A182626 -id:A182626 - cp's OEIS frontend

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

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

Peter Luschny, Mar 20 2015

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.

			[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, ...

Seiichi Manyama, Rows n = 0..139, flattened
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).

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).
Cf. A000142, A001286, A001754, A001755, A001777, A182626, A266732 (row k=3), A008316, A033282, A063007, A345013, A269944.

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));
# gives A000142, A001286, A001754, A001755, A001777, ...
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:

Table[Binomial[n, k] Binomial[n + k - 1, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)

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

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)

A226994 Number of lattice paths from (0,0) to (n,n) consisting of steps U=(1,1), H=(1,0) and S=(0,1) such that the first step leaving the diagonal (if any) is an H step.

Original entry on oeis.org

1, 2, 7, 32, 161, 842, 4495, 24320, 132865, 731282, 4048727, 22523360, 125797985, 704966810, 3961924127, 22321190912, 126027618305, 712917362210, 4039658528935, 22924714957472, 130271906898721, 741188107113962, 4221707080583087, 24070622500965632

Offset: 0

Alois P. Heinz, Jun 26 2013

nonn

a(n) is also the n-th order truncated expansion in x and y of 1/(1-x*y+x+y) evaluated at x=1, y=1 (see Mathematica code). - Benedict W. J. Irwin, Oct 06 2016

			a(0) = 1: the empty path.
a(1) = 2: HS, U.
a(2) = 7: HHSS, HSHS, HSSH, HSU, HUS, UHS, UU.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Column k=2 of A330942.

Cf. A001850 (unrestricted paths), A006318 (subdiagonal paths), A047665, A182626, A226995, A226996.

a:= proc(n) option remember; `if`(n<3, n*(2*n-1)+1,
     ((n-2)*(2*n-1) *a(n-3) -(7*n-4)*(2*n-3) *a(n-2)
      +(2*n-1)*(7*n-10) *a(n-1))/ (n*(2*n-3)))
    end:
seq(a(n), n=0..25);

Table[CoefficientList[Series[1/(1-x*y+x+y), {x, 0, n}, {y, 0, n}], z][[1]] /.x -> 1 /. y -> 1, {n, 0, 10}] (* Benedict W. J. Irwin, Oct 06 2016 *)

a(n) = 1/2 + pollegendre(n, 3)/2; \\ Michel Marcus, Oct 06 2016

G.f.: 1/(2-2*x) + 1/(2*sqrt(1-6*x+x^2)).
a(n) = A001850(n) - A047665(n).
a(n) = 1/2 + LegendreP(n, 3)/2. - Benedict W. J. Irwin, Oct 06 2016
a(n) ~ sqrt(3*sqrt(2) + 4) * (3 + 2*sqrt(2))^n / (4*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 07 2016
a(n) = Sum_{k=0..n} (-1)^k * A182626(k). - J. Conrad, Apr 08 2018
a(n) = 1 + Sum_{k=1..n} binomial(n,k)^2 * 2^(k-1). - Ilya Gutkovskiy, Nov 15 2021
a(n) = 1 + A047665(n). - Alois P. Heinz, Nov 15 2021

A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 8, 10, 1, 15, 45, 35, 1, 24, 126, 224, 126, 1, 35, 280, 840, 1050, 462, 1, 48, 540, 2400, 4950, 4752, 1716, 1, 63, 945, 5775, 17325, 27027, 21021, 6435, 1, 80, 1540, 12320, 50050, 112112, 140140, 91520, 24310, 1, 99, 2376, 24024, 126126, 378378, 672672, 700128, 393822, 92378

Offset: 0

Alford Arnold, May 30 2010

Antidiagonal sums are given by A113682. - Johannes W. Meijer, Mar 24 2013
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial binomial(x+n,n)*binomial(x+n,n-1) in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022
Chapoton's observation above is correct: the precise expansion is  binomial(x+n,n)*binomial(x+n,n-1) = Sum_{k = 0..n-1} (-1)^k*T(n-1,n-1-k)*binomial(x+2*n-1-k,2*n-1-k), as can be verified using the WZ algorithm. For example, n = 4 gives binomial(x+4,4)*binomial(x+4,3)  = 35*binomial(x+7,7) - 45*binomial(x+6,6) + 15*binomial(x+5,5) - binomial(x+4,4). - Peter Bala, Jun 24 2023

			n=0: 1;
n=1: 1,  3;
n=2: 1,  8,  10;
n=3: 1, 15,  45,   35;
n=4: 1, 24, 126,  224,   126;
n=5: 1, 35, 280,  840,  1050,   462;
n=6: 1, 48, 540, 2400,  4950,  4752,  1716;
n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;

David A. Corneth, Table of n, a(n) for n = 0..10010
Author?, Norm of a continuous function, dxdy.ru (in Russian).

Cf. A007318, A047781 (row sums), A178300, A182626, A123160, A132813.

A178301 := proc(n,k)
        binomial(n,k)*binomial(n+k+1,n+1) ;
end proc: # R. J. Mathar, Mar 24 2013
R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k,2*x+1)/(n+1), k=0..n) end:
for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # Peter Luschny, Aug 25 2021

Flatten[Table[Binomial[n,k]Binomial[n+k+1,n+1],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 23 2014 *)

create_list(binomial(n,k)*binomial(n+k+1,n+1),n,0,12,k,0,n); /* Emanuele Munarini, Dec 16 2016 */

R(n,x) = sum(k=0,n, (-1)^(n+k) * (2*k+1) * pollegendre(k,2*x+1)) / (n+1); \\ Max Alekseyev, Aug 25 2021

T(n,k) = A007318(n,k) * A178300(n+1,k+1).
From Peter Bala, Jun 18 2015: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k+1,n+1)*x^k = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,n+1)*(1 + x)^k.
Recurrence: (2*n - 1)*(n + 1)*R(n,x) = 2*(4*n^2*x + 2*n^2 - x - 1)*R(n-1,x) - (2*n + 1)(n - 1)*R(n-2,x) with R(0,x) = 1, R(1,x) = 1 + 3*x.
A182626(n) = -R(n-1,-2) for n >= 1. (End)
From Peter Bala, Jul 20 2015: (Start)
n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).
(1 + x)*R(n,x) gives the row polynomials of A123160. (End)
G.f.: (1+x-sqrt(1-2*x+x^2-4*x*y))/(2*(1+y)*x*sqrt(1-2*x+x^2-4*x*y)). - Emanuele Munarini, Dec 16 2016
R(n,x) = Sum_{k=0..n} (-1)^(n+k)*(2*k+1)*P(k,2*x+1)/(n+1), where P(k,x) is the k-th Legendre polynomial (cf. A100258) and P(k,2*x+1) is the k-th shifted Legendre polynomial (cf. A063007). - Max Alekseyev, Jun 28 2018; corrected by Peter Bala, Aug 08 2021
Polynomial g(n,x) = R(n,-x)/(n+1) delivers the maximum of f(1)^2/(Integral_{x=0..1} f(x)^2 dx) over all polynomials f(x) with real coefficients and deg(f(x)) <= n. This maximum equals (n+1)^2. See dxdy.ru link. - Max Alekseyev, Jun 28 2018

A268138 a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.

Original entry on oeis.org

1, 5, 51, 747, 13245, 264329, 5721415, 131425079, 3159389817, 78729848397, 2019910325499, 53087981674275, 1423867359013749, 38855956977763857, 1076297858301372687, 30203970496501504239, 857377825323716359665, 24586286492003180067989, 711463902659879056604995, 20756358426519694831851227

Offset: 1

Zhi-Wei Sun, Jan 26 2016

nonn

Conjecture: (i) All the terms are odd integers. Also, p | a(p) for any odd prime p.
(ii) Let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k = Sum_{k=0..n} binomial(n,k)^2*x^k*(x+1)^(n-k) for n >= 0, and s_n(x) = Sum_{k=1..n} (binomial(n,k)*binomial(n,k-1)/n)*x^(k-1)*(x+1)^(n-k) = (Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(k+1))/(x+1) for n > 0. Then, for any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*s_{k+1}(x) are integral and the polynomial is irreducible over the field of rational numbers.
The conjecture was essentially proved by the author in arXiv:1602.00574, except for the irreducibility of (Sum_{k=0..n-1} D_k(x)*s_{k+1}(x))/n. - Zhi-Wei Sun, Feb 01 2016

			a(3) = 51 since (A001850(0)*A001003(1) + A001850(1)*A001003(2) + A001850(2)*A001003(3))/3 = (1*1 + 3*3 + 13*11)/3 = 153/3 = 51.

Zhi-Wei Sun, Table of n, a(n)for n = 1..100
Zhi-Wei Sun, On Delannoy numbers and Schroder numbers, J. Number Theory 131(2011), no.12, 2387-2397.
Zhi-Wei Sun, Arithmetic properties of Delannoy numbers and Schröder numbers, preprint, arXiv:1602.00574 [math.CO], 2016.

Cf. A001003, A001850, A006318, A268136, A268137, A182626, A358388.

A001850 := n -> LegendreP(n, 3); seq(((3*(2*n+1)*A001850(n)*A001850(n-1)-n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4, n=1..20); # Mark van Hoeij, Nov 12 2022
# Alternative (which also gives an integer for n = 0):
f := n -> hypergeom([-n, -n], [1], 2):          # A001850
h := n -> hypergeom([-n,  n], [1], 2):          # A182626
g := n -> hypergeom([-n,  n, 1/2], [1, 1], -8): # A358388
a := n -> (f(n)*((3*n + 1)*f(n) - (-1)^n*(6*n + 3)*h(n)) - n*g(n))/(2*n + 2):
seq(simplify(a(n)), n = 1..20); # Peter Luschny, Nov 13 2022

d[n_]:=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]
s[n_]:=Sum[Binomial[n,k]Binomial[n,k-1]/n*2^(k-1),{k,1,n}]
a[n_]:=Sum[d[k]s[k+1],{k,0,n-1}]/n
Table[a[n],{n,1,20}]

a(n) = ((3*(2*n+1)*A001850(n)*A001850(n-1) - n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4. - Mark van Hoeij, Nov 12 2022

G.f.: (1-(1+1/x)*Int((1-34*x+x^2)^(1/2) * hypergeom([-1/2,1/2],[1], -32*x/(1-34*x+x^2))/((1-x)*(1+x)^2),x))/4. - Mark van Hoeij, Nov 28 2024

A360282 Triangle read by rows. T(n, k) = (1/2) * binomial(2(n - k + 1), n - k + 1) binomial(2*n - k, k - 1) for n > 0, T(0, 0) = 1.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 0, 10, 12, 3, 0, 35, 60, 30, 4, 0, 126, 280, 210, 60, 5, 0, 462, 1260, 1260, 560, 105, 6, 0, 1716, 5544, 6930, 4200, 1260, 168, 7, 0, 6435, 24024, 36036, 27720, 11550, 2520, 252, 8

Offset: 0

Peter Luschny, Feb 11 2023

Inspired by Vladimir Kruchinin's A360546.

			Triangle T(n, k) starts:
[0] 1;
[1] 0,     1;
[2] 0,     3,      2;
[3] 0,    10,     12,      3;
[4] 0,    35,     60,     30,      4;
[5] 0,   126,    280,    210,     60,     5;
[6] 0,   462,   1260,   1260,    560,   105,     6;
[7] 0,  1716,   5544,   6930,   4200,  1260,   168,    7;
[8] 0,  6435,  24024,  36036,  27720, 11550,  2520,  252,   8;
[9] 0, 24310, 102960, 180180, 168168, 90090, 27720, 4620, 360, 9;

Cf. A135503, A088218 (and A001700), A182626 (row sums apart from sign).

Family: A172431 and unsigned A053123 (case 1), this sequence is case 2, A360546 (case 3).

T := proc(n, k) if n = 0 then 1 else m := n - k + 1; (1/2) * binomial(2*m, m) * binomial(m + n - 1, k - 1) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..8);
# With Vladimir Kruchinin's g.f.:
gf := 1 + (y - x*y^2)/(2*sqrt((x*y - 1)^2 - 4*x)) ;
serx := series(gf, x, 10): poly := n -> simplify(coeff(serx, x, n)):
seq(print(seq(coeff(poly(n), y, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 14 2023

The sequence is member of a family of sequences defined by T(0, 0, m) = 1 and for m > 0, n > 0 as T(n, k, m) = (1/m)*binomial(m*u, u)*binomial(u + n - 1, k - 1), where u = n - k + 1. Here the case m = 2 is considered.

G.f.: (y*(1 - x*y)) / (2*sqrt(x*(y*(x*y - 2) - 4) + 1)) - y/2 + 1. - Vladimir Kruchinin, Feb 14 2023

cp's OEIS Frontend

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

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A226994 Number of lattice paths from (0,0) to (n,n) consisting of steps U=(1,1), H=(1,0) and S=(0,1) such that the first step leaving the diagonal (if any) is an H step.

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A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

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A268138 a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.

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A360282 Triangle read by rows. T(n, k) = (1/2) * binomial(2*(n - k + 1), n - k + 1) * binomial(2*n - k, k - 1) for n > 0, T(0, 0) = 1.

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A360282 Triangle read by rows. T(n, k) = (1/2) * binomial(2(n - k + 1), n - k + 1) binomial(2*n - k, k - 1) for n > 0, T(0, 0) = 1.