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A253283
Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.
9
1, 0, 1, 0, 2, 3, 0, 3, 12, 10, 0, 4, 30, 60, 35, 0, 5, 60, 210, 280, 126, 0, 6, 105, 560, 1260, 1260, 462, 0, 7, 168, 1260, 4200, 6930, 5544, 1716, 0, 8, 252, 2520, 11550, 27720, 36036, 24024, 6435, 0, 9, 360, 4620, 27720, 90090, 168168, 180180, 102960, 24310
OFFSET
0,5
COMMENTS
The rows give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1) / (n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022
This is related to the cluster fans of type B (see Fomin and Zelevinsky reference) - F. Chapoton, Nov 17 2022.
LINKS
F. Chapoton Enumerative Properties of Generalized Associahedra, Sém. Loth. Comb. B51b (2004).
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, Electronic Journal Of Combinatorics, 23(1) (2016), #P1.45.
S. Fomin and A. Zelevinsky Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3.
Yunlong Shen and Lixin Shen, Orthogonal Fourier-Mellin moments for invariant pattern recognition, J. Opt. Soc. Am. A 11 (6) (1994) p 1748-1757, eq. (6).
FORMULA
The exponential generating functions for the rows of the square array L(n,k) = ((n+k)!/n!)*C(n+k-1,n-1) (associated to the unsigned Lah numbers) are given by R_n(x) = Sum_{k=0..n} T(n,k)/(x-1)^(n+k).
T(n,k) = C(n,k)*C(n+k-1,k-1).
Sum_{k=0..n} T(n,k) = (-1)^n*hypergeom([-n,n],[1],2) = (-1)^n*A182626(n).
Row generating function: Sum_{k>=1} T(n,k)*z^k = z*n* 2F1(1-n,n+1 ; 2; -z). - R. J. Mathar, Dec 18 2016
From Peter Bala, Feb 22 2017: (Start)
G.f.: (1/2)*( 1 + (1 - t)/sqrt(1 - 2*(2*x + 1)*t + t^2) ) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n,x) = (1/2)*(LegendreP(n, 2*x + 1) - LegendreP(n-1, 2*x + 1)) for n >= 1.
The row polynomials are the black diamond product of the polynomials x^n and x^(n+1) (see Dukes and White 2016 for the definition of this product).
exp(Sum_{n >= 1} R(n,x)*t^n/n) = 1 + x*t + x*(1 + 2*x)*t^2 + x*(1 + 5*x + 5*x^2)*t^3 + ... is a g.f. for A033282, but with a different offset.
The polynomials P(n,x) := (-1)^n/n!*x^(2*n)*(d/dx)^n(1 + 1/x)^n begin 1, 3 + 2*x , 10 + 12*x + 3*x^2, ... and are the row polynomials for the row reverse of this triangle. (End)
Let Q(n, x) = Sum_{j=0..n} (-1)^(n - j)*A269944(n, j)*x^(2*j - 1) and P(x, y) = (LegendreP(x, 2*y + 1) - LegendreP(x-1, 2*y + 1)) / 2 (see Peter Bala above). Then n!*(n - 1)!*[y^n] P(x, y) = Q(n, x) for n >= 1. - Peter Luschny, Oct 31 2022
From Peter Bala, Apr 18 2024: (Start)
G.f.: Sum_{n >= 0} binomial(2*n-1, n)*(x*t)^n/(1 - t)^(2*n) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n, x) = [t^n] ( (1 - t)/(1 - (1 + x)*t) )^n.
It follows that for integer x, the sequence {R(n, x) : n >= 0} satisfies the Gauss congruences: R(n*p^r, x) == R(n*p^(r-1), x) (mod p^r) for all primes p and positive integers n and r.
R(n, -2) = (-1)^n * A002003(n) for n >= 1.
R(n, 3) = A299507(n). (End)
EXAMPLE
[1]
[0, 1]
[0, 2, 3]
[0, 3, 12, 10]
[0, 4, 30, 60, 35]
[0, 5, 60, 210, 280, 126]
[0, 6, 105, 560, 1260, 1260, 462]
[0, 7, 168, 1260, 4200, 6930, 5544, 1716]
.
R_0(x) = 1/(x-1)^0.
R_1(x) = 0/(x-1)^1 + 1/(x-1)^2.
R_2(x) = 0/(x-1)^2 + 2/(x-1)^3 + 3/(x-1)^4.
R_3(x) = 0/(x-1)^3 + 3/(x-1)^4 + 12/(x-1)^5 + 10/(x-1)^6.
Then k!*[x^k] R_n(x) is A001286(k+2) and A001754(k+3) for n = 2, 3 respectively.
.
Seen as an array A(n, k) = binomial(n + k, k)*binomial(n + 2*k - 1, n + k):
[0] 1, 1, 3, 10, 35, 126, 462, ...
[1] 0, 2, 12, 60, 280, 1260, 5544, ...
[2] 0, 3, 30, 210, 1260, 6930, 36036, ...
[3] 0, 4, 60, 560, 4200, 27720, 168168, ...
[4] 0, 5, 105, 1260, 11550, 90090, 630630, ...
[5] 0, 6, 168, 2520, 27720, 252252, 2018016, ...
[6] 0, 7, 252, 4620, 60060, 630630, 5717712, ...
MAPLE
T_row := proc(n) local egf, k, F, t;
if n=0 then RETURN(1) fi;
egf := (x/(1-x))^n/n!; t := diff(egf, [x$n]);
F := convert(t, parfrac, x);
# print(seq(k!*coeff(series(F, x, 20), x, k), k=0..7));
seq(coeff(F, (x-1)^(-k)), k=n..2*n) end:
seq(print(T_row(n)), n=0..7);
# 2nd version by R. J. Mathar, Dec 18 2016:
A253283 := proc(n, k)
binomial(n, k)*binomial(n+k-1, k-1) ;
end proc:
MATHEMATICA
Table[Binomial[n, k] Binomial[n + k - 1, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)
PROG
(PARI) T(n, k) = binomial(n, k)*binomial(n+k-1, k-1);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Apr 29 2018
CROSSREFS
T(n, n) = C(2*n-1, n) = A001700(n-1).
T(n, n-1) = A005430(n-1) for n >= 1.
T(n, n-2) = A051133(n-2) for n >= 2.
T(n, 2) = A027480(n-1) for n >= 2.
T(2*n, n) = A208881(n) for n >= 0.
A002002 (row sums).
Sequence in context: A316607 A194365 A216217 * A261719 A137663 A370983
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Mar 20 2015
STATUS
approved