Search: id:A002105 Results 1-1 of 1 results found. %I A002105 M3655 N1487 %S A002105 1,1,4,34,496,11056,349504,14873104,819786496,56814228736, %T A002105 4835447317504,495812444583424,60283564499562496,8575634961418940416, %U A002105 1411083019275488149504,265929039218907754399744 %N A002105 Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers. %C A002105 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 A002105 "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 A002105 "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 A002105 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 %C A002105 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 %D A002105 R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386. %D A002105 M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466. %D A002105 C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374. %D A002105 H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67. %D A002105 Dominique Foata and Guo-Niu Han, Dimers and new q-tangent numbers, Preprint, 2008. %D A002105 Dominique Foata and Guo-Niu Han, The dimer polynomial triangle, Preprint, 2008. %D A002105 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). %D A002105 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). %H A002105 T. D. Noe, Table of n, a(n) for n=1..100 %H A002105 Index entries for sequences related to Bernoulli numbers. %H A002105 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] %F A002105 E.g.f.: 2*log(sec(x/sqrt(2))) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!. %F A002105 A000182(n)=2^(n-1)*a(n). %F A002105 a(n) = 2^(n-1)/n * A110501(n). - D. E. Knuth, Jan 16 2007 %F A002105 a(n+1) = Sum_{k =0..n} A094665(n, k) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jun 11 2004 %F A002105 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 %F A002105 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 A002105 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] %t A002105 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 %o A002105 (PARI) a(n)=if(n<1,0,((-2)^n-(-8)^n)*bernfrac(2*n)/n) %o A002105 (PARI) a(n)=if(n<0,0,(2*n)!*polcoeff(-2*log(cos(x/quadgen(8)+O(x^(2*n+1)))), 2*n)) %o A002105 (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) %Y A002105 Row sums of A008301. %Y A002105 A000364, A000464, A002439, A079144, A158690. [From Peter Bala (pbala(AT)talktalk.net), Mar 24 2009] %Y A002105 Sequence in context: A151919 A156325 A111169 this_sequence A081972 A158961 A134354 %Y A002105 Adjacent sequences: A002102 A002103 A002104 this_sequence A002106 A002107 A002108 %K A002105 easy,nonn,nice %O A002105 1,3 %A A002105 N. J. A. Sloane (njas(AT)research.att.com). %E A002105 Additional comments from Michael Somos, Jun 25, 2002 Search completed in 0.002 seconds