1,3

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

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.

T. D. Noe, Table of n, a(n) for n=1..100

Index entries for sequences related to Bernoulli numbers.

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

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

(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)

Row sums of A008301.

Sequence in context: A071213 A052629 A111169 this_sequence A081972 A134354 A030243

Adjacent sequences: A002102 A002103 A002104 this_sequence A002106 A002107 A002108

easy,nonn,nice

njas

Additional comments from Michael Somos, Jun 25, 2002