%I #144 Nov 15 2023 08:03:19
%S 1,1,2,7,35,226,1787,16717,180560,2211181,30273047,458186752,
%T 7596317885,136907048461,2665084902482,55726440112987,
%U 1245661569161135,29642264728189066,748158516941653967,19962900431638852297,561472467839585937560,16602088291822017588121
%N Expansion of the e.g.f. sqrt(exp(x) / (2 - exp(x))).
%C The Hankel transform of this sequence is A121835. - _Philippe Deléham_, Aug 31 2006
%C 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
%C 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
%C Ren gives seven combinatorial interpretations for this sequence. - _Peter Bala_, Feb 01 2013
%C Number of left-right arrangements of [n] [Crane, 2015]. - _N. J. A. Sloane_, Nov 21 2014
%C 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
%H Seiichi Manyama, <a href="/A014307/b014307.txt">Table of n, a(n) for n = 0..424</a> (terms 0..100 from Vincenzo Librandi)
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Barry1/barry263.html">Generalized Stirling Numbers, Exponential Riordan Arrays, and Toda Chain Equations</a>, Journal of Integer Sequences, 17 (2014), #14.2.3.
%H Harry Crane, <a href="https://ajc.maths.uq.edu.au/pdf/61/ajc_v61_p057.pdf">Left-right arrangements, set partitions, and pattern avoidance</a>, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
%H D. Dominici, <a href="http://arxiv.org/abs/math/0501052">Nested derivatives: A simple method for computing series expansions of inverse functions</a>, arXiv:math/0501052 [math.CA], 2005.
%H F. J. Dyson, N. E. Frankel and M. L. Glasser, <a href="http://arxiv.org/abs/1009.4274">Lehmer's Interesting Series</a>, arXiv:1009.4274 [math-ph], 2010-2011. (See Table IV on p. 14.)
%H F. J. Dyson, N. E. Frankel and M. L. Glasser, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.120.02.116">Lehmer's interesting series</a>, Amer. Math. Monthly, 120 (2013), 116-130. (See Table 2.)
%H M. Klazar, <a href="http://dx.doi.org/10.1006/eujc.1995.0095">Twelve countings with rooted plane trees</a>, European Journal of Combinatorics 18 (1997), 195-210; Addendum, 18 (1997), 739-740.
%H D. H. Lehmer, <a href="https://www.jstor.org/stable/2322496">Interesting series involving the central binomial coefficient</a>, Amer. Math. Monthly, 92(7) (1985), 449-457.
%H Feng Qi and Mark Daniel Ward, <a href="https://arxiv.org/abs/2110.08576">Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function</a>, arXiv:2110.08576 [math.CO], 2021.
%H Q. Ren, <a href="http://arxiv.org/abs/1301.6327">Ordered partitions and drawings of rooted plane trees</a> arXiv:1301.6327 [math.CO], 2013-2014.
%H Andrew T. Wilson, <a href="https://arxiv.org/abs/1602.00068">Torus link homology and the nabla operator</a>, arXiv preprint arXiv:1606.00764 [cond-mat.str-el], 2016.
%F 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
%F The Hankel transform of this sequence is A121835. - _Philippe Deléham_, Aug 31 2006
%F 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]
%F From _Vladimir Kruchinin_, May 10 2011: (Start)
%F 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)).
%F E.g.f. B(x) = Integral_{t = 0..x} A(t) dt satisfies B'(x) = tan(B(x)) + sec(B(x)). (End)
%F From _Peter Bala_, Aug 25 2011: (Start)
%F It follows from _Vladimir Kruchinin_'s formula above that
%F 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! + ....
%F 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.
%F (End)
%F 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
%F a(n) ~ sqrt(2) * n^n / (exp(n) * (log(2))^(n+1/2)). - _Vaclav Kotesovec_, Jan 07 2014
%F 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
%F 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
%F a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n, k)*(2*k-1)!! (see Qi/Ward). - _Peter Luschny_, Oct 19 2021
%F 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
%p seq(coeff(series(1/sqrt(2*exp(-x)-1), x, n+1)*n!, x, n), n = 0..20); # _G. C. Greubel_, Oct 20 2019
%p a := n -> add((-1)^(n-k)*Stirling2(n,k)*doublefactorial(2*k-1), k=0..n):
%p seq(a(n), n = 0..21); # _Peter Luschny_, Oct 19 2021
%t 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_ *)
%t CoefficientList[Series[Sqrt[E^x/(2-E^x)], {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Jan 07 2014 *)
%t 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 *)
%o (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
%o (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
%o (Maxima)
%o 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 */
%o (Magma) m:=20; R<x>:=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
%o (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
%o (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
%Y Cf. A000110, A180875.
%Y Variants: A136727, A136728, A136729. A190392.
%Y Row sums of triangle A156920 (row sums (n) = a(n+1)). - _Johannes W. Meijer_, Feb 20 2009
%K nonn
%O 0,3
%A _N. J. A. Sloane_
%E Name edited by _Petros Hadjicostas_, May 14 2020