A090816 -id:A090816 - 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

A003506 Triangle of denominators in Leibniz's Harmonic Triangle a(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 12, 12, 4, 5, 20, 30, 20, 5, 6, 30, 60, 60, 30, 6, 7, 42, 105, 140, 105, 42, 7, 8, 56, 168, 280, 280, 168, 56, 8, 9, 72, 252, 504, 630, 504, 252, 72, 9, 10, 90, 360, 840, 1260, 1260, 840, 360, 90, 10, 11, 110, 495, 1320, 2310, 2772, 2310, 1320, 495, 110, 11

Offset: 1

N. J. A. Sloane

Array 1/Beta(n,m) read by antidiagonals. - Michael Somos, Feb 05 2004
a(n,3) = A027480(n-2); a(n,4) = A033488(n-3). - Ross La Haye, Feb 13 2004
a(n,k) = total size of all of the elements of the family of k-size subsets of an n-element set. For example, a 2-element set, say, {1,2}, has 3 families of k-size subsets: one with 1 0-size element, one with 2 1-size elements and one with 1 2-size element; respectively, {{}}, {{1},{2}}, {{1,2}}. - Ross La Haye, Dec 31 2006
Second slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with a(1,0,0)=1 and a(m<>1=0,n>=0,0>=o)=0, for which the first slice is Pascal's triangle (slice read by antidiagonals). - Thomas Wieder, Aug 06 2006
Triangle, read by rows, given by [2,-1/2,1/2,0,0,0,0,0,0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 07 2007
This sequence * [1/1, 1/2, 1/3, ...] = (1, 3, 7, 15, 31, ...). - Gary W. Adamson, Nov 14 2007
n-th row = coefficients of first derivative of corresponding Pascal's triangle row. Example: x^4 + 4x^3 + 6x^2 + 4x + 1 becomes (4, 12, 12, 4). - Gary W. Adamson, Dec 27 2007
From Paul Curtz, Jun 03 2011: (Start)
Consider
   1     1/2   1/3    1/4    1/5
  -1/2  -1/6  -1/12  -1/20  -1/30
   1/3   1/12  1/30   1/60   1/105
  -1/4  -1/20 -1/60  -1/140 -1/280
   1/5   1/30  1/105  1/280  1/630
This is an autosequence (the inverse binomial transform is the sequence signed) of the second kind: the main diagonal is 2 times the first upper diagonal.
Note that 2, 12, 60, ... = A005430(n+1), Apery numbers = 2*A002457(n). (End)
From Louis Conover (for the 9th grade G1c mathematics class at the Chengdu Confucius International School), Mar 02 2015: (Start)
The i-th order differences of n^-1 appear in the (i+1)th row.
  1,    1/2,   1/3,   1/4,    1/5,    1/6,    1/7,     1/8, ...
  1/2,  1/6,  1/12,  1/20,   1/30,   1/42,   1/56,    1/72, ...
  1/3, 1/12,  1/30,  1/60,  1/105,  1/168,  1/252,   1/360, ...
  1/4, 1/20,  1/60, 1/140,  1/280,  1/504,  1/840,  1/1320, ...
  1/5, 1/30, 1/105, 1/280,  1/630, 1/1260, 1/2310,  1/3960, ...
  1/6, 1/42, 1/168, 1/504, 1/1260, 1/2772, 1/5544, 1/12012, ...
(End)
T(n,k) is the number of edges of distance k from a fixed vertex in the n-dimensional hypercube. - Simon Burton, Nov 04 2022

			The triangle begins:
  1;
  1/2, 1/2;
  1/3, 1/6, 1/3;
  1/4, 1/12, 1/12, 1/4;
  1/5, 1/20, 1/30, 1/20, 1/5;
  ...
The triangle of denominators begins:
   1
   2   2
   3   6   3
   4  12  12    4
   5  20  30   20    5
   6  30  60   60   30    6
   7  42 105  140  105   42    7
   8  56 168  280  280  168   56    8
   9  72 252  504  630  504  252   72   9
  10  90 360  840 1260 1260  840  360  90  10
  11 110 495 1320 2310 2772 2310 1320 495 110 11

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, see 130.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 38.
G. Boole, A Treatise On The Calculus of Finite Differences, Dover, 1960, p. 26.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25.
M. Elkadi and B. Mourrain, Symbolic-numeric methods for solving polynomial equations and applications, Chap 3. of A. Dickenstein and I. Z. Emiris, eds., Solving Polynomial Equations, Springer, 2005, pp. 126-168. See p. 152.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 35.

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
F. S. Al-Kharousi, A. Umar, and M. M. Zubairu, On injective partial Catalan monoids, arXiv:2501.00285 [math.GR], 2024. See pp. 8-9.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993.
H. J. Brothers, Pascal's Prism: Supplementary Material.
D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
Yilmaz Simsek, Construction of a generalization of the Leibnitz numbers and their properties, arXiv:2011.13701 [math.NT], 2020.
Eric Weisstein's World of Mathematics, Leibniz Harmonic Triangle

Cf. A007622, A128064, A164555/A027642, A356546.
Cf. A094305, A121547, A121306, A119800, A002378, A007318.
Row sums are in A001787. Central column is A002457. Half-diagonal is in A090816. A116071, A215652.
Denominators of i-th order differences of n^-1 are given in: (1st) A002378, (2nd) A027480, (3rd) A033488, (4th) A174002, (5th) A253946. - Louis Conover, Mar 02 2015
Columns k >= 1 (offset 1): A000027, A002378, A027480, A033488, A174002, A253946(n+4), ..., with sum of reciprocals: infinity, 1, 1/2, 1/3, 1/4, 1/5, ..., respectively. - Wolfdieter Lang, Jul 20 2022

a003506 n k = a003506_tabl !! (n-1) !! (n-1)
a003506_row n = a003506_tabl !! (n-1)
a003506_tabl = scanl1 (\xs ys ->
   zipWith (+) (zipWith (+) ([0] ++ xs) (xs ++ [0])) ys) a007318_tabl
a003506_list = concat a003506_tabl
-- Reinhard Zumkeller, Nov 14 2013, Nov 17 2011

with(combstruct):for n from 0 to 11 do seq(m*count(Combination(n), size=m), m = 1 .. n) od; # Zerinvary Lajos, Apr 09 2008
A003506 := (n,k) -> k*binomial(n,k):
seq(print(seq(A003506(n,k),k=1..n)),n=1..7); # Peter Luschny, May 27 2011

L[n_, 1] := 1/n; L[n_, m_] := L[n, m] = L[n - 1, m - 1] - L[n, m - 1]; Take[ Flatten[ Table[ 1 / L[n, m], {n, 1, 12}, {m, 1, n}]], 66]
t[n_, m_] = Gamma[n]/(Gamma[n - m]*Gamma[m]); Table[Table[t[n, m], {m, 1, n - 1}], {n, 2, 12}]; Flatten[%] (* Roger L. Bagula and Gary W. Adamson, Sep 14 2008 *)
Table[k*Binomial[n,k],{n,1,7},{k,1,n}] (* Peter Luschny, May 27 2011 *)
t[n_, k_] := Denominator[n!*k!/(n+k+1)!]; Table[t[n-k, k] , {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013 *)

A(i,j)=if(i<1||j<1,0,1/subst(intformal(x^(i-1)*(1-x)^(j-1)),x,1))

A(i,j)=if(i<1||j<1,0,1/sum(k=0,i-1,(-1)^k*binomial(i-1,k)/(j+k)))

{T(n, k) = (n + 1 - k) * binomial( n, k - 1)} /* Michael Somos, Feb 06 2011 */

T_row = lambda n: (n*(x+1)^(n-1)).list()
for n in (1..10): print(T_row(n)) # Peter Luschny, Feb 04 2017
# Assuming offset 0:
def A003506(n, k):
    return falling_factorial(n+1,n)//(factorial(k)*factorial(n-k))
for n in range(9): print([A003506(n, k) for k in range(n+1)]) # Peter Luschny, Aug 13 2022

a(n, 1) = 1/n; a(n, k) = a(n-1, k-1) - a(n, k-1) for k > 1.
Considering the integer values (rather than unit fractions): a(n, k) = k*C(n, k) = n*C(n-1, k-1) = a(n, k-1)*a(n-1, k-1)/(a(n, k-1) - a(n-1, k-1)) = a(n-1, k) + a(n-1, k-1)*k/(k-1) = (a(n-1, k) + a(n-1, k-1))*n/(n-1) = k*A007318(n, k) = n*A007318(n-1, k-1). Row sums of integers are n*2^(n-1) = A001787(n); row sums of the unit fractions are A003149(n-1)/A000142(n). - Henry Bottomley, Jul 22 2002
From Vladeta Jovovic, Nov 01 2003: (Start)
G.f.: x*y/(1-x-y*x)^2.
E.g.f.: x*y*exp(x+x*y). (End)
T(n,k) = n*binomial(n-1,k-1) = n*A007318(n-1,k-1). - Philippe Deléham, Aug 04 2006
Binomial transform of A128064(unsigned). - Gary W. Adamson, Aug 29 2007
From Roger L. Bagula and Gary W. Adamson, Sep 14 2008: (Start)
t(n,m) = Gamma(n)/(Gamma(n - m)*Gamma(m)).
f(s,n) = Integral_{x=0..oo} exp(-s*x)*x^n dx = Gamma(n)/s^n; t(n,m) = f(s,n)/(f(s,n-m)*f(s,m)) = Gamma(n)/(Gamma(n - m)*Gamma(m)); the powers of s cancel out. (End)
From Reinhard Zumkeller, Mar 05 2010: (Start)
T(n,5) = T(n,n-4) = A174002(n-4) for n > 4.
T(2*n,n) = T(2*n,n+1) = A005430(n). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(1,1) = 1 and, for n > 1, T(n,k) = 0 if k <= 1 or if k > n. - Philippe Deléham, Mar 17 2012
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+1-k,k+1-i). - Mircea Merca, Apr 11 2012
If we include a main diagonal of zeros so that the array is in the form
  0
  1   0
  2   2   0
  3   6   3   0
  4  12  12   4   0
...
then we obtain the exponential Riordan array [x*exp(x),x], which factors as [x,x]*[exp(x),x] = A132440*A007318. This array is the infinitesimal generator for A116071. A signed version of the array is the infinitesimal generator for A215652. - Peter Bala, Sep 14 2012
a(n,k) = (n-1)!/((n-k)!(k-1)!) if k > n/2 and a(n,k) = (n-1)!/((n-k-1)!k!) otherwise. [Forms 'core' for Pascal's recurrence; gives common term of RHS of T(n,k) = T(n-1,k-1) + T(n-1,k)]. - Jon Perry, Oct 08 2013
Assuming offset 0: T(n, k) = FallingFactorial(n + 1, n) / (k! * (n - k)!). The counterpart using the rising factorial is A356546. - Peter Luschny, Aug 13 2022

Edited by N. J. A. Sloane, Oct 07 2007

A090957 a(n) = 1/(Integral_{x=0..1} (x^4 - x^5)^n dx).

Original entry on oeis.org

1, 30, 495, 7280, 101745, 1381380, 18407025, 242082720, 3153092085, 40763504210, 523886186670, 6700599687600, 85360889543475, 1083790852008480, 13721016740550360, 173280964190422080, 2183615911571190525

Offset: 0

Al Hakanson (hawkuu(AT)excite.com), Feb 27 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..915

Cf. A002457, A090816, A090969.

[Factorial(5*n+1)/(Factorial(n)*Factorial(4*n)): n in [0..20]]; // G. C. Greubel, Feb 03 2019

seq(factorial(5*n+1)/(factorial(n)*factorial(4*n)), n = 0 .. 16); # Emeric Deutsch, Jul 03 2009

Table[1/Integrate[(x^4-x^5)^n,{x,0,1}],{n,0,20}] (* Harvey P. Dale, Jan 02 2013 *)
Table[1/Beta[4*n+1,n+1], {n,0,20}] (* G. C. Greubel, Feb 03 2019 *)

for (n = 0, 20, pol = (x^4 - x^5)^n; s = 0; for (i = 4*n, 5*n, s += polcoeff(pol, i)/(i + 1)); print(1/s)); \\ David Wasserman, Feb 22 2006

vector(20, n, n--; (5*n+1)!/(n!*(4*n)!)) \\ G. C. Greubel, Feb 03 2019

[1/beta(4*n+1,n+1) for n in range(20)] # G. C. Greubel, Feb 03 2019

a(n) = 1/B(4*n+1,n+1) = (5*n+1)!/(n! * (4*n)!), where B(p,q) is Euler's beta function. - Emeric Deutsch, Jul 03 2009

a(n) ~ sqrt(n)*5^(5*n+3/2) / (sqrt(Pi)*2^(8*n+3/2)). - Vaclav Kotesovec, Aug 15 2017

More terms from David Wasserman, Feb 22 2006

A090969 a(n) = 1/Integral_{x=0..1} (x^5 - x^6)^n.

Original entry on oeis.org

1, 42, 858, 15504, 265650, 4417686, 72068304, 1160068104, 18490100706, 292486494300, 4599035681526, 71963547329856, 1121519754006288, 17419158268943970, 269767427275060200, 4167406330765934256

Offset: 0

Al Hakanson (hawkuu(AT)excite.com), Feb 29 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..845

Cf. A002457, A004355, A016921, A090816, A090957.

[Factorial(6*n+1)/(Factorial(n)*Factorial(5*n)): n in [0..20]]; // G. C. Greubel, Feb 03 2019

seq(factorial(6*n+1)/(factorial(n)*factorial(5*n)), n = 0 .. 16); # Emeric Deutsch, Jun 29 2009

Table[1/Beta[5*n+1, n+1], {n,0,20}] (* G. C. Greubel, Feb 03 2019 *)

vector(20, n, n--; (6*n+1)!/(n!*(5*n)!)) \\ G. C. Greubel, Feb 03 2019

[1/beta(5*n+1,n+1) for n in range(20)] # G. C. Greubel, Feb 03 2019

a(n) = A016921(n)*A004355(n). - R. J. Mathar, Jun 21 2009
a(n) = 1/B(5*n+1,n+1) = (6*n+1)!/(n! * (5*n)!), where B(p,q) is Euler's beta function (basically identical with R. J. Mathar's comment). - Emeric Deutsch, Jun 29 2009
a(n) ~ 2^(6*n+1) * 3^(6*n+3/2) * sqrt(n) / (sqrt(Pi) * 5^(5*n+1/2)). - Vaclav Kotesovec, Aug 15 2017

Extended by Emeric Deutsch, Jun 29 2009

A306290 a(n) = 1/(Integral_{x=0..1} (x^3 - x^4)^n dx).

Original entry on oeis.org

1, 20, 252, 2860, 30940, 325584, 3364900, 34337160, 347103900, 3483301360, 34754081648, 345120260940, 3413758188932, 33655718658800, 330869721936600, 3244839440755920, 31754250910172700, 310165459118369712, 3024542552887591120, 29449493278116018800, 286360607519186119920

Offset: 0

G. C. Greubel, Feb 03 2019

Necdet Batir, On certain series involving reciprocals of binomial coefficients, Journal of Classical Analysis, Vol. 2, No. 1 (2013), pp. 1-8.

Cf. A002457, A005810, A016813, A090816, A090957, A090969.

List([0..30], n -> Factorial(4*n+1)/(Factorial(n)*Factorial(3*n)));

[Factorial(4*n+1)/(Factorial(n)*Factorial(3*n)): n in [0..20]];

Mathematica
```
Table[1/Beta[3*n+1, n+1], {n, 0, 20}]
```

vector(20, n, n--; (4*n+1)!/(n!*(3*n)!))

Sage
```
[1/beta(3*n+1,n+1) for n in range(20)]
```

a(n) = 1/Beta(3*n+1,n+1) = (4*n+1)!/(n! * (3*n)!).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (3*n + 2*k + 1)*binomial(3*n+k, k).  This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (3*n + 2*k + 1) * binomial(3*n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+3)*n+1, 3*n). - Peter Bala, Nov 02 2024
From Amiram Eldar, Dec 09 2024: (Start)
a(n) = (4*n + 1) * binomial(4*n, n) = A016813(n) * A005810(n).
Formulas from Batir (2013):
Sum_{n>=0} 1/a(n) = f(c) = 1.05435362585114283076..., where f(x) = (x*(x^2-1)/(2*(2*x^2+1))) * log(abs((x+1)/(x-1))) + ((x-1)*(x^3+1)/(4*x*(2*x^2+1))) * sqrt(x/(x-2)) * (arctan(sqrt(x/(x-2))) + arctan(((3-x)/(x+1))*sqrt(x/(x-2)))) + ((x+1)*(x^3-1)/(4*x*(2*x^2+1))) * sqrt(x/(x+2)) * (arctan(((x+3)/(x-1))*sqrt(x/(x+2))) - arctan(sqrt(x/(x+2)))), and c = sqrt(1 + (16/sqrt(3))*cos(arctan(sqrt(229/27))/3)).
Sum_{n>=0} (-1)^n/a(n) = f(d) = 0.953648123517883351708..., where f(x) is defined above, and d = sqrt(1 + 16*(2/(3*(9+sqrt(849))))^(1/3) - 2*(2/3)^(2/3)*(9+sqrt(849))^(1/3)). (End)

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 21, 84, 84, 1, 36, 270, 660, 495, 1, 55, 660, 2860, 5005, 3003, 1, 78, 1365, 9100, 27300, 37128, 18564, 1, 105, 2520, 23800, 107100, 244188, 271320, 116280, 1, 136, 4284, 54264, 339150, 1139544, 2089164, 1961256, 735471, 1, 171, 6840, 111720, 921690, 4239774, 11306064

Offset: 0

Peter Bala, Feb 13 2024

Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).

			Triangle begins
n\k| 0    1     2      3       4       5       6       7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 0 | 1
 1 | 1    3
 2 | 1   10    15
 3 | 1   21    84     84
 4 | 1   36   270    660     495
 5 | 1   55   660   2860    5005    3003
 6 | 1   78  1365   9100   27300   37128   18564
 7 | 1  105  2520  23800  107100  244188  271320  116280
 ...

A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).

Cf. A026000, A063007, A102537, A110608, A339710.

seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);

n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
(x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
Analog of Rodrigues' formula for the shifted Legendre polynomials:
R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
Analog of Rodrigues' formula for the Legendre polynomials:
R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
Orthogonality properties:
Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
Recurrence for row polynomials:
2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
(2^n)*R(n, -1/2) = A234839(n).

cp's OEIS Frontend

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

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A003506 Triangle of denominators in Leibniz's Harmonic Triangle a(n,k), n >= 1, 1 <= k <= n.

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A090957 a(n) = 1/(Integral_{x=0..1} (x^4 - x^5)^n dx).

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A090969 a(n) = 1/Integral_{x=0..1} (x^5 - x^6)^n.

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A306290 a(n) = 1/(Integral_{x=0..1} (x^3 - x^4)^n dx).

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.