A132813 -id:A132813 - cp's OEIS frontend

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

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

A134287 Fifth column of triangle A103371 (without leading zeros).

Original entry on oeis.org

1, 30, 315, 1960, 8820, 31752, 97020, 261360, 637065, 1431430, 3006003, 5962320, 11262160, 20391840, 35581680, 60093504, 98590905, 157608990, 246142435, 376372920, 564559380, 832117000, 1206913500, 1724814000, 2431508625

Offset: 0

Wolfdieter Lang, Nov 13 2007

Kekulé numbers for certain benzenoids.
a(n) = K(L(n))*K(O(2,4,n)) with the Cyvin and Gutman Kekulé number notation. See p. 62 for the L(n) structure with K(L(n))=n+1 and p. 105 (i) for the O(k,m,n) structure and its Kekulé number. This corresponds to an essentially disconnected 7-tier benzenoid structure similar to the 6-tier structure shown on p. 230, nr. 23 (see A108647).
a(n-5), n >= 5, is the number of ways to put n identical objects into m=5 of altogether n distinguishable boxes (n-5 boxes stay empty).

			a(2)=315 because n=7 identical balls can be put into m=5 of n=7 distinguishable boxes in binomial(7,5)*(5!/(4!*1!)+ 5!/(3!*2!)) = 21*(5+10) = 315 ways. The m=5 part partitions of 7, namely (1^4,3) and (1^3,2^2) specify the filling of each of the 21 possible five box choices. - _Wolfdieter Lang_, Nov 13 2007

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A108647 (fourth column of triangle A103371).

a134287 = flip a103371 4 . (+ 4)  -- Reinhard Zumkeller, Apr 04 2014

[5*Binomial(n+5, 5)^2/(n+5): n in [0..30]]; // G. C. Greubel, Oct 28 2022

seq(binomial(n+4,4)^2*(n+5)/5, n=0..24); # Peter Luschny, Jan 13 2014

CoefficientList[Series[(1 + 20 x + 60 x^2 + 40 x^3 + 5 x^4)/(1 - x)^10, {x, 0, 24}], x]

5*binomial(n+5,5)^2/(n+5) $ n = 0..35; // Zerinvary Lajos, May 09 2008

a(n) = 5*binomial(n+5, 5)^2/(n+5); \\ Michel Marcus, Jan 07 2014

[5*binomial(n+5,5)^2/(n+5) for n in range(31)] # G. C. Greubel, Oct 28 2022

a(n) = A103371(n+4,4), n >= 0.
a(n) = ((n+1)*(n+2)*(n+3)*(n+4))^2*(n+5)/2880, n >= 0. 2880 = 4!*5! = A010790(4).
G.f.: (1+20*x+60*x^2+40*x^3+5*x^4)/(1-x)^10. Numerator polynomial from fifth row of triangle A132813.
a(n) = 5*C(n+5,5)^2/(n+5), n >= 0. - Zerinvary Lajos, May 09 2008
a(n) = (C(n+6,6)*C(n+5,4)+5*C(n+5,6)*C(n+5,4))/(n+5). - Gary Detlefs, Jan 06 2014
From Amiram Eldar, May 31 2022: (Start)
Sum_{n>=0} 1/a(n) = 350*Pi^2/3 - 13805/12.
Sum_{n>=0} (-1)^n/a(n) = 5*Pi^2 + 640*log(2)/3 - 785/4. (End)
E.g.f.: (2880 + 83520*x + 368640*x^2 + 529920*x^3 + 330120*x^4 + 102024*x^5 + 16616*x^6 + 1432*x^7 + 61*x^8 + x^9)*exp(x)/2880. - G. C. Greubel, Oct 28 2022

cp's OEIS Frontend

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

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