cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

References

  • 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]

Crossrefs

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

Programs

Formula

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

A005259 Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.

Original entry on oeis.org

1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485
Offset: 0

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Author

Keywords

Comments

Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021

Examples

			G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - _Michael B. Porter_, Jul 14 2016
		

References

  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 137-153.
  • Wolfram Koepf, Hypergeometric Identities. Ch. 2 in Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. Braunschweig, Germany: Vieweg, pp. 55, 119 and 146, 1998.
  • Maxim Kontsevich and Don Zagier, Periods, pp. 771-808 of B. Engquist and W. Schmid, editors, Mathematics Unlimited - 2001 and Beyond, 2 vols., Springer-Verlag, 2001.
  • Leonard Lipshitz and Alfred van der Poorten, "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), pp. 339-358.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Related to diagonal of rational functions: A268545-A268555.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Cf. A092826 (prime terms).

Programs

  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k)^2)); # Muniru A Asiru, Sep 28 2018
    
  • Haskell
    a005259 n = a005259_list !! n
    a005259_list = 1 : 5 : zipWith div (zipWith (-)
       (tail $ zipWith (*) a006221_list a005259_list)
       (zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
    -- Reinhard Zumkeller, Mar 13 2014
    
  • Magma
    [&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in  [0..17]]; // Marius A. Burtea, Jan 20 2020
    
  • Maple
    a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
    # Alternative:
    a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
    seq(simplify(a(n)), n=0..17); # Peter Luschny, Jan 19 2020
  • Mathematica
    Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1,1,1}, 1],{n,0,13}] (* Jean-François Alcover, Apr 01 2011 *)
    Table[Sum[(Binomial[n,k]Binomial[n+k,k])^2,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 15 2011 *)
    a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)
  • PARI
    a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k,k))^2) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    def A005259(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= ((n+k+1)*(n-k))**2
            m //=(k+1)**4
        return g # Chai Wah Wu, Oct 02 2022

Formula

D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
a(n) = Sum_{k >= 0} A063007(n, k)*A000172(k). A000172 = Franel numbers. - Philippe Deléham, Aug 14 2003
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Peter Bala, Jan 18 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,k)^2 * C(n,j)^2 * C(k+j,k), see Labelle et al. link. - Max Alekseyev, Mar 12 2025

A005258 Apéry numbers: a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,k).

Original entry on oeis.org

1, 3, 19, 147, 1251, 11253, 104959, 1004307, 9793891, 96918753, 970336269, 9807518757, 99912156111, 1024622952993, 10567623342519, 109527728400147, 1140076177397091, 11911997404064793, 124879633548031009, 1313106114867738897, 13844511065506477501
Offset: 0

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Author

Keywords

Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Equals the main diagonal of square array A108625. - Paul D. Hanna, Jun 14 2005
This sequence is t_5 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Conjecture: For each n=1,2,3,... the polynomial a_n(x) = Sum_{k=0..n} C(n,k)^2*C(n+k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Diagonal of rational functions 1/(1 - x - x*y - y*z - x*z - x*y*z), 1/(1 + y + z + x*y + y*z + x*z + x*y*z), 1/(1 - x - y - z + x*y + x*y*z), 1/(1 - x - y - z + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 07 2018

Examples

			G.f. = 1 + 3*x + 19*x^2 + 147*x^3 + 1251*x^4 + 11253*x^5 + 104959*x^6 + ...
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • S. Melczer, An Invitation to Analytic Combinatorics, 2021; p. 129.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A007318.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • GAP
    a:=n->Sum([0..n],k->(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)^2);;
    A005258:=List([0..20],n->a(n));; # Muniru A Asiru, Feb 11 2018
    
  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k))); # Muniru A Asiru, Jul 29 2018
    
  • Haskell
    a005258 n = sum [a007318 n k ^ 2 * a007318 (n + k) k | k <- [0..n]]
    -- Reinhard Zumkeller, Jan 04 2013
    
  • Magma
    [&+[Binomial(n,k)^2 * Binomial(n+k,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Nov 28 2018
    
  • Maple
    with(combinat): seq(add((multinomial(n+k,n-k,k,k))*binomial(n,k), k=0..n), n=0..18); # Zerinvary Lajos, Oct 18 2006
    a := n -> binomial(2*n, n)*hypergeom([-n, -n, -n], [1, -2*n], 1):
    seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 10 2018
  • Mathematica
    a[n_] := HypergeometricPFQ[ {n+1, -n, -n}, {1, 1}, 1]; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Jan 20 2012, after Vladeta Jovovic *)
    Table[Sum[Binomial[n,k]^2 Binomial[n+k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 25 2019 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-1-n), sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)))} /* Michael Somos, Sep 18 2013 */
    
  • Python
    def A005258(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= (n+k+1)*(n-k)**2
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 02 2022

Formula

a(n) = hypergeom([n+1, -n, -n], [1, 1], 1). - Vladeta Jovovic, Apr 24 2003
D-finite with recurrence: (n+1)^2 * a(n+1) = (11*n^2+11*n+3) * a(n) + n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Let b(n) be the solution to the above recurrence with b(0) = 0, b(1) = 5. Then the b(n) are rational numbers with b(n)/a(n) -> zeta(2) very rapidly. The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^(n-1)*5/n^2 leads to a series acceleration formula: zeta(2) = 5 * Sum_{n >= 1} 1/(n^2*a(n)*a(n-1)) = 5*(1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...). Similar results hold for the constant e: see A143413. - Peter Bala, Aug 14 2008
G.f.: hypergeom([1/12, 5/12],[1], 1728*x^5*(1-11*x-x^2)/(1-12*x+14*x^2+12*x^3+x^4)^3) / (1-12*x+14*x^2+12*x^3+x^4)^(1/4). - Mark van Hoeij, Oct 25 2011
a(n) ~ ((11+5*sqrt(5))/2)^(n+1/2)/(2*Pi*5^(1/4)*n). - Vaclav Kotesovec, Oct 05 2012
1/Pi = 5*(sqrt(47)/7614)*Sum_{n>=0} (-1)^n a(n)*binomial(2n,n)*(682n+71)/15228^n. [Cooper, equation (4)] - Jason Kimberley, Nov 26 2012
a(-1 - n) = (-1)^n * a(n) if n>=0. a(-1 - n) = -(-1)^n * a(n) if n<0. - Michael Somos, Sep 18 2013
0 = a(n)*(a(n+1)*(+4*a(n+2) + 83*a(n+3) - 12*a(n+4)) + a(n+2)*(+32*a(n+2) + 902*a(n+3) - 147*a(n+4)) + a(n+3)*(-56*a(n+3) + 12*a(n+4))) + a(n+1)*(a(n+1)*(+17*a(n+2) + 374*a(n+3) - 56*a(n+4)) + a(n+2)*(+176*a(n+2) + 5324*a(n+3) - 902*a(n+4)) + a(n+3)*(-374*a(n+3) + 83*a(n+4))) + a(n+2)*(a(n+2)*(-5*a(n+2) - 176*a(n+3) + 32*a(n+4)) + a(n+3)*(+17*a(n+3) - 4*a(n+4))) for all n in Z. - Michael Somos, Aug 06 2016
a(n) = binomial(2*n, n)*hypergeom([-n, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+k,k)^2. - Peter Bala, Feb 10 2018
G.f. y=A(x) satisfies: 0 = x*(x^2 + 11*x - 1)*y'' + (3*x^2 + 22*x - 1)*y' + (x + 3)*y. - Gheorghe Coserea, Jul 01 2018
From Peter Bala, Jan 15 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(j+k)*C(n,k)*C(n+k,k)^2*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)^2*C(n+k,k)*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^j*C(n,k)^2*C(n,j)*C(3*n-j-k,2*n). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 1. At m = 2 we get the Apéry numbers A005259. - Peter Bala, Dec 22 2020
a(n) = (-1)^n*Sum_{j=0..n} (1 - 5*j*H(j) + 5*j*H(n - j))*binomial(n, j)^5, where H(n) denotes the n-th harmonic number, A001008/A002805. (Paule/Schneider). - Peter Luschny, Jul 23 2021
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=(3 + x)*T(x) + (-1 + 22*x + 3*x^2)*T'(x) + x*(-1 + 11*x + x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(1 - 12*x + 14*x^2 + 12*x^3 + x^4);
g3 = 1 - 18*x + 75*x^2 + 75*x^4 + 18*x^5 + x^6;
which determine an elliptic surface with four singular fibers. (End)
Conjecture: a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Jul 08 2024

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

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Comments

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

Examples

			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
		

References

  • 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.

Crossrefs

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

Programs

  • Haskell
    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
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    A(i,j)=if(i<1||j<1,0,1/subst(intformal(x^(i-1)*(1-x)^(j-1)),x,1))
    
  • PARI
    A(i,j)=if(i<1||j<1,0,1/sum(k=0,i-1,(-1)^k*binomial(i-1,k)/(j+k)))
    
  • PARI
    {T(n, k) = (n + 1 - k) * binomial( n, k - 1)} /* Michael Somos, Feb 06 2011 */
    
  • SageMath
    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

Formula

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

Extensions

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

A001803 Numerators in expansion of (1 - x)^(-3/2).

Original entry on oeis.org

1, 3, 15, 35, 315, 693, 3003, 6435, 109395, 230945, 969969, 2028117, 16900975, 35102025, 145422675, 300540195, 9917826435, 20419054425, 83945001525, 172308161025, 1412926920405, 2893136075115, 11835556670925
Offset: 0

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Comments

a(n) is the denominator of the integral from 0 to Pi of (sin(x))^(2*n+1). - James R. Buddenhagen, Aug 17 2008
a(n) is the denominator of (2n)!!/(2*n + 1)!! = 2^(2*n)*n!*n!/(2*n + 1)! (see Andersson). - N. J. A. Sloane, Jun 27 2011
a(n) = (2*n + 1)*A001790(n). A046161(n)/a(n) = 1, 2/3, 8/15, 16/35, 128/315, 256/693, ... is binomial transform of Madhava-Gregory-Leibniz series for Pi/4 (i.e., 1 - 1/3 + 1/5 - 1/7 + ... ). See A173384 and A173396. - Paul Curtz, Feb 21 2010
a(n) is the denominator of Integral_{x=-oo..oo} sech(x)^(2*n+2) dx. The corresponding numerator is A101926(n). - Mohammed Yaseen, Jul 25 2023

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
  • G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:9 at page 51.

Crossrefs

The denominator is given in A046161.
Largest odd divisors of A001800, A002011, A002457, A005430, A033876, A086228.
Bisection of A004731, A004735, A086116.
Second column of triangle A100258.
Cf. A002596 (numerators in expansion of (1-x)^(1/2)).
Cf. A161198 (triangle related to the series expansions of (1-x)^((-1-2*n)/2)).
A163590 is the odd part of the swinging factorial, A001790 at even indices. - Peter Luschny, Aug 01 2009

Programs

  • Julia
    A001803(n) = sum(<<(A001790(k), A005187(n) - A005187(k)) for k in 0:n) # Peter Luschny, Oct 03 2019
    
  • Magma
    A001803:= func< n | Numerator(Binomial(n+2,2)*Catalan(n+1)/4^n) >;
    [A001803(n): n in [0..30]]; // G. C. Greubel, Apr 27 2025
    
  • Maple
    swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
    sigma := n -> 2^(add(i,i= convert(iquo(n,2),base,2))):
    a := n -> swing(2*n+1)/sigma(2*n+1); # Peter Luschny, Aug 01 2009
    A001803 := proc(n) (2*n+1)*binomial(2*n,n)/4^n ; numer(%) ; end proc: # R. J. Mathar, Jul 06 2011
    a := n -> denom(Pi*binomial(n, -1/2)): seq(a(n), n = 0..22); # Peter Luschny, Dec 06 2024
  • Mathematica
    Numerator/@CoefficientList[Series[(1-x)^(-3/2),{x,0,25}],x]  (* Harvey P. Dale, Feb 19 2011 *)
    Table[Denominator[Beta[1, n + 1, 1/2]], {n, 0, 22}] (* Gerry Martens, Nov 13 2016 *)
  • PARI
    a(n) = numerator((2*n+1)*binomial(2*n,n)/(4^n)); \\ Altug Alkan, Sep 06 2018
    
  • SageMath
    def A001803(n): return numerator((n+1)*binomial(2*n+2,n+1)/2^(2*n+1))
    print([A001803(n) for n in range(31)]) # G. C. Greubel, Apr 27 2025

Formula

a(n) = (2*n + 1)! /(n!^2*2^A000120(n)) = (n + 1)*binomial(2*n+2,n+1)/2^(A000120(n)+1). - Ralf Stephan, Mar 10 2004
From Johannes W. Meijer, Jun 08 2009: (Start)
a(n) = numerator( (2*n+1)*binomial(2*n,n)/(4^n) ).
(1 - x)^(-3/2) = Sum_{n>=0} ((2*n+1)*binomial(2*n,n)/4^n)*x^n. (End)
Truncations of rational expressions like those given by the numerator or denominator operators are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n+1)$ / sigma(2*n+1) = A056040(2*n+1) / A060632(2*n+2). Simply said: This sequence gives the odd part of the swinging factorial at odd indices. - Peter Luschny, Aug 01 2009
a(n) = denominator(Pi*binomial(n, -1/2)). - Peter Luschny, Dec 06 2024

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

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Comments

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

References

  • 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).

Crossrefs

Programs

  • Magma
    [n^2*Binomial(2*n, n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014
    
  • Maple
    seq(n^2*binomial(2*n,n), n=0..50); # Robert Israel, Aug 07 2014
  • Mathematica
    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 *)
  • MuPAD
    combinat::catalan(n)*(n+1)*n^2 $ n = 0..36 // Zerinvary Lajos, Apr 17 2007
    
  • PARI
    my(x='x+O('x^100)); concat(0, Vec(x*(4*x+2)/((1-4*x)^(5/2)))) \\ Altug Alkan, Mar 21 2016
    
  • PARI
    a(n) = n^2*binomial(2*n, n); \\ Michel Marcus, Mar 21 2016
    
  • Sage
    [n^2*(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Mar 23 2022

Formula

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

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

Original entry on oeis.org

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

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Author

Peter Luschny, Mar 20 2015

Keywords

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.

Examples

			[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, ...
		

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).

Programs

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

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)

A000150 Number of dissections of an n-gon, rooted at an exterior edge, asymmetric with respect to that edge.

Original entry on oeis.org

0, 0, 1, 2, 7, 20, 66, 212, 715, 2424, 8398, 29372, 104006, 371384, 1337220, 4847208, 17678835, 64821680, 238819350, 883629164, 3282060210, 12233125112, 45741281820, 171529777432, 644952073662, 2430973096720, 9183676536076
Offset: 0

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Author

Keywords

Comments

Number of Dyck paths of length 2n having an odd number of peaks at even height. Example: a(3)=2 because we have UDU(UD)D and U(UD)DUD, where U=(1,1), D=(1,-1) and the peaks at even height are shown between parentheses. - Emeric Deutsch, Nov 13 2004
For n>=1, a(n) is the number of unordered binary trees with n internal nodes in which the left subtree is distinct from the right subtree. - Geoffrey Critzer, Feb 21 2013
Assuming offset -1 this is an analog of A275166: pairs of distinct Catalan numbers with index sum n. - R. J. Mathar, Jul 19 2016

References

  • S. J. Cyvin, J. Brunvoll, E. Brendsdal, B. N. Cyvin and E. K. Lloyd, Enumeration of polyene hydrocarbons: a complete mathematical solution, J. Chem. Inf. Comput. Sci., 35 (1995) 743-751
  • R. K. Guy, "Dissecting a polygon into triangles," Bull. Malayan Math. Soc., Vol. 5, pp. 57-60, 1958.
  • R. K. Guy, Dissecting a polygon into triangles, Research Paper #9, Math. Dept., Univ. Calgary, 1967.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 78, (3.5.26).
  • 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).
  • P. K. Stockmeyer, The charm bracelet problem and its applications, pp. 339-349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974.

Crossrefs

a(n) = T(2n+2, n), array T as in A051168, a count of Lyndon words.
Cf. A007595.
A diagonal of the square array described in A051168.

Programs

  • Mathematica
    nn=20;CoefficientList[Series[x/2(((1-(1-4x)^(1/2))/(2x))^2-(1-(1-4x^2)^(1/2))/(2x^2)),{x,0,nn}],x]  (* Geoffrey Critzer, Feb 21 2013 *)

Formula

Let c(x) = (1-sqrt(1-4*x))/(2*x) = g.f. for Catalan numbers (A000108), let d(x) = 1+x*c(x^2). Then g.f. is (c(x)-d(x))/2.
G.f.: (sqrt(1-4*z^2) - sqrt(1-4*z) - 2*z)/(4*z). - Emeric Deutsch, Nov 13 2004
With c(x) defined as above: g.f. = x*(c(x)^2/2 - c(x^2)/2). - Geoffrey Critzer, Feb 21 2013
a(n) = ( 2^(n-3)/sqrt(Pi) ) * ( 4*2^n*GAMMA(n+1/2)/GAMMA(n+2) + ((-1)^n - 1)*GAMMA(n/2)/GAMMA(n/2 + 3/2) ) for n>0. - Mark van Hoeij, Nov 11 2009
a(n) ~ 2^(2*n-1) / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 10 2014
a(2n) = A000108(2n) / 2; a(2n+1) = ( A000108(2n+1) - A000108(n) ) / 2. - John Bodeen, Jun 24 2015
D-finite with recurrence +n*(n+1)*(n-2)^2*a(n) -2*n*(2*n-5)*(n-1)^2*a(n-1) -4*n*(n-2)^3*a(n-2) +8*(2*n-5)*(n-3)*(n-1)^2*a(n-3)=0. - R. J. Mathar, Oct 28 2021

Extensions

Additional comments from Clark Kimberling

A001814 Coefficient of H_2 when expressing x^{2n} in terms of Hermite polynomials H_m.

Original entry on oeis.org

1, 12, 180, 3360, 75600, 1995840, 60540480, 2075673600, 79394515200, 3352212864000, 154872234316800, 7771770303897600, 420970891461120000, 24481076457277440000, 1521324036987955200000, 100610229646136770560000
Offset: 1

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Author

Keywords

Comments

a(n) = A126804(n)/2. - Zerinvary Lajos, Sep 21 2007
a(n) is the number of ways to partition a set of 2n elements into parts of size 2 and then multiply by the number n of parts. - Alain Goupil, Jul 27 2025

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 801.
  • 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).

Crossrefs

a(n) = A048854(n, 1) = A067147(2n, 2).
Cf. A001879.
Cf. A005430.

Programs

  • Magma
    [Factorial(2*n)/(2*Factorial(n-1)): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011
  • Maple
    with(combinat):for n from 1 to 16 do printf(`%d, `,n!/2*sum(binomial(2*n, n), k=1..n)) od: # Zerinvary Lajos, Mar 13 2007
    a:=n->sum((count(Permutation(n*2+2),size=n+1)),j=0..n)/2: seq(a(n), n=0..15); # Zerinvary Lajos, May 03 2007
    seq(1/2*mul((n+k), k=1..n), n=0..16); # Zerinvary Lajos, Sep 21 2007
  • Mathematica
    Table[(2*n)!/(2*(n-1)!),{n,1,20}] (* Vincenzo Librandi, Nov 22 2011 *)
  • MuPAD
    combinat::catalan(n)*binomial(n+1,2)*n! $ n = 1..16; // Zerinvary Lajos, Feb 15 2007
    

Formula

E.g.f.: x/(1 - 4*x)^(3/2). - corrected by Alain Goupil, Jul 28 2025
a(n) = (2*n)!/(2*(n-1)!).
(n!/2)*binomial(2*n,n)*n or n!/2*A005430. - Zerinvary Lajos, Jun 06 2006
Sum_{n>=0} a(n)*x^(2n)/(2n)! = (x^2/2)*exp(x^2). - Alain Goupil, Jul 28 2025

Extensions

More terms and new description from Christian G. Bower, Dec 18 2001

A005429 Apéry numbers: n^3*C(2n,n).

Original entry on oeis.org

0, 2, 48, 540, 4480, 31500, 199584, 1177176, 6589440, 35443980, 184756000, 938929992, 4672781568, 22850118200, 110079950400, 523521630000, 2462025277440, 11465007358860, 52926189069600, 242433164404200, 1102772230560000, 4984806175188840, 22404445765690560
Offset: 0

Views

Author

Keywords

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.3.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [Binomial(2*n,n)*n^3 : n in [0..30]]; // Wesley Ivan Hurt, Oct 21 2014
    
  • Mathematica
    Table[n^3 Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Apr 08 2012 *)
    CoefficientList[Series[(2*x*(2*x*(2*x+5)+1))/(1-4*x)^(7/2), {x,0,30}], x] (* Vincenzo Librandi, Oct 22 2014 *)
  • SageMath
    [n^3*binomial(2*n,n) for n in range(31)] # G. C. Greubel, Nov 19 2022

Formula

Sum_{n>=1} (-1)^(n+1) / a(n) = 2 * zeta(3) / 5.
G.f.: (2*x*(2*x*(2*x + 5) + 1))/(1-4*x)^(7/2). - Harvey P. Dale, Apr 08 2012
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*n^(5/2)/sqrt(Pi).
Sum_{n>=1} 1/a(n) = (1/2)*4F3(1,1,1,1; 3/2,2,2; 1/4) = A145438. (End)

Extensions

Entry revised by N. J. A. Sloane, Apr 06 2004
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