A026641 Number of nodes of even outdegree (including leaves) in all ordered trees with n edges.

1, 1, 4, 13, 46, 166, 610, 2269, 8518, 32206, 122464, 467842, 1794196, 6903352, 26635774, 103020253, 399300166, 1550554582, 6031074184, 23493410758, 91638191236, 357874310212, 1399137067684, 5475504511858, 21447950506396

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Number of lattice paths from (0,0) to (n,n) using steps (1,0),(0,2),(1,1). - Joerg Arndt, Jun 30 2011

From Emeric Deutsch, Jan 25 2004: (Start)

Let B = 1/sqrt(1-4*z) = g.f. for central binomial coeffs (A000984); F = (1-sqrt(1-4*z))/(z*(3-sqrt(1-4*z))) = g.f. for (A000957).

B = 1 + 2*z + 6*z^2 + 20*z^3 + ... gives the number of nodes in all ordered trees with 0,1,2,3,... edges. On p. 288 of the Deutsch-Shapiro paper one finds that z*B*F = z + 2*z^2 + 7*z^3 + 24*z^4 + ... gives the number of nodes of odd outdegree in all ordered trees with 1,2,3,... edges (cf. A014300).

Consequently, B - z*B*F = 2/(3*sqrt(1-4*z)-1+4*z) = 1 + z + 4*z^2 + 13*z^3 + 46*z^4 + ... gives the total number of nodes of even degree in all ordered trees with 0,1,2,3,4,... edges. (End)

Main diagonal of the following array: first column is filled with 1's, first row is filled alternatively with 1's or 0's: m(i,j) = m(i-1,j) + m(i,j-1): 1 0 1 0 1 ... / 1 1 2 2 3 ... / 1 2 4 6 9 ... / 1 3 7 13 22 ... / 1 4 11 24 46 ... - Benoit Cloitre, Aug 05 2002

The Hankel transform of [1,1,4,13,46,166,610,2269,...] is 3^n. - Philippe Deléham, Mar 08 2007

Second binomial transform of A127361. - Philippe Deléham, Mar 14 2007

Starting with offset 1, generated from iterates of M * [1,1,1,...]; where M = a tridiagonal matrix with (0,2,2,2,...) in the main diagonal and (1,1,1,...) in the super and subdiagonals. - Gary W. Adamson, Jan 04 2009

Equals left border of triangle A158815. - Gary W. Adamson, Mar 27 2009

Equals the INVERTi transform of A101850: (1, 2, 7, 26, 100, ...). - Gary W. Adamson, Jan 10 2012

Diagonal of rational function 1/(1 - (x + x*y + y^2)). - Gheorghe Coserea, Aug 06 2018

Let A(i, j) denote the infinite array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i times to the function (-1)^(n+1) for n > 0. Then A(n, n) equals a(n-1) for all n > 0. - John M. Campbell, Jan 20 2019

These numbers have the same parity as the Catalan numbers A000108; that is, a(n) is odd if and only if n = 2^k - 1 for some nonnegative integer k. It appears that if a(n) is odd then a(n) == 1 (mod 4). - Peter Bala, Feb 07 2024

The number a(n)/(n+1) is the coefficient of x^(n+1) in log(1+(1-sqrt(1-4*x))/2), the generating series of the Sabinin operad. - F. Chapoton, Mar 14 2024

Examples

			From _Joerg Arndt_, Jul 01 2011: (Start)
The triangle of number of lattice paths from (0,0) to (n,k) using steps (1,0),(0,2),(1,1) begins
  1;
  1, 1;
  1, 2,  4;
  1, 3,  7, 13;
  1, 4, 11, 24,  46;
  1, 5, 16, 40,  86, 166;
  1, 6, 22, 62, 148, 314,  610;
  1, 7, 29, 91, 239, 553, 1163, 2269;
This sequence is the diagonal. (End)
G.f. = 1 + x + 4*x^2 + 13*x^3 + 46*x^4 + 166*x^5 + 610*x^6 + 2269*x^7 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
• Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
• Emeric Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
• Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1) = 1, arXiv:1909.11222 [math.NT], 2019.
• Filippo Disanto, Andrea Frosini and Simone Rinaldi, Renzo Pinzani, The Combinatorics of Convex Permutominoes, Southeast Asian Bulletin of Mathematics (2008) 32: 883-912.

Crossrefs

Cf. A101850, A120588, A158815.

Cf. A091526 (k=-2), A072547 (k=-1), this sequence (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

GAP

List([0..25],n->(-1)^n*Sum([0..n],k->(-1)^k*Binomial(n+k,k))); # Muniru A Asiru, Aug 06 2018

Magma

[(-1)^n*(&+[(-1)^k*Binomial(n+k, k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 12 2019

Maple

seq(add((binomial(k+n, n-k)*binomial(n-k, k)),k=0..floor(n/2)),n=0..30);
# From Richard Choulet, Jan 22 2010: (Start)
a:= n -> add(binomial(2*n-k, k)*binomial(k, n-k), k=floor(n/2)..n):
a:= n -> `if`(n<2, 1, (3/(2))*binomial(2*n-1, n-1)-(1/2)*a(n-1)):
a:= n -> (-1/2)^(n+2)+(2/3)*add(4^(n-k)*(binomial(2*k, k)*(1/(1-2*k))
        *(1-(-1/8)^(n-k+1))), k=0..n):
a:= n -> (-1/2)^(n+2)+(3/4)*add(((-1/2)^(n-k))*(binomial(2*k, k)), k=0..n):
seq(a(n), n=0..30); # (End)
gf := log(1 + (1 - sqrt(1 - 4*x))/2) / x: ser := series(gf, x, 30):
seq((n + 1)*coeff(ser, x, n), n = 0..24);  # Peter Luschny, Mar 16 2024

Mathematica

f[n_]:= Sum[ Binomial[n+k, k]*Cos[Pi*(n+k)], {k, 0, n}]; Array[f, 25, 0] (* Robert G. Wilson v, Apr 02 2012 *)
CoefficientList[Series[2/(3*Sqrt[1-4*x]-1+4*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
a[ n_]:= SeriesCoefficient[ D[ Log[1+(1-Sqrt[1-4x])/2], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)

PARI

a(n)=(-1)^n*sum(k=0,n,(-1)^k*binomial(n+k,k))

PARI

/* same as in A092566 but use */
steps=[[1,0], [0,2], [1,1]]; /* Joerg Arndt, Jun 30 2011 */

Sage

[(-1)^n*sum((-1)^k*binomial(n+k, k) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Feb 12 2019

Formula

G.f. is logarithmic derivative of the generating function for the Catalan numbers A000108. So this sequence might be called the "log-Catalan" numbers. - Murray R. Bremner, Jan 25 2004

a(n) = Sum_{k=0..floor(n/2)} binomial(k+n, n-k)*binomial(n-k, k). - Detlef Pauly (dettodet(AT)yahoo.de), Nov 15 2001

G.f.: 2/(3*sqrt(1-4*z)-1+4*z). - Emeric Deutsch, Jul 09 2002

a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*C(n+k, k). - Benoit Cloitre, Aug 20 2002

a(n) = Sum_{j=0..floor(n/2)} binomial(2*n-2*j-1, n-1). - Emeric Deutsch, Jan 28 2004

From Paul Barry, Dec 18 2004: (Start)

A Catalan transform of the Jacobsthal numbers A001045(n+1) under the mapping G(x)-> G(xc(x)), c(x) the g.f. of A000108. The inverse mapping is H(x)->H(x(1-x)).

a(n) = Sum_{k=0..n} (k/(2*n-k))*binomial(2*n-k, n-k)*A001045(k+1). (End)

a(n) = Sum_{k=0..n} binomial(2*n-k, k)*binomial(k, n-k). - Paul Barry, Jul 25 2005

a(n) = Sum_{k=0..n-1} A126093(n,k). - Philippe Deléham, Mar 08 2007

a(n) = (-1/2)^(n+2) + (2/3)*Sum_{k=0..n} ( (4^n-k)*binomial(2*k,k)*(1/(1-2*k))*(1-(-1/8)^(n-k+1)) ). - Yalcin Aktar, Jul 06 2007

a(n) = (-1/2)^(n+2) + (3/4)*Sum_{k=0..n} (-1/2)^(n-k)*binomial(2*k,k). - Yalcin Aktar, Jul 06 2007

From Richard Choulet, Jan 22 2010: (Start)

a(n) = (3*binomial(2*n-1,n-1) - d(n-1))/2, where d(n) = Sum_{k=floor(n/2)..n} binomial(2*n-k, k)*binomial(k, n-k).

a(n) = a(n-1) + (3/2)*Sum_{k=2..n} (1/(2*k-1))*binomial(2*k,k)*a(n-k).

a(n) = (2/3)*binomial(2*n,n) + (2/9)*((-2)^n/n!)*Sum_{k>=0} ( Product_{p=0..n-1} (k-2*p) /3^k).

a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n-k,n).

a(n) = ( Sum_{k=0..n} (1/2)^(n-k+1)*binomial(n+k,k) )^2*(-1/2)^(n+2). (End)

From Gary W. Adamson, Nov 22 2011: (Start)

a(n) is the upper left term of M^n, M = an infinite square production matrix as follows:

1, 3, 0, 0, 0, ...

1, 1, 1, 0, 0, ...

1, 1, 1, 1, 0, ...

1, 1, 1, 1, 1, ...

...

Also, a(n+1) is the sum of top row terms of M^n; e.g. top row of M^3 = (13, 21, 9, 3), sum = 46 = a(4), a(3) = 13. (End)

D-finite with recurrence: 2n*a(n) + (4-7n)*a(n-1) + 2*(1-2n)*a(n-2) = 0. - R. J. Mathar, Dec 17 2011 [The recurrence is proved with the Wilf-Zeilberger (WZ) method applied to Sum_{k=0..floor(n/2)} binomial(k+n, n-k)*binomial(n-k, k). - T. Amdeberhan, Jul 23 2012]

a(n) = A035317(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jul 19 2012

a(n) ~ 2^(2*n+1) / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014

a(n) = binomial(2*n,n)*hypergeom([1, -n], [-2*n], -1). - Peter Luschny, May 22 2014

G.f. is the derivative of the logarithm of the g.f. for A120588. - Michael Somos, May 18 2015

a(n) = [x^n] 1/((1 - x^2)*(1 - x)^n). - Ilya Gutkovskiy, Oct 25 2017

From Peter Bala, Feb 25 2019: (Start)

a(n) = Sum_{k = 0..n} binomial(2*n + 1, n + k + 1)*(-2)^k.

a(n-1) = (1/2)*binomial(2*n,n)*( 1 - 2*(n-1)/(n+1) + 4*(n-1)*(n-2)/((n+1)*(n+2)) - 8*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...) = (1/2)*binomial(2*n,n)*hypergeom([1 - n, 1], [n + 1], 2). (End)

a(0)=1, a(1)=1, and a(n) = (2 - 1/n)*a(n-2) + (7/2 - 2/n)*a(n-1) for n > 1. - Reginald Robson, Nov 01 2022