A172061 -id:A172061 - cp's OEIS frontend

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

Clark Kimberling

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

			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 + ...

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.

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).

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

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

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

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 *)

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

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

[(-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

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

A072547 Main diagonal of the array in which first column and row are filled alternatively with 1's or 0's and then T(i,j) = T(i-1,j) + T(i,j-1).

Original entry on oeis.org

1, 0, 2, 6, 22, 80, 296, 1106, 4166, 15792, 60172, 230252, 884236, 3406104, 13154948, 50922986, 197519942, 767502944, 2987013068, 11641557716, 45429853652, 177490745984, 694175171648, 2717578296116, 10648297329692, 41757352712480

Offset: 1

Benoit Cloitre, Aug 05 2002

nonn

A Catalan transform of A078008 under the mapping g(x)->g(xc(x)). - Paul Barry, Nov 13 2004
Number of positive terms in expansion of (x_1 + x_2 + ... + x_{n-1} - x_n)^n. - Sergio Falcon, Feb 08 2007
Hankel transform is A088138(n+1). - Paul Barry, Feb 17 2009
Without the beginning "1", we obtain the first diagonal over the principal diagonal of the array notified by B. Cloitre in A026641 and used by R. Choulet in A172025, and from A172061 to A172066. - Richard Choulet, Jan 25 2010
Also central terms of triangles A108561 and A112465. - Reinhard Zumkeller, Jan 03 2014
With offset 0 and for p prime, the p-th term is divisible by p. - F. Chapoton, Dec 03 2021

			The array begins:
  1 0 1 0 1..
  0 0 1 1 2..
  1 1 2 3 5..
  0 1 3 6 11..
so sequence begins : 1, 0, 2, 6, ...

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
David Anderson, E. S. Egge, M. Riehl, L. Ryan, R. Steinke, and Y. Vaughan, Pattern Avoiding Linear Extensions of Rectangular Posets, arXiv:1605.06825 [math.CO], 2016.
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015. [It is only a conjecture that this is the same sequence. It would be nice to have a proof.]
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Colin Defant, Proofs of Conjectures about Pattern-Avoiding Linear Extensions, arXiv:1905.02309 [math.CO], 2019.
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017)

Cf. A014300, A014301, A026641, A092785, A000957.

Cf. A026641, A172025, A172061, A172062, A172063, A172064, A172065, A172066. - Richard Choulet, Jan 25 2010

a072547 n = a108561 (2 * (n - 1)) (n - 1)
-- Reinhard Zumkeller, Jan 03 2014

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( x*(1 + Sqrt(1-4*x))/(Sqrt(1-4*x)*(3-Sqrt(1-4*x))) )); // G. C. Greubel, Feb 17 2019

taylor( (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-1),z=0,42); for n from -1 to 40 do a(n):=sum('(-1)^(p)*binomial(2n-p+1,1+n-p)',p=0..n+1): od:seq(a(n),n=-1..40):od; # Richard Choulet, Jan 25 2010

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x]) /(2*x))^(-1), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
a[n_] := Binomial[2 n - 2, n] Hypergeometric2F1[1, 2 - n, n + 1, 1/2] / 2 + (-2)^(1 - n); Table[a[n], {n, 1, 26}] (* Peter Luschny, Dec 03 2021 *)

a(n) = (-1)^n*sum(k=0, n, binomial(-n, k));
vector(100, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

a=(x*(1+sqrt(1-4*x))/(sqrt(1-4*x)*(3-sqrt(1-4*x)))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Feb 17 2019

If offset is 0, a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k). - Vladeta Jovovic, Feb 18 2003
G.f.: x*(1-x*C)/(1-2*x*C)/(1+x*C), where C = (1-sqrt(1-4*x))/(2*x) is g.f. for Catalan numbers (A000108). - Vladeta Jovovic, Feb 18 2003
a(n) = Sum_{j=0..floor((n-1)/2)} binomial(2*n-2*j-4, n-3). - Emeric Deutsch, Jan 28 2004
a(n) = A108561(2*(n-1),n-1). - Reinhard Zumkeller, Jun 10 2005
a(n) = (-1)^n*Sum_{k=0..n} binomial(-n,k) (offset 0). - Paul Barry, Feb 17 2009
Other form of the G.f: f(z) = (2/(3*sqrt(1-4*z) -1 +4*z))*((1 -sqrt(1-4*z))/(2*z))^(-1). - Richard Choulet, Jan 25 2010
D-finite with recurrence 2*(-n+1)*a(n) + (9*n-17)*a(n-1) + (-3*n+19)*a(n-2) + 2*(-2*n+7)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
From Peter Bala, Oct 01 2015: (Start)
a(n) = [x^n] ((1 - x)^2/(1 - 2*x))^n.
Exp( Sum_{n >= 1} a(n+1)*x^n/n ) = 1 + x^2 + 2*x^3 + 6*x^4 + 18*x^5 + ... is the o.g.f for A000957. (End)
a(n) = binomial(2*n-2, n)*hypergeom([1, 2-n], [n+1], 1/2) / 2 + (-2)^(1-n). - Peter Luschny, Dec 03 2021
a(n) = 2 * A014301(n-1) for n>=3. - Alois P. Heinz, Dec 27 2023

Corrected and extended by Vladeta Jovovic, Feb 17 2003

A091526 Coefficient of x^n in 1/((1+x)*(1-x)^(n-1)).

Original entry on oeis.org

1, -1, 1, 2, 9, 34, 130, 496, 1897, 7274, 27966, 107788, 416394, 1611908, 6251596, 24287212, 94499689, 368202778, 1436458486, 5610483532, 21936442894, 85852554748, 336300861436, 1318441228432, 5172792817834, 20309402206084

Offset: 0

Michael Somos, Jan 18 2004

sign

Number of positive terms in expansion of (x_1+x_2+...+x_{n-1}-x_n)^(n+1). - Sergio Falcon, Feb 08 2007

Without the beginning "1" and "-1", we obtain the second diagonal over the principal diagonal of the array notified by B. Cloitre in A026641 and used by R. Choulet in A172025, and from A172061 to A172066. - Richard Choulet, Jan 25 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A172025, A172061-A172066. - Richard Choulet, Jan 25 2010

Cf. A072547 (k=-1), A026641 (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).

k:=-2; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 18 2019

for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+2,2+n-p)',p=0..n+2): od:seq(a(n),n=0..40):od; taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-2),z=0,42); # Richard Choulet, Jan 25 2010

Table[Sum[Binomial[n+i-2, i]*(-1)^(n-i),{i,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 19 2014 *)
Table[(-1)^n 2^(1-n)+Binomial[-1+2 n,1+n] Hypergeometric2F1[1,2 n,2+n,-1],{n,0,20}] (* Vaclav Kotesovec, Apr 19 2014 *)
With[{k = -2}, CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1 - Sqrt[1-4*x])/(2*x))^k, {x, 0, 30}], x]] (* G. C. Greubel, Feb 18 2019 *)

a(n)=sum(i=0,n,binomial(n+i-2,i)*(-1)^(n-i));

k=-2; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 18 2019

k=-2; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019

From Richard Choulet, Jan 25 2010: (Start)
G.f: f such as: f(z)=(2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-2).
a(n) = Sum_{j=0..n+2} (-1)^j*binomial(2*n-j+2, 2+n-j). (End)
Recurrence: 2*n*(3*n-7)*a(n) = (21*n^2 - 61*n + 48)*a(n-1) + 2*(2*n-3)*(3*n-4)*a(n-2). - Vaclav Kotesovec, Apr 19 2014
a(n) ~ 2^(2*n-1)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014

A172025 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=3.

Original entry on oeis.org

1, 4, 16, 62, 239, 920, 3544, 13672, 52834, 204528, 793092, 3080226, 11980667, 46662704, 181971248, 710454896, 2776717742, 10863073784, 42537035408, 166704021596, 653827252022, 2566222449104, 10079023179536, 39611016586832

Offset: 0

Richard Choulet, Jan 23 2010

This sequence is the third diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows:
  1,  0,  1,  0,  1,  0,  1,  0,  1,  0,  1,  0,  1,  0,
  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,
  1,  1,  2,  2,  3,  3,  4,  4,  5,  5,  6,  6,  7,  7,
  1,  2,  4,  6,  9, 12, 16, 20, 25, 30,
  1,  3,  7, 13, 22, 34, 50, 70, 95.
The Maple programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+3 that are 1 step higher than the preceding peak. - David Scambler, Apr 22 2013

			a(4) = C(11,4) - C(10,3) + C(9,2) - C(8,1) + C(7,0) = 330 - 120 + 36 - 8 + 1 = 239.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=3; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 16 2019

a:= n-> add((-1)^(p)*binomial(2*n+3-p,n-p), p=0..n):
seq(a(n), n=0..30);
# second Maple program:
gf:= (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^3:
a:= n-> coeff(series(gf,z,n+10),z,n):
seq(a(n), n=0..30);

a[n_] := Binomial[2*n+3, n+3]*Hypergeometric2F1[1, -n, -3-2*n, -1]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Dec 17 2013 *)

k=3; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 16 2019

k=3; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 16 2019

G.f.: (2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k with k=3.
a(n) = Sum_{p=0..n} (-1)^(p)*binomial(2*n+k-p,n-p), with k=3.
a(n) ~ 2^(2*n+4)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+3)*a(n) + (-7*n^2 - 17*n - 8)*a(n-1) -2*(n+2)*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Feb 19 2016
a(n) = [x^n] 1/((1 - x^2)*(1 - x)^(n+3)). - Ilya Gutkovskiy, Oct 25 2017

A172062 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=5.

Original entry on oeis.org

1, 6, 29, 128, 541, 2232, 9076, 36568, 146446, 584082, 2322967, 9220544, 36548573, 144732176, 572756312, 2265577184, 8959034798, 35421613196, 140035644602, 553606049024, 2188652065586, 8653317051056, 34216118389384

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 5th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+5 that are 3 steps higher than the preceding peak. - David Scambler, Apr 22 2013
Apparently half the sum of all height differences between adjacent peaks in all Dyck paths of semilength n+3. - David Scambler, Apr 22 2013

			a(4) = C(13,4) - C(12,3) + C(11,2) - C(10,1) + C(9,0) = 13*11*5 - 20*11 + 55 - 10 + 1 = 541.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=5; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^5, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=5; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=5; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j*binomial(2*n+k-j, n-j), with k=5.
a(n) ~ 2^(2*n+6)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+5)*(3*n+7)*a(n) - (n+3)*(21*n^2+79*n+80)*a(n-1) - 2*(3*n+10)*(2*n+3)*(n+2)*a(n-2) = 0. - R. J. Mathar, Feb 19 2016

A172063 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=6.

Original entry on oeis.org

1, 7, 37, 174, 771, 3300, 13820, 57044, 233108, 945793, 3817351, 15347362, 61520899, 246052888, 982365976, 3916739872, 15599504614, 62076995998, 246866382826, 981218764540, 3898442536366, 15483778158792, 61482966826992

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 6th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

Apparently the number of peaks in all Dyck paths of semilength n+6 that are 4 steps higher than the preceding peak. - David Scambler, Apr 22 2013

			a(4) = C(14,4) - C(13,3) + C(12,2) - C(11,1) + C(10,0) = 7*13*11 - 26*11 + 66 - 11 + 1 = 771.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=6; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^6, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=6; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=6; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=6.
a(n) ~ 2^(2*n+7)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+6)*(n+3)*a(n) -(7*n^3+59*n^2+166*n+160)*a(n-1) -2*(2*n+5)*(n+4)*(n+2)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

Original entry on oeis.org

1, 8, 46, 230, 1068, 4744, 20476, 86662, 361711, 1494384, 6126818, 24972326, 101320712, 409609664, 1651162688, 6640469816, 26655382802, 106830738224, 427612715516, 1709790470780, 6830461107736, 27266848437608

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 7th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

Apparently the number of peaks in all Dyck paths of semilength n+7 that are 5 steps higher than the preceding peak. - David Scambler, Apr 22 2013

			a(4) = C(15,4) - C(14,3) + C(13,2) - C(12,1) + C(11,0) = 7*13*15 - 14*13*2 + 78 - 12 + 1 = 1068.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=7; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^7, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=7; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=7; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=7.
a(n) ~ 2^(2*n+8)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+7)*(3*n+11)*a(n) -(21*n^3+212*n^2+719*n+840)*a(n-1) -2*(2*n+5)*(n+3)*(3*n+14)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

Original entry on oeis.org

1, 9, 56, 297, 1444, 6656, 29618, 128603, 548591, 2309467, 9624964, 39799813, 163556776, 668796712, 2723729944, 11055878188, 44753742226, 180746332690, 728571706240, 2932018571370, 11783070278816, 47297147250204

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 8th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(16,4) - C(15,3) + C(14,2) - C(13,1) + C(12,0) = 20*91 - 35*13 + 91 - 13 + 1 = 1820 - 455 + 79 = 1444.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172066 (k=9), A172067 (k=10)

k:=8; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

a:= n-> add((-1)^(p)*binomial(2*n-p+8, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^8, z, n+20), z, n):
seq(a(n), n=0..40);

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^8, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=8; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=8; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j *binomial(2*n+k-j, n-j), with k=8.
a(n) ~ 2^(2*n+9)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+8)*(3*n+13)*a(n) -(21*n^3 + 247*n^2 + 980*n + 1344)*a(n-1) - 2*(n+3)*(3*n+16)*(2*n+7)*a(n-2) = 0. - R. J. Mathar, Feb 29 2016

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

Original entry on oeis.org

1, 10, 67, 376, 1912, 9142, 41941, 186880, 815083, 3498146, 14827487, 62236064, 259187048, 1072567256, 4415408372, 18098359424, 73915594466, 300958990724, 1222228100590, 4952609171080, 20030298812596, 80876902778482

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 9th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(17,4) - C(16,3) + C(15,2) - C(14,1) + C(13,0) = 17*4*5*7 - 16*5*7 + 105 - 14 + 1 = 5*7*(68-16) + 92 = 1912.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (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), A172067 (k=10).

k:=9; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^9, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=9; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=9; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j,n-j), with k=9.
a(n) ~ 2^(2*n+10)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+9)*(n+5)*a(n) -(7*n^3+94*n^2+427*n+672)*a(n-1) -2*(2*n+7)*(n+6)*(n+4)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.

Original entry on oeis.org

1, 11, 79, 468, 2486, 12323, 58277, 266492, 1188679, 5202523, 22436251, 95630272, 403770544, 1691678428, 7042481236, 29161852240, 120212658034, 493656394350, 2020590599710, 8247228533780, 33579755528278, 136434358356201

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 10th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(18,4) - C(17,3) + C(16,2) - C(15,1) + C(14,0) = 60*51 - 680 + 120 - 15 + 1 = 2486.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (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), this sequence (k=10).

k:=10; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 27 2019

a:= n-> add((-1)^(p)*binomial(2*n-p+10, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^10, z, n+20), z, n):
seq(a(n), n=0..40);

With[{k=10}, CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^k, {x, 0, 30}], x]] (* G. C. Greubel, Feb 27 2019 *)

k=10; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 27 2019

k=10; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 27 2019

a(n) = Sum_{j=0..n} (-1)^j*binomial(2*n+k-j,n-j), with k=10.

Conjecture: 2*n*(n+10)*(3*n+17)*a(n) - (21*n^3 + 317*n^2 + 1622*n + 2880)*a(n-1) - 2*(3*n+20)*(n+4)*(2*n+9)*a(n-2) = 0. - R. J. Mathar, Feb 21 2016

cp's OEIS Frontend

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

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A072547 Main diagonal of the array in which first column and row are filled alternatively with 1's or 0's and then T(i,j) = T(i-1,j) + T(i,j-1).

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A091526 Coefficient of x^n in 1/((1+x)*(1-x)^(n-1)).

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A172025 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=3.

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A172062 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=5.

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A172025 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=3.

A172062 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=5.

A172063 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=6.

A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.