(n, k) = k! * [x^k] exp(x)/cosh(n*x).
A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 1/2) *(-2*n)^j.
EXAMPLE
Array starts:
[0] 1, 1, 1, 1, 1, 1, 1, ... A000012
[1] 1, 1, 0, -2, 0, 16, 0, ... A155585
[2] 1, 1, -3, -11, 57, 361, -2763, ... A188458
[3] 1, 1, -8, -26, 352, 1936, -38528, ... A000810
[4] 1, 1, -15, -47, 1185, 6241, -230895, ... A000813
[5] 1, 1, -24, -74, 2976, 15376, -906624, ... A378065
[6] 1, 1, -35, -107, 6265, 32041, -2749355, ...
[7] 1, 1, -48, -146, 11712, 59536, -6997248, ...
MAPLE
A := (n, k) -> ifelse(n = 0, 1, (-2*n)^k * euler(k, (n - 1) / (2*n))):
for n from 0 to 7 do seq(A(n, k), k = 0..9) od; # row by row
# Alternative:
A := proc(n, k) local j; add(binomial(k, j)*euler(j, 1/2)*(-2*n)^j, j = 0..k) end: seq(seq(A(n - k, k), k = 0..n), n = 0..10);
# Using generating functions:
egf := n -> exp(x)/cosh(n*x): ser := n -> series(egf(n), x, 14):
row := n -> local k; seq(k!*coeff(ser(n), x, k), k = 0..7):
seq(lprint(row(n)), n = 0..7);
CROSSREFS
Columns: A005563 (k=2), A080663 (k=3), A378064 (k=4).
Cf. A378063 (main diagonal), A377666 (secant), A081658 (column generating polynomials).
KEYWORD
sign,tabl,new
AUTHOR
Peter Luschny, Nov 15 2024
STATUS
approved

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