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%I M3655 N1487 #212 Sep 30 2023 09:20:59
%S 1,1,4,34,496,11056,349504,14873104,819786496,56814228736,
%T 4835447317504,495812444583424,60283564499562496,8575634961418940416,
%U 1411083019275488149504,265929039218907754399744,56906245479134057176170496,13722623393637762299131396096,3704005473270641755597685653504
%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 token 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
%C Number of bilabeled unordered increasing trees with 2n labels. - _Markus Kuba_, Nov 18 2014
%C Conjecture: taking the sequence modulo an integer k gives an eventually purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 10 begins [1, 1, 4, 4, 6, 6, 4, 4, 6, 6, 4, 4, 6, 6, ...] with an apparent period [4, 4, 6, 6] of length 4 = phi(10) beginning at a(3). - _Peter Bala_, May 08 2023
%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).
%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, and K. Ono, <a href="http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/055.pdf">q-series identities and values of certain L-functions</a>, Duke Math J. Volume 108, Number 3 (2001), 395-419.
%H R. C. Archibald, <a href="http://dx.doi.org/10.1090/S0025-5718-45-99088-0">Review of Terrill-Terrill paper</a>, Math. Comp., 1 (1945), pp. 385-386.
%H Peter Bala, <a href="/A002105/a002105.pdf">A continued fraction expansion related to A002105</a>
%H Paul Barry, <a href="https://arxiv.org/abs/1803.06408">Three Études on a sequence transformation pipeline</a>, arXiv:1803.06408 [math.CO], 2018.
%H R. B. Brent, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Brent/brent5.html">Generalizing Tuenter's Binomial Sums</a>, J. Int. Seq. 18 (2015) # 15.3.2.
%H M. P. Develin and S. P. Sullivant, <a href="http://dx.doi.org/10.1007/s00026-003-0196-9">Markov Bases of Binary Graph Models</a>, Annals of Combinatorics 7 (2003) 441-466.
%H Dominique Foata and Guo-Niu 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 Dominique Foata and Guo-Niu Han, <a href="http://www-irma.u-strasbg.fr/~foata/paper/pub120DeltaMatrices.pdf">Tree Calculus for Bivariable Difference Equations, 2012</a>. - From _N. J. A. Sloane_, Feb 02 2013
%H R. L. Graham and Nan Zang, <a href="http://dx.doi.org/10.1016/j.jcta.2007.06.003">Enumerating split-pair arrangements</a>, J. Combin. Theory Ser. A 115 (2008), no. 2, 293-303.
%H Aoife Hennessy, <a href="http://repository.wit.ie/1693/1/AoifeThesis.pdf">A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths</a>, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
%H Markus Kuba and Alois Panholzer, <a href="http://arxiv.org/abs/1411.4587">Combinatorial families of multilabelled increasing trees and hook-length formulas</a>, arXiv:1411.4587 [math.CO], 2014.
%H B. A. Kupershmidt, <a href="https://doi.org/10.1007/s10958-008-9027-1">On skew-symmetric and general deformations of Lax pseudodifferential operators</a>, J. Math. Sci. 151 (4) (2008) 3139, eq. (1.13).
%H Zhicong Lin, Shi-Mei Ma, David G. L. Wang, and Liuquan Wang, <a href="https://arxiv.org/abs/2011.02685">Positivity and divisibility of alternating descent polynomials</a>, arXiv:2011.02685 [math.CO], 2020.
%H E. Norton, <a href="http://arxiv.org/abs/1302.5411">Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions</a>, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
%H Ronald Orozco López, <a href="https://www.researchgate.net/publication/350397609_Solution_of_the_Differential_Equation_ykeay_Special_Values_of_Bell_Polynomials_and_ka-Autonomous_Coefficients">Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients</a>, Universidad de los Andes (Colombia 2021).
%H C. Poupard, <a href="http://dx.doi.org/10.1016/S0195-6698(89)80009-5">Deux proprietes des arbres binaires ordonnes stricts</a>, European J. Combin., 10 (1989), 369-374.
%H M. Safaryan, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.120.07.660">Problem 11725</a>, Amer. Math. Monthly, 120 (2013), 661.
%H H. M. Terrill and E. M. Terrill, <a href="https://ur.booksc.eu/ireader/2106189">Tables of numbers related to the tangent coefficients</a>, J. Franklin Inst., 239 (1945), 66-67.
%H H. M. Terrill and E. M. Terrill, <a href="/A001469/a001469.pdf">Tables of numbers related to the tangent coefficients</a>, J. Franklin Inst., 239 (1945), 64-67. [Annotated scanned copy]
%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 [math.NT], 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)!. - _Michael Somos_, Jun 22 2002
%F A000182(n) = 2^(n-1) * a(n). - _Michael Somos_, Jun 22 2002
%F a(n) = 2^(n-1)/n * A110501(n). - _Don 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} 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. - _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!. - _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. - _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)). - _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
%F G.f.: (1/G(0))/sqrt(x) - 1/sqrt(x), where G(k) = 1 - sqrt(x)*(2*k+1)/(1 + sqrt(x)*(2*k+1)/(1 + sqrt(x)*(k+1)/(1 - sqrt(x)*(k+1)/G(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Jul 07 2013
%F log(2) - 1/1 + 1/2 - 1/3 + ... + (-1)^n / n = (-1)^n / 2 * (1/n - 1 / (2*n^2) + 1 / (2*n^2)^2 - 4 / (2*n^2)^3 + ... + (-1)^k * a(k) / (2*n^2)^k + ...) asymptotic expansion. - _Michael Somos_, Sep 07 2013
%F G.f.: T(0), where T(k) = 1-x*(k+1)*(k+2)/(x*(k+1)*(k+2)-2/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Oct 24 2013
%F a(n) ~ 2^(3*n+2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - _Vaclav Kotesovec_, Nov 03 2014
%F From _Peter Bala_, Sep 10 2015: (Start)
%F The e.g.f. A(x) = sqrt(2)*tan(x/sqrt(2)) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 0, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0,1,0,1,0,4,0,34,0,496,...].
%F Note that the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence [1,1,1,2,5,16,61,272,...] = A000111. (End)
%F a(n) = polygamma(2*n-1, 1/2)*2^(2-n)/Pi^(2*n). - _Vladimir Reshetnikov_, Oct 18 2015
%F E.g.f.: sqrt(2)*tan(x/sqrt(2)) = Sum_{n>0} a(n) * x^(2*n-1) / (2*n-1)!. - _Michael Somos_, Mar 05 2017
%F From _Peter Bala_, May 05 2017: (Start)
%F Let B(x) = A(x)/x = 1 + x + 4*x^2 + 34*x^3 + ... denote the shifted o.g.f. Then B(x) = 1/(1 + 2*x - 3*x/(1 - x/(1 + 2*x - 10*x/(1 - 6*x/(1 + 2*x - 21*x/(1 - 15*x/(1 + 2*x - 36*x/(1 - 28*x/(1 + 2*x - ...))))))))), where the coefficient sequence [3, 1, 10, 6, 21, 15, 36, 28, ...] in the partial numerators of the continued fraction is obtained by swapping adjacent triangular numbers. Cf. A079144.
%F It follows (by means of an equivalence transformation) that the second binomial transform of B(x), with g.f. equal to 1/(1 - 2*x)*B(x/(1 - 2*x)), has the S-fraction representation 1/(1 - 3*x/(1 - x/(1 - 10*x/(1 - 6*x/(1 - 21*x/(1 - 15*x/(1 - 36*x/(1 - 28*x/(1 - ...))))))))). Compare with the S-fraction representation of the g.f. A(x) given above by Hanna, dated Oct 7 2005. (End)
%F The computation can be based on the triangular numbers, a(n) = T(n, n) where T(n, k) = A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, k-1) if k > 0. This is equivalent to _Paul D. Hanna_'s continued fraction 2005. - _Peter Luschny_, Sep 30 2023
%e G.f. = x + x^2 + 4*x^3 + 34*x^4 + 496*x^5 + 11056*x^6 + 349504*x^7 + ...
%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): seq(A002105(i),i=1..16); # _Peter Luschny_, Jul 08 2012
%p # The above function written as a formula: a(n) = A008281(2*n-1, 2*n-1)/2^(n-1).
%p # Alternatively, based on the triangular numbers A000217:
%p T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else
%p A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
%p a := n -> T(n, n): seq(a(n), n = 0..18); # _Peter Luschny_, Sep 30 2023
%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 *)
%t Table[(-1)^(n+1)*2^n*(2^(2n)-1)*BernoulliB[2n]/n,{n,1,20}] (* _Vaclav Kotesovec_, Nov 03 2014 *)
%t eulerCF[f_, len_] := Module[{g}, g[len-1]=1; g[k_]:=g[k]=1-f[k]/(f[k]-1/g[k+1]); CoefficientList[g[0] + O[x]^len, x]]; A002105List[len_] := eulerCF[(1/2) x (#+1) (#+2)&, len]; A002105List[19] (* _Peter Luschny_, Aug 08 2015 after _Sergei N. Gladkovskii_ *)
%t Table[PolyGamma[2n-1, 1/2] 2^(2-n)/Pi^(2n), {n, 1, 10}] (* _Vladimir Reshetnikov_, Oct 18 2015 *)
%t Table[EulerE[2n-1, 0] (-2)^n, {n, 1, 10}] (* _Vladimir Reshetnikov_, Oct 21 2015 *)
%o (PARI) {a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* _Michael Somos_, Jun 22 2002 */
%o (PARI) {a(n) = if( n<0, 0, (2*n)! * polcoeff( -2 * log( cos(x / quadgen(8) + O(x^(2*n + 1)))), 2*n))}; /* _Michael Somos_, Jul 17 2003 */
%o (PARI) {a(n) = if( n<0, 0, -(-2)^(n+1) * sum(i=1, 2*n, 2^-i * sum(j=1, i, (-1)^j * binomial( i-1, j-1) * j^(2*n - 1))))}; /* _Michael Somos_, Sep 07 2013 */
%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
%o (Python)
%o from sympy import bernoulli
%o def A002105(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<<n-1)//n) # _Chai Wah Wu_, Apr 14 2023
%Y Row sums of A008301.
%Y Left edge of triangle A210108.
%Y Cf. A000111, A000217, A000364, A000464, A002439, A008281, A079144, A158690.
%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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