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.

A014307 Expansion of the e.g.f. sqrt(exp(x) / (2 - exp(x))).

Original entry on oeis.org

1, 1, 2, 7, 35, 226, 1787, 16717, 180560, 2211181, 30273047, 458186752, 7596317885, 136907048461, 2665084902482, 55726440112987, 1245661569161135, 29642264728189066, 748158516941653967, 19962900431638852297, 561472467839585937560, 16602088291822017588121
Offset: 0

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Comments

The Hankel transform of this sequence is A121835. - Philippe Deléham, Aug 31 2006
a(n) is the moment of order (n-1) for the discrete measure associated to the weight rho(j + 1/2) = 2^(j + 1/2)/(Pi*binomial(2*j + 1, j + 1/2)), with j integral. So we have a(n) = Sum_{j >= 0} (j + 1/2)^(n-1)*rho(j + 1/2). - Groux Roland, Jan 05 2009
Let f(n) = Sum_{j >= 1} j^n*2^j/binomial(2*j, j) = r_n*Pi/2 + s_n; sequence gives r_{n-1}. For example, f(0) through f(5) are [1 + (1/2)*Pi, 3 + Pi, 11 + (7/2)*Pi, 55 + (35/2)*Pi, 355 + 113*Pi, 2807 + (1787/2)*Pi]. For s_n, see A180875. - N. J. A. Sloane, following a suggestion from Herb Conn, Feb 08 2011
Ren gives seven combinatorial interpretations for this sequence. - Peter Bala, Feb 01 2013
Number of left-right arrangements of [n] [Crane, 2015]. - N. J. A. Sloane, Nov 21 2014
In Dyson et al. (2010-2011, 2013), we have S_n(2) = Sum_{j>=1} j^n*2^j/binomial(2*j, j) = A014307(n+1)*Pi/2 + A180875(n) for n >= 1 (and S_0(2) is not defined). This series was originally defined by Lehmer (1985). - Petros Hadjicostas, May 14 2020

Crossrefs

Row sums of triangle A156920 (row sums (n) = a(n+1)). - Johannes W. Meijer, Feb 20 2009

Programs

  • GAP
    Concatenation([1], List([1..20], n-> Sum([1..n], k-> Sum([k..n], m-> Stirling2(n,m)*Factorial(m)*Binomial(m-1,k-1)*Binomial(2*k-2,k-1)*(-2)^(1-k)/k )))); # G. C. Greubel, Oct 20 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( 1/Sqrt(2*Exp(-x)-1) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jun 30 2019
    
  • Maple
    seq(coeff(series(1/sqrt(2*exp(-x)-1), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Oct 20 2019
    a := n -> add((-1)^(n-k)*Stirling2(n,k)*doublefactorial(2*k-1), k=0..n):
    seq(a(n), n = 0..21); # Peter Luschny, Oct 19 2021
  • Mathematica
    a[n_] := Sum[ Sum[ StirlingS2[n, k]*k!*Binomial[k-1, m-1], {k, m, n}]/m*Binomial[2*m-2, m-1]*(-1)^(m-1)/2^(m-1), {m, 1, n}]; a[0]=1; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Sep 10 2012, after Vladimir Kruchinin *)
    CoefficientList[Series[Sqrt[E^x/(2-E^x)], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Jan 07 2014 *)
    A014307 = ConstantArray[0,20]; A014307[[1]]=1; Do[A014307[[n+1]] = 1 + Sum[(-1+Binomial[n+1,j])*A014307[[j]],{j,1,n}],{n,1,19}]; Flatten[{1,A014307}] (* Vaclav Kotesovec after Jon Perry, Jan 07 2014 *)
  • Maxima
    a(n):=sum(sum(stirling2(n,k)*k!*binomial(k-1,m-1),k,m,n)/(m)* binomial(2*m-2,m-1)*(-1)^(m-1)/2^(m-1),m,1,n); /* Vladimir Kruchinin, May 10 2011 */
    
  • PARI
    {a(n)=n!*polcoeff((exp(x +x*O(x^n))/(2-exp(x +x*O(x^n))))^(1/2),n)} \\ Paul D. Hanna, Jan 24 2008
    
  • PARI
    /* As solution to integral equation: */ {a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A^3*exp(-x+x*O(x^n))));n!*polcoeff(A,n)} \\ Paul D. Hanna, Jan 24 2008
    
  • Sage
    m = 20; T = taylor(1/sqrt(2*exp(-x)-1), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # G. C. Greubel, Jun 30 2019
    

Formula

a(n+1) = 1 + Sum_{j=1..n} (-1 + binomial(n+1,j))*a(j). - Jon Perry, Apr 25 2005, corrected by Vaclav Kotesovec, Jan 07 2014
The Hankel transform of this sequence is A121835. - Philippe Deléham, Aug 31 2006
E.g.f. A(x) satisfies A(x) = 1 + Integral_{t=0..x} (A(t)^3 * exp(-t)) dt. - Paul D. Hanna, Jan 24 2008 [Edited by Petros Hadjicostas, May 14 2020]
From Vladimir Kruchinin, May 10 2011: (Start)
a(n) = Sum_{m=1..n} (Sum_{k=m..n} Stirling2(n,k)*k!*binomial(k-1,m-1))*(1/m)*binomial(2*m-2,m-1)*(-1)^(m-1)/2^(m-1), n > 0.
E.g.f. B(x) = Integral_{t = 0..x} A(t) dt satisfies B'(x) = tan(B(x)) + sec(B(x)). (End)
From Peter Bala, Aug 25 2011: (Start)
It follows from Vladimir Kruchinin's formula above that
Sum_{n>=1} a(n-1)*x^n/n! = series reversion (Integral_{t = 0..x} 1/(sec(t)+tan(t)) dt) = series reversion (Integral_{t = 0..x} (sec(t)-tan(t)) dt) = series reversion (x - x^2/2! + x^3/3! - 2*x^4/4! + 5*x^5/5! - 16*x^6/6! + ...) = x + x^2/2! + 2*x^3/3! + 7*x^4/4! + 35*x^5/5! + 226*x^6/6! + ....
Let f(x) = sec(x) + tan(x). Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = (d/dx)(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n) = D^n[f](0). Compare with A190392.
(End)
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013
a(n) ~ sqrt(2) * n^n / (exp(n) * (log(2))^(n+1/2)). - Vaclav Kotesovec, Jan 07 2014
G.f.: R(0)/(1-x), where R(k) = 1 - x^2*(k+1)*(2*k+1)/(x^2*(k+1)*(2*k+1) - (3*x*k+x-1)*(3*x*k+4*x-1)/R(k+1)); (continued fraction). - Sergei N. Gladkovskii, Jan 30 2014
a(0) = 1 and a(n) = a(n-1) + Sum_{k=1..n-1} binomial(n-1, k-1)*a(k) for n > 0. - Seiichi Manyama, Oct 20 2019
a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n, k)*(2*k-1)!! (see Qi/Ward). - Peter Luschny, Oct 19 2021
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^k * (k/n - 2) * binomial(n,k) * a(n-k). - Seiichi Manyama, Nov 15 2023
Conjecture from Mikhail Kurkov, Jun 24 2025: (Start)
a(n) = R(n,0,2) where
R(0,0,m) = 1,
R(n,0,m) = Sum_{j=0..n-1} R(n-1,j,m),
R(n,k,m) = m*R(n,0,m) - Sum_{j=0..k-1} R(n-1,j,m) for 0 < k <= n.
More generally, R(n,0,m) gives expansion of the e.g.f. (exp(x) / (m - (m-1)*exp(x)))^(1/m) for any m>0. (End)

Extensions

Name edited by Petros Hadjicostas, May 14 2020