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%I M3655 N1487
%S 1,1,4,34,496,11056,349504,14873104,819786496,56814228736,
%T 4835447317504,495812444583424,60283564499562496,8575634961418940416,
%U 1411083019275488149504,265929039218907754399744
%N Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers.
%C Comments from R. L. Graham, Apr 25 2006 and Jun 08 2006: "This sequence also gives the number of ways of arranging 2n tokens in a row, with 2 copies of each token from 1 through n, such that the first token is a 1 and between every pair of tokens labeled i (i=1..n-1) there is exactly one taken labeled i+1.
%C "For example, for n=3, there are 4 possibilities: 123123, 121323, 132312 and 132132 and indeed a(3) = 4. This is the work of my Ph. D. student Nan Zang. See also A117513, A117514, A117515.
%C "Develin and Sullivant give another occurrence of this sequence and show that their numbers have the same generating function, although they were unable to find a 1-1-mapping between their problem and Poupard's."
%C The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with e.g.f. sec^2(x/sqrt 2). - _Wenjin Woan_, Oct 03 2007
%C a(n) = number of increasing full binary trees on vertex set [2n-1] with the left-largest property: the largest descendant of each non-leaf vertex occurs in its left subtree (Poupard). The first Mathematica recurrence below counts these trees by number 2k-1 of vertices in the left subtree of the root: the root is necessarily labeled 1 and n necessarily occurs in the left subtree and so there are Binomial[2n-3,2k-2] ways to choose the remaining labels for the left subtree. - _David Callan_, Nov 29 2007
%D R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
%D M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
%D D. Foata and G.-N. Han, Tree Calculus for Bivariable Difference Equations, 2012, http://www-irma.u-strasbg.fr/~foata/paper/pub120DeltaMatrices.pdf. - From _N. J. A. Sloane_, Feb 02 2013
%D Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011; http://repository.wit.ie/1693/1/AoifeThesis.pdf
%D E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411, 2013
%D C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67.
%H T. D. Noe, <a href="/A002105/b002105.txt">Table of n, a(n) for n=1..100</a>
%H G. E. Andrews, J. Jimenez-Urroz, K. Ono, <a href="http://www.math.wisc.edu/~ono/reprints/055.pdf">q-series identities and values of certain L-functions</a>, Duke Math J. Volume 108, Number 3 (2001), 395-419. [From _Peter Bala_, Mar 24 2009]
%H D. Foata, G-N. Han, <a href="http://dx.doi.org/10.1007/s11139-009-9194-9">The doubloon polynomial triangle</a>, Ram. J. 23 (2010), 107-126
%H Dominique Foata and Guo-Niu Han, <a href="http://dx.doi.org/10.1093/qmath/hap043">Doubloons and new q-tangent numbers</a>, Quart. J. Math. 62 (2) (2011) 417-432
%H A. Vieru, <a href="http://arxiv.org/abs/1107.2938">Agoh's conjecture: its proof, its generalizations, its analogues</a>, Arxiv preprint arXiv:1107.2938, 2011.
%H <a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers.</a>
%F E.g.f.: 2*log(sec(x/sqrt(2))) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
%F A000182(n)=2^(n-1)*a(n).
%F a(n) = 2^(n-1)/n * A110501(n). - D. E. Knuth, Jan 16 2007
%F a(n+1) = Sum_{k =0..n} A094665(n, k) . - _Philippe Deléham_, Jun 11 2004
%F O.g.f.: A(x) = x/(1-x/(1-3*x/(1-6*x/(1-10*x/(1-15*x/(... -n*(n+1)/2*x/(1 - ...))))))) (continued fraction). - _Paul D. Hanna_, Oct 07 2005
%F sqrt(2) tan( x/sqrt(2)) = Sum_(n>=0) (x^(2n+1)/(2n+1)!) a_n. - Dominique Foata and Guo-Niu Han, Oct 24 2008
%F Basic hypergeometric generating function: Sum {n = 0..inf} Product {k = 1..n} (1-exp(-2*k*t))/Product {k = 1..n} (1+exp(-2*k*t)) = 1 + t + 4*t^2/2! + 34*t^3/3! + 496*t^4/4! + ... [Andrews et al., Theorem 4]. For other sequences with generating functions of a similar type see A000364, A000464, A002439, A079144 and A158690. [From _Peter Bala_, Mar 24 2009]
%F E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*x) = Sum_{n>=0} a(n)*x^n/n!. [From _Paul D. Hanna_, May 11 2010]
%F a(n)=(-1)^(n+1)*sum(j!*stirling2(2*n+1,j)*2^(n+1-j)*(-1)^(j),j,1,2*n+1), n>=0. [From Vladimir Kruchinin, Aug 23 2010]
%F a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; where M = the infinite square production matrix:
%F 1, 3, 0, 0, 0, 0, 0,...
%F 1, 3, 6, 0, 0,, 0, 0,...
%F 1, 3, 6, 10, 0, 0, 0,...
%F 1, 3, 6, 10, 15, 0, 0,... - _Gary W. Adamson_, Jul 14 2011
%F E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). [From _Vladimir Kruchinin_, Nov 18 2011]
%F E.g.f.: For E(x)=sqrt(2)* tan( x/sqrt(2))=x/G(0) ; G(k)= 4*k + 1 - x^2/(8*k + 6 - x^2/G(k+1)); (from continued fraction Lambert's, 2-step). - _Sergei N. Gladkovskii_, Jan 14 2012
%F a(n) = (-1)^n*2^(n+1)*Li_{1-2*n}(-1). (See also the Mathematica prog. by Vladimir Reshetnikov.) - _Peter Luschny_, Jun 28 2012
%F G.f.: 1/G(0) where G(k) = 1 - x*( 4*k^2 + 4*k + 1 ) - x^2*(k+1)^2*( 4*k^2 + 8*k + 3)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Jan 14 2013
%F G.f.: 1/Q(0), where Q(k)= 1 - (k+1)*(k+2)/2*x/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, May 03 2013
%p S := proc(n, k) option remember; if k=0 then `if`(n=0, 1, 0) else S(n, k-1) + S(n-1, n-k) fi end: A002105 := n -> S(2*n-1, 2*n-1)/2^(n-1):
%p seq(A002105(i),i=1..16); # _Peter Luschny_, Jul 08 2012
%t u[1] = 1; u[n_]/;n>=2 := u[n] = Sum[Binomial[2n-3,2k-2]u[k]u[n-k],{k,n-1}]; Table[u[n],{n,8}] (* Poupard and also Develin and Sullivant, give a different recurrence that involves a symmetric sum: v[1] = 1; v[n_]/;n>=2 := v[n] = 1/2 Sum[Binomial[2n-2,2k-1]v[k]v[n-k],{k,n-1}] *) - _David Callan_, Nov 29 2007
%t a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* _Vladimir Reshetnikov_, Jan 23 2011 *)
%o (PARI) a(n)=if(n<1,0,((-2)^n-(-8)^n)*bernfrac(2*n)/n)
%o (PARI) a(n)=if(n<0,0,(2*n)!*polcoeff(-2*log(cos(x/quadgen(8)+O(x^(2*n+1)))),2*n))
%o (PARI) {a(n)=local(CF=1+x*O(x^n));if(n<1,return(0), for(k=1,n,CF=1/(1-(n-k+1)*(n-k+2)/2*x*CF));return(Vec(CF)[n]))} /* _Paul D. Hanna_ */
%o (PARI) {a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,tanh(k*X)));n!*polcoeff(Egf,n)} /* _Paul D. Hanna_, May 11 2010 */
%o (Sage) # Algorithm of L. Seidel (1877)
%o # n -> [a(1), ..., a(n)] for n >= 1.
%o def A002105_list(n) :
%o D = [0]*(n+2); D[1] = 1
%o R = []; z = 1/2; b = True
%o for i in(0..2*n-1) :
%o h = i//2 + 1
%o if b :
%o for k in range(h-1, 0, -1) : D[k] += D[k+1]
%o z *= 2
%o else :
%o for k in range(1, h+1, 1) : D[k] += D[k-1]
%o b = not b
%o if b : R.append(D[h]*z/h)
%o return R
%o A002105_list(16) # _Peter Luschny_, Jun 29 2012
%Y Row sums of A008301.
%Y Cf. A000364, A000464, A002439, A079144, A158690. [From _Peter Bala_, Mar 24 2009]
%Y Left edge of triangle A210108.
%K easy,nonn,nice
%O 1,3
%A _N. J. A. Sloane_.
%E Additional comments from _Michael Somos_, Jun 25, 2002
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