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A002105
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Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers.
(Formerly M3655 N1487)
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27
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1, 1, 4, 34, 496, 11056, 349504, 14873104, 819786496, 56814228736, 4835447317504, 495812444583424, 60283564499562496, 8575634961418940416, 1411083019275488149504, 265929039218907754399744
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| 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.
"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.
"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."
The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with g.f. sec^2(x sqrt 2). - Wenjin Woan (wjwoan(AT)hotmail.com), Oct 03 2007
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 (callan(AT)stat.wisc.edu), Nov 29 2007
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REFERENCES
| R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67.
Dominique Foata and Guo-Niu Han, Dimers and new q-tangent numbers, Preprint, 2008.
Dominique Foata and Guo-Niu Han, The dimer polynomial triangle, Preprint, 2008.
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).
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, Arxiv preprint arXiv:1107.2938, 2011.
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..100
Index entries for sequences related to Bernoulli numbers.
G. E. Andrews, J. Jimenez-Urroz, K. Ono, q-series identities and values of certain L-functions, Duke Math J. Volume 108, Number 3 (2001), 395-419. [From Peter Bala (pbala(AT)talktalk.net), Mar 24 2009]
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FORMULA
| E.g.f.: 2*log(sec(x/sqrt(2))) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
A000182(n)=2^(n-1)*a(n).
a(n) = 2^(n-1)/n * A110501(n). - D. E. Knuth, Jan 16 2007
a(n+1) = Sum_{k =0..n} A094665(n, k) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jun 11 2004
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 (pauldhanna(AT)juno.com), Oct 07 2005
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
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 (pbala(AT)talktalk.net), Mar 24 2009]
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 (pauldhanna(AT)juno.com), May 11 2010]
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]
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:
1, 3, 0, 0, 0, 0, 0,...
1, 3, 6, 0, 0,, 0, 0,...
1, 3, 6, 10, 0, 0, 0,...
1, 3, 6, 10, 15, 0, 0,... - Gary W. Adamson, Jul 14 2011
E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). [From Vladimir Kruchinin, Nov 18 2011]
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
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MATHEMATICA
| 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 (callan(AT)stat.wisc.edu), Nov 29 2007
a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* From Vladimir Reshetnikov, Jan 23 2011 *)
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PROG
| (PARI) a(n)=if(n<1, 0, ((-2)^n-(-8)^n)*bernfrac(2*n)/n)
(PARI) a(n)=if(n<0, 0, (2*n)!*polcoeff(-2*log(cos(x/quadgen(8)+O(x^(2*n+1)))), 2*n))
(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]))} (Hanna)
(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)} [From Paul D. Hanna (pauldhanna(AT)juno.com), May 11 2010]
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CROSSREFS
| Row sums of A008301.
A000364, A000464, A002439, A079144, A158690. [From Peter Bala (pbala(AT)talktalk.net), Mar 24 2009]
Sequence in context: A198976 A156325 A111169 * A198717 A198908 A081972
Adjacent sequences: A002102 A002103 A002104 * A002106 A002107 A002108
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KEYWORD
| easy,nonn,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Additional comments from Michael Somos, Jun 25, 2002
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