A331431 -id:A331431 - cp's OEIS frontend

A002457 a(n) = (2n+1)!/n!^2.

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900

Offset: 0

N. J. A. Sloane

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

G. C. Greubel, Table of n, a(n) for n = 0..1000 [Terms 0 to 200 computed by T. D. Noe; terms 201 to 1000 by G. C. Greubel, Jan 14 2017]
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Alexander Barg, Stolarsky's invariance principle for finite metric spaces, arXiv:2005.12995 [math.CO], 2020.
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables) [Annotated scanned copy]
Sara C. Billey, Matjaž Konvalinka, and Joshua P. Swanson, Asymptotic normality of the major index on standard tableaux, arXiv:1905.00975 [math.CO], 2019.See p. 15, Remark 4.2
R. Chapman, Moments of Dyck paths, Discrete Math., 204 (1999), 113-117.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
F. Disanto, A. Frosini, R. Pinzani and S. Rinaldi, A closed formula for the number of convex permutominoes, arXiv:math/0702550 [math.CO], 2007.
Luca Ferrari and Emanuele Munarini, Enumeration of edges in some lattices of paths, arXiv preprint arXiv:1203.6792 [math.CO], 2012 and J. Int. Seq. 17 (2014) #14.1.5.
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
P.-Y. Huang, S.-C. Liu, and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
C. Jordan, Calculus of Finite Differences, Budapest, 1939. [Annotated scans of pages 448-450 only]
Cemil Karaçam and Alper Vural, Enumerating 2D and 3D lattice paths with arbitrary steps, Mathematica Bohemica, pp. 1-13 (2025). See p. 4.
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135. [Annotated scanned copy]
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
L. W. Shapiro, W.-J. Woan and S. Getu, Runs, slides and moments, SIAM J. Alg. Discrete Methods, 4 (1983), 459-466.
Andrei K. Svinin, On some class of sums, arXiv:1610.05387 [math.CO], 2016. See p. 5.
T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.
Eric Weisstein's World of Mathematics, Central Beta Function
Eric Weisstein's World of Mathematics, Pi Formulas
Y. Q. Zhao, Introduction to Probability with Applications

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
Cf. A000108, A000217.
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

a002457 n = a116666 (2 * n + 1) (n + 1)
-- Reinhard Zumkeller, Nov 02 2013

[Factorial(2*n+1)/Factorial(n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 12 2015

A002457:=n->(n+1) * binomial(2*(n+1),(n+1)) / 2; seq(A002457(n), n=0..50);
seq((2*n)!*coeff(series(HeunC(0,0,-2,-1/4,7/4,4*x^2),x,2*n+1),x,2*n),n=0..22); # Peter Luschny, Nov 22 2013

a[n_]:=(2*n+1)!/n!^2; Array[f, 23, 0] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

{a(n) = if( n<0, 0, (2*n + 1)! / n!^2)}; /* Michael Somos, Dec 09 2002 */

a(n) = (2*n+1)*binomial(2*n, n); \\ Altug Alkan, Apr 16 2018

A002457 = lambda n: binomial(n+1/2,1/2)<<2*n
[A002457(n) for n in range(23)] # Peter Luschny, Sep 22 2014

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k).  This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A002736 Apéry numbers: a(n) = n^2*C(2n,n).

Original entry on oeis.org

0, 2, 24, 180, 1120, 6300, 33264, 168168, 823680, 3938220, 18475600, 85357272, 389398464, 1757701400, 7862853600, 34901442000, 153876579840, 674412197580, 2940343837200, 12759640231800, 55138611528000, 237371722628040, 1018383898440480

Offset: 0

N. J. A. Sloane

Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n-1 of B equals -a(n-1). - T. D. Noe, May 01 2011

J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933, p. 93.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Kunle Adegoke, Robert Frontczak, and Taras Goy, Fibonacci-Catalan Series, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 22 (2022), #A110.
Paul S. Bruckman, Problem B-871, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 37, No. 1 (1999), p. 85.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 (Annotated scans of some selected pages).
Indulis Strazdins, Solution to problem B-871, Fibonacci Quartely, Vol. 38, No. 1 (2000), pp. 86-87.
Hans J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, Vol. 40, No. 2 (May 2002), pp. 175-180.
Alfred J. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3), Math. Intelligencer, Vol. 1 (1978/1979), pp. 195-203.

Cf. A000108, A002544, A005258, A005259, A005429, A005430, A086467.

A diagonal of A331431.

[n^2*Binomial(2*n, n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014

seq(n^2*binomial(2*n,n), n=0..50); # Robert Israel, Aug 07 2014

CoefficientList[ Series[x (4 x + 2)/(1 - 4 x)^(5/2), {x, 0, 20}], x] (* Robert G. Wilson v, Aug 08 2011 *)
Table[n^2 Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Jun 21 2017 *)

combinat::catalan(n)*(n+1)*n^2 $ n = 0..36 // Zerinvary Lajos, Apr 17 2007

my(x='x+O('x^100)); concat(0, Vec(x*(4*x+2)/((1-4*x)^(5/2)))) \\ Altug Alkan, Mar 21 2016

a(n) = n^2*binomial(2*n, n); \\ Michel Marcus, Mar 21 2016

[n^2*(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Mar 23 2022

G.f.: x*(4*x+2)/((1-4*x)^(5/2)). - Marco A. Cisneros Guevara, Jul 25 2011
Sum_{n>=1} 1/a(n) = Pi^2/18 (Euler). - Benoit Cloitre, Apr 07 2002
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*n^(3/2)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)^2 = A086467, where phi is the golden ratio. (End)
D-finite with recurrence: (-n+1)*a(n) +2*(n+4)*a(n-1) +4*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 21 2020
a(n) = (2n)!/(Gamma(n))^2. - Diego Rattaggi, Mar 30 2020
a(n) = Sum_{k=0..2*n} binomial(2*n,k)*abs(n-k)^3 (Bruckman, 1999; Strazdins, 2000). - Amiram Eldar, Jan 12 2022
Sum_{n>=1} x^n/a(n) = 2*arcsin(sqrt(x)/2)^2, for abs(x) < 4 (Adegoke et al., 2022, section 5, p. 10). - Amiram Eldar, Dec 07 2024
From Peter Bala, Aug 02 2025: (Start)
For n >= 1,
a(n) = 2*n*(2*n-1)/(n-1)^2 * a(n-1) with a(1) = 2 and
1/a(n) = Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n, k)*binomial(n+k, k)/(n+k)^2. (End)
a(n) = 2 * A002544(n-1) for n>=1. - Alois P. Heinz, Aug 03 2025

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).

Original entry on oeis.org

-1, -1, 2, -1, 6, -6, -1, 12, -30, 20, -1, 20, -90, 140, -70, -1, 30, -210, 560, -630, 252, -1, 42, -420, 1680, -3150, 2772, -924, -1, 56, -756, 4200, -11550, 16632, -12012, 3432, -1, 72, -1260, 9240, -34650, 72072, -84084, 51480, -12870, -1, 90, -1980, 18480, -90090, 252252, -420420, 411840, -218790, 48620, -1, 110, -2970, 34320, -210210, 756756, -1681680, 2333760, -1969110, 923780, -184756

Offset: 0

N. J. A. Sloane, Jan 17 2020

This is Table I of Ser (1933), page 92.
From Petros Hadjicostas, Jul 09 2020: (Start)
Essentially Ser (1933) in his book (and in particular for Tables I-IV) finds triangular arrays that allow him to express the coefficients of various kinds of series in terms of the coefficients of other series.
He uses Newton's series (or some variation of it), factorial series, and inverse factorial series. Unfortunately, he uses unusual notation, and as a result it is difficult to understand what he is actually doing.
Rivoal (2008, 2009) essentially uses factorial series and transformations to other kinds of series to provide new proofs of the irrationality of log(2), zeta(2), and zeta(3). As a result, the triangular array T(n,k) appears in various parts of his papers.
We believe Table I (p. 92) in Ser (1933), regarding the numbers T(n,k), corresponds to four different formulas. We have deciphered the first two of them. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins:
  -1;
  -1,  2;
  -1,  6,   -6;
  -1, 12,  -30,   20;
  -1, 20,  -90,  140,    -70;
  -1, 30, -210,  560,   -630,   252;
  -1, 42, -420, 1680,  -3150,  2772,   -924;
  -1, 56, -756, 4200, -11550, 16632, -12012, 3432;
  ...
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92) is
  -1;
  -1/2, 1/2;
  -1/3, 1/2,   -1/6;
  -1/4, 9/20,  -1/4,  1/20;
  -1/5, 2/5,   -2/7,  1/10, -1/70;
  -1/6, 5/14, -25/84, 5/36, -1/28,  1/252;
  -1/7, 9/28, -25/84, 1/6,  -9/154,  1/84, -1/924;
   ... (End)

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, pp. 92-93.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Buhl, Book review: J. Ser - Les calculs formels des séries de factorielles, L'Enseignement Mathématique, 32 (1933), p. 275.
L. A. MacColl, Review: J. Ser, Les calculs formels des séries de factorielles, Bull. Amer. Math. Soc., 41(3) (1935), p. 174.
L. M. Milne-Thomson, Review of Les calculs formels des séries de factorielles. By J. Ser. Pp. vii, 98. 20 fr. 1933. (Gauthier-Villars), The Mathematical Gazette, Vol. 18, No. 228 (May, 1934), pp. 136-137.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, preprint (2008); see pp. 1 and 15.
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, Int. J. Number Theory 5.2 (2009), 185-208; see pp. 185 and 199.

A063007 is the same triangle without the minus signs, and has much more information.
Columns 1 and 2 are A002378 and A033487; the last three diagonals are A002544, A002457, A000984.
Cf. A331431, A331432.

/* As triangle: */ [[(-1)^(k+1) * Factorial(n+k) / (Factorial(k) * Factorial(k) * Factorial(n-k)): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 19 2020

Table[CoefficientList[-Hypergeometric2F1[-n, n + 1, 1, x], x], {n, 0, 9}] // Flatten (* Georg Fischer, Jan 18 2020 after Peter Luschny in A063007 *)

def T(n,k): return (-1)^(k+1)*falling_factorial(n+k,2*k)/factorial(k)^2
flatten([[T(n,k) for k in (0..n)] for n in (0..10)]) # Peter Luschny, Jul 09 2020

T(n,k) can also be written as (-1)^(k+1)*(n+k)!/(k!*k!*(n-k)!).
From Petros Hadjicostas, Jul 09 2020: (Start)
Ser's first formula from his Table I (p. 92) is the following:
Sum_{k=0..n} T(n,k)*k!/(x*(x+1)*...*(x+k)) = -(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, Sum_{k=0..n} T(n,k)/binomial(m+k, k) = 0 for m = 1..n.
Ser's second formula from his Table I appears also in Rivoal (2008, 2009) in a slightly different form:
Sum_{k=0..n} T(n,k)/(x + k) = (-1)^(n+1)*(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, for m = 1..n, Sum_{k=0..n} T(n,k)/(m + k) = 0. (End)
T(n,k) = (-1)^(k+1)*FallingFactorial(n+k,2*k)/(k!)^2. - Peter Luschny, Jul 09 2020
From Petros Hadjicostas, Jul 10 2020: (Start)
Peter Luschny's formula above is essentially the way the numbers T(n,k) appear in Eq. (7) on p. 86 of Ser's (1933) book. Eq. (7) is essentially equivalent to the first formula above (related to Table I on p. 92).
By inverting that formula (in some way), he gets
n!/(x*(x+1)*...*(x+n)) = Sum_{p=0..n} (-1)^p*(2*p+1)*f_p(n+1)*f_p(x), where f_p(x) = (x-1)*...*(x-p)/(x*(x+1)*...*(x+p)). This is equivalent to Eq. (8) on p. 86 of Ser's book.
The rational coefficients A(n,p) = (2*p+1)*f_p(n+1) = (2*p+1)*(n*(n-1)*...*(n+1-p))/((n+1)*...*(n+1+p)) appear in Table II on p. 92 of Ser's book.
If we consider the coefficients T(n,k) and (-1)^(p+1)*A(n,p) as infinite lower triangular matrices, then they are inverses of one another (see the example below). This means that, for m >= s,
Sum_{k=s..m} T(m,k)*(-1)^(s+1)*A(k,s) = I(s=m) = Sum_{k=s..m} (-1)^(k+1)*A(m,k)*T(k,s), where I(s=m) = 1, if s = m, and = 0, otherwise.
Without the (-1)^p, we get the formula
1/(x+n) = Sum_{p=0..n} (2*p+1)*f_p(n+1)*f_p(x),
which apparently is the inversion of the second of Ser's formulas (related to Table I on p. 92).
In all of the above formulas, an empty product is by definition 1, so f_0(x) = 1/x. (End)

Thanks to Bob Selcoe, who noticed a typo in one of the entries, which, when corrected, led to an explicit formula for the whole of Ser's Table I.

A331432 Triangle T(n,k) (n >= k >= 0) read by rows: T(n,0) = (1+(-1)^n)/2; for k>=1, set T(0,k) = 0, S(n,k) = binomial(n,k)*binomial(n+k+1,k), and for n>=1, T(n,k) = S(n,k)-T(n-1,k).

Original entry on oeis.org

1, 0, 3, 1, 5, 10, 0, 10, 35, 35, 1, 14, 91, 189, 126, 0, 21, 189, 651, 924, 462, 1, 27, 351, 1749, 4026, 4290, 1716, 0, 36, 594, 4026, 13299, 22737, 19305, 6435, 1, 44, 946, 8294, 36751, 89375, 120835, 85085, 24310, 0, 55, 1430, 15730, 89375, 289003, 551837, 615043, 369512, 92378, 1, 65, 2080, 27950, 197275, 811733, 2047123, 3203837, 3031678, 1587222, 352716

Offset: 0

N. J. A. Sloane, Jan 17 2020

The scanned pages of Ser are essentially illegible, and the book is out of print and hard to locate.

For Table IV on page 93, it is simplest to ignore the minus signs. The present triangle then matches all the given terms in that triangle, so it seems best to define the triangle by the recurrences given here, and to conjecture (strongly) that this is the same as Ser's triangle.

			Triangle begins:
  1;
  0,  3;
  1,  5,   10;
  0, 10,   35,    35;
  1, 14,   91,   189,    126;
  0, 21,  189,   651,    924,    462;
  1, 27,  351,  1749,   4026,   4290,    1716;
  0, 36,  594,  4026,  13299,  22737,   19305,    6435;
  1, 44,  946,  8294,  36751,  89375,  120835,   85085,   24310;
  0, 55, 1430, 15730,  89375, 289003,  551837,  615043,  369512,   92378;
  1, 65, 2080, 27950, 197275, 811733, 2047123, 3203837, 3031678, 1587222, 352716;

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)

Columns 1 and 2 are A176222 and A331429; the last three diagonals are A002739, A002737, A001700.
Taking the component-wise sums of the rows by pairs give the triangle in A178303.
Ser's tables I and III are A331430 and A331431 (both are still mysterious).

SS := (n,k)->binomial(n,k)*binomial(n+k+1,k);
T4:=proc(n,k) local i; global SS; option remember;
if k=0 then return((1+(-1)^n)/2); fi;
if n=0 then 0 else SS(n,k)-T4(n-1,k); fi; end;
rho:=n->[seq(T4(n,k),k=0..n)];
for n from 0 to 14 do lprint(rho(n)); od:

T[n_, k_]:= T[n, k]= If[n<0, 0, If[k==0, (1 + (-1)^n)/2, Binomial[n, k]*Binomial[n+k+1, k] - T[n-1, k]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 21 2022 *)

def T(n,k): # A331432
    if (n<0): return 0
    elif (k==0): return ((n+1)%2)
    else: return binomial(n,k)*binomial(n+k+1,k) - T(n-1,k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 21 2022

T(n, k) = binomial(n,k)*binomial(n+k+1,k) - T(n-1, k), with T(n, 0) = (1 + (-1)^n)/2.
T(n, 0) = A000035(n+1).
T(n, 1) = A176222(n).
T(n, 2) = A331429(n).
T(n, n-2) = A002739(n).
T(n, n-1) = A002737(n).
T(n, n) = A001700(n).

A002738 Coefficients for extrapolation.

Original entry on oeis.org

3, 60, 630, 5040, 34650, 216216, 1261260, 7001280, 37413090, 193993800, 981608628, 4867480800, 23728968900, 114011377200, 540972351000, 2538963567360, 11802213457650, 54396360988200, 248812984520100, 1130341536324000, 5103492036502860, 22913637714910800

Offset: 0

N. J. A. Sloane

nonn

Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n-2 of B equals a(n-3). - T. D. Noe, May 01 2011

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)

Cf. A002802, A007744, A020918.

A diagonal of A331431.

[3*Binomial(2*n+3,n)*Binomial(n+3,3): n in [0..30]]; // G. C. Greubel, Mar 21 2022

Table[Total[Inverse[HilbertMatrix[n]][[n - 2]]], {n, 3, 25}] (* T. D. Noe, May 02 2011 *)

[3*binomial(2*n+3,3)*binomial(2*n,n) for n in (0..30)] # G. C. Greubel, Mar 21 2022

From Alois P. Heinz, May 02 2011: (Start)
a(n) = 3*binomial(2*n+3,n)*binomial(n+3,n).
G.f.: 3*(1 + 6*x)/(1-4*x)^(7/2). (End)
a(n) = binomial(2*n+3,n)*(n^3 + 6*n^2 + 11*n+6)/2. - Charles R Greathouse IV, May 02 2011
a(n) = 3*A007744(n). - R. J. Mathar, Jan 21 2020
a(n) = (3/2)*( 5*A020918(n) - 3*A002802(n)). - G. C. Greubel, Mar 21 2022

Extended by T. D. Noe, May 01 2011

A331323 a(n) = [x^n] (1 - 2x)/(1 - 8x + 4*x^2)^(3/2).

Original entry on oeis.org

1, 10, 90, 772, 6430, 52524, 423220, 3375880, 26720118, 210195100, 1645295212, 12825551160, 99633196780, 771702434104, 5961969066600, 45958506432016, 353585912577190, 2715647948258940, 20824876515839932, 159474192002499160, 1219708190630800836, 9318143974952519080

Offset: 0

Peter Luschny, Jan 18 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Column 5 of A331511.

Cf. A331431.

[(&+[3^k*(k+1)*Binomial(n+1,k+1)^2: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Mar 22 2022

gf := (1-2*x)/(4*x^2-8*x+1)^(3/2): ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0..21); # Or:
a := proc(n) option remember; if n<3 then [1, 10, 90][n+1] else
(10*n*a(n-1) + 20*(1-n)*a(n-2) + 8*(n-1)*a(n-3))/n fi end:
seq(a(n), n=0..21);

a[n_] := Sum[3^k * (k + 1) * Binomial[n + 1, k + 1]^2, {k, 0, n}]; Array[a, 22, 0] (* Amiram Eldar, Jan 20 2020 *)

{a(n) = 2^n*sum(k=0, n, (n+k+1)*binomial(n, k)*binomial(n+k, k)/2^k)} \\ Seiichi Manyama, Jan 18 2020

N=20; x='x+O('x^N); Vec((1-2*x)/(4*x^2-8*x+1)^(3/2)) \\ Seiichi Manyama, Jan 18 2020

{a(n) = sum(k=0, n, 3^k*(k+1)*binomial(n+1, k+1)^2)} \\ Seiichi Manyama, Jan 20 2020

[sum(2^(n-k)*(n+k+1)*binomial(2*k,k)*binomial(n+k,2*k)  for k in (0..n)) for n in (0..30)] # G. C. Greubel, Mar 22 2022

a(n) = (-2)^n*Sum_{k=0..n} A331431(n, k)/(-2)^k.
a(n) = (10*n*a(n-1) + 20*(1-n)*a(n-2) + 8*(n-1)*a(n-3))/n.
a(n) = 2^(n-1)*(n+1)*(n*hypergeom([1-n, n+2], [2], -1/2) + 2*hypergeom([-n, n+1], [1], -1/2)).
a(n) = Sum_{k=0..n} 3^k * (k+1) * binomial(n+1,k+1)^2. - Seiichi Manyama, Jan 20 2020
a(n) = (n + 1)^2*hypergeom([-n, -n], [2], 3). - Peter Luschny, Jan 20 2020
n * (2*n-1) * a(n) = 2 * (8 * n^2 - 3) * a(n-1) - 4 * n * (2*n+1) * a(n-2) for n>1. - Seiichi Manyama, Jan 25 2020

A331322 a(n) = (3*n + 1)!/(n!)^3.

Original entry on oeis.org

1, 24, 630, 16800, 450450, 12108096, 325909584, 8779605120, 236637794250, 6380456082000, 172080900531540, 4641917845743360, 125235075213284400, 3379123922914656000, 91184624634161304000, 2460769070127233057280, 66411927755894739034170, 1792432652235221330334000

Offset: 0

Peter Luschny, Jan 18 2020

nonn

Diagonal of the rational function 1 / (1 - x - y - z)^2. - Ilya Gutkovskiy, Apr 23 2025

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A000108, A006480, A117671, A331431.

[(n+1)^2*Binomial(3*n+1,n+1)*Catalan(n): n in [0..25]]; // G. C. Greubel, Mar 22 2022

a := n -> (3*n+1)!/(n!)^3: seq(a(n), n=0..17); # Or:
hypergeom([2/3, 4/3], [1], 27*x): ser := series(%, x, 20):
seq(coeff(%, x, n), n=0..17); # Or:
a := proc(n) option remember; if n=0 then 1 else 3*(9 - n^(-2))*a(n-1) fi end:
# 4th Maple program:
W:=proc(x)sqrt(3)*MeijerG([[], [0, 0]], [[1/3, -1/3], []], x/27)/(18*Pi);end;
a:=proc(n) round(evalf[32](int(x^n*W(x),x=0..27)));end;
seq(a(n),n=0..17);
# Karol A. Penson, Jul 28 2023

Table[(3*n+1)*Binomial[3*n,n]*Binomial[2*n,n], {n,0,25}] (* G. C. Greubel, Mar 22 2022 *)

[(3*n+1)*binomial(2*n,n)*binomial(3*n,n) for n in (0..25)] # G. C. Greubel, Mar 22 2022

a(n) = [x^n] hypergeom([2/3, 4/3], [1], 27*x).
a(n) = 3*(9 - n^(-2))*a(n-1) for n > 0.
a(n) = (-1)^n*A331431(2*n, n).
a(n) = (n+1)^2*A117671(n)*A000108(n). - G. C. Greubel, Mar 22 2022
From Karol A. Penson, Jul 28 2023: (Start)
a(n) = Integral_{x=0..27} x^n*W(x) dx, where the weight function W(x) is defined on (0, 27) and it can be expressed with the Meijer G-function MeijerG as: W(x) = (sqrt(3)/(18*Pi))*MeijerG([[],[0,0]],[[-1/3,1/3],[]],x/27). The function W(x) is positive on its support (0, 27), is singular at x=0, and decreases monotonically to zero at x = 27.
The function W(x) is unique as it is the solution of the Hausdorff moment problem with the moments a(n). Due to the presence of two equal parameters (0,0) in MeijerG, it is not certain if W(x) can be represented by other known special functions. (End)
From  Peter Bala, Oct 10 2024: (Start)
a(n) = (3*n + 1)*A006480(n).
a(n-1) = 1/(8*n^3) * Sum_{k = 0..2*n} (-1)^(n+k) * k*(2*n-k)^3 * binomial(2*n, k)^3 for n >= 1.
a(n-1) = 1/(4*n^2) * Sum_{k = 0..2*n-1} (-1)^(n+k) * k^3 * binomial(2*n, k)^2 * binomial(2*n-1, k) for n >= 1. (End)

cp's OEIS Frontend

A331433 Column 1 of triangle in A331431.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A331434 Column 2 of triangle in A331431.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A002457 a(n) = (2n+1)!/n!^2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Sage

Formula

A002736 Apéry numbers: a(n) = n^2*C(2n,n).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

MuPAD

PARI

PARI

Sage

Formula

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)*binomial(n,k)*binomial(n+k,k) (n >= k >= 0).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A331432 Triangle T(n,k) (n >= k >= 0) read by rows: T(n,0) = (1+(-1)^n)/2; for k>=1, set T(0,k) = 0, S(n,k) = binomial(n,k)*binomial(n+k+1,k), and for n>=1, T(n,k) = S(n,k)-T(n-1,k).

Views

Author

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).

A331323 a(n) = [x^n] (1 - 2x)/(1 - 8x + 4*x^2)^(3/2).