A005043 -id:A005043 - cp's OEIS frontend

A358451 Inverse Euler transform of the Riordan numbers, (A005043).

1, 0, 1, 1, 2, 5, 11, 28, 68, 174, 445, 1166, 3068, 8190, 21994, 59585, 162360, 445145, 1226376, 3394654, 9434260, 26317865, 73661588, 206809307, 582255448, 1643536725, 4650250254, 13186484316, 37468566744, 106666821221, 304200399505, 868977304140, 2486163857424

Offset: 0

Peter Luschny, Nov 20 2022

nonn

OEIS Wiki, Euler transform

Cf. A005043.

EulerInvTransform := proc(f) local c, b;
    c := proc(n) option remember;
    ifelse(n = 0, f(0), f(n) - b(n, n-1)) end:
    b := proc(n, k) option remember;
    if n = 0 then return 1 elif k < 1 then return 0 fi;
    add(binomial(c(k) + j - 1, j)*b(n-k*j, k-1), j=0..n/k) end:
c end:
a := EulerInvTransform(A005043): seq(a(n), n = 0..32);

EulerInvTransform[seq_List] := Module[{final = {}}, Do[AppendTo[final, i*seq[[i]] - Sum[final[[d]]*seq[[i-d]], {d, i-1}]], {i, Length[seq]}]; Table[Sum[MoebiusMu[i/d]*final[[d]], {d, Divisors[i]}]/i, {i, Length[seq] }]];
A005043[n_] := A005043[n] = If[n <= 1, 1-n, (n-1)*(2*A005043[n-1] + 3*A005043[n-2])/(n+1)];
Join[{1}, EulerInvTransform[Array[A005043, 32]]] (* Jean-François Alcover, Jun 15 2024 *)

from typing import Callable
from functools import cache
from math import comb
# Define 'binomial' for compatibility with Maple.
def binomial(n: int, k: int) -> int:
    if 0 <= k <= n: return comb(n, k)
    if k <= n <  0: return comb(-k-1, n-k)*(-1)**(n-k)
    if n <  0 <= k: return comb(-n+k-1, k)*(-1)**k
    return 0
def EulerInvTransform(f: Callable) -> Callable:
    @cache
    def h(n: int, k: int) -> int:
        if n == 0: return 1
        if k <  1: return 0
        return sum(binomial(b(k)+j-1, j) * h(n-k*j, k-1)
               for j in range(1 + n // k))
    @cache
    def b(n: int) -> int:
        if n == 0: return f(0)
        return f(n) - h(n, n - 1)
    return b
a = EulerInvTransform(A005043)
print([a(n) for n in range(33)])

z = PowerSeriesRing(ZZ, 'z').gen().O(33)
g = 1 + z + sqrt(1 - 2*z - 3*z**2)
f = -z * g.derivative() / g
print([1] + [sum(moebius(n // d) * f[d]
      for d in divisors(n)) // n for n in range(1, 33)])

a(n) ~ 3^(n + 1/2) / (4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 15 2024

A126930 Inverse binomial transform of A005043.

Original entry on oeis.org

1, -1, 2, -3, 6, -10, 20, -35, 70, -126, 252, -462, 924, -1716, 3432, -6435, 12870, -24310, 48620, -92378, 184756, -352716, 705432, -1352078, 2704156, -5200300, 10400600, -20058300, 40116600, -77558760, 155117520, -300540195, 601080390, -1166803110

Offset: 0

Philippe Deléham, Mar 17 2007

Successive binomial transforms are A005043, A000108, A007317, A064613, A104455. Hankel transform is A000012.
Moment sequence of the trace of the square of a random matrix in USp(2)=SU(2). If X=tr(A^2) is a random variable (a distributed with Haar measure) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
From Tom Copeland, Nov 08 2014: (Start)
This array is one of a family of Catalan arrays related by compositions of the special fractional linear (Mobius) transformation P(x,t) = x/(1-t*x); its inverse Pinv(x,t) = P(x,-t); an o.g.f. of the Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2; and its inverse Cinv(x) = x*(1-x). The Motzkin sums, or Riordan numbers, A005043 are generated by Mot(x)=C[P(x,1)]. One could, of course, choose the Riordan numbers as the parent sequence.
O.g.f.: G(x) = C[P[P(x,1),1]1] = C[P(x,2)] = (1-sqrt(1-4*x/(1+2*x)))/2 = x - x^2 + 2 x^3 - ... = Mot[P(x,1)].
Ginv(x) =  Pinv[Cinv(x),2] = P[Cinv(x),-2] = x(1-x)/[1-2x(1-x)] = (x-x^2)/[1-2(x-x^2)] = x*A146559(x).
Cf. A091867 and A210736 for an unsigned version with a leading 1. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 100.

Cf. A126120, A126869.

Cf. A000108, A005043, A146559, A091867, A210736.

egf := BesselI(0,2*x) - BesselI(1,2*x):
seq(n!*coeff(series(egf,x,34),x,n),n=0..33); # Peter Luschny, Dec 17 2014

CoefficientList[Series[(1 + 2 x - Sqrt[1 - 4 x^2])/(2 x (1 + 2 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)
Table[2^n Hypergeometric2F1[3/2, -n, 2, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 02 2015 *)

x='x+O('x^50); Vec((1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x))) \\ Altug Alkan, Nov 03 2015

a(n) = (-1)^n*C(n, floor(n/2)) = (-1)^n*A001405(n).
a(2*n) = A000984(n), a(2*n+1) = -A001700(n).
a(n) = (1/Pi)*Integral_{t=0..Pi}(2cos(2t))^n*2sin^2(t) dt. - Andrew V. Sutherland, Feb 29 2008, Mar 09 2008
a(n) = (-2)^n + Sum_{k=0..n-1} a(k)*a(n-1-k), a(0)=1. - Philippe Deléham, Dec 12 2009
G.f.: (1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x)). - Philippe Deléham, Mar 01 2013
O.g.f.: (1 + x*c(x^2))/(1 + 2*x), with the o.g.f. c(x) for the Catalan numbers A000108. From the o.g.f. of the Riordan type Catalan triangle A053121. This is the rewritten g.f. given in the previous formula. This is G(-x) with the o.g.f. G(x) of A001405. - Wolfdieter Lang, Sep 22 2013
D-finite with recurrence (n+1)*a(n) +2*a(n-1) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Dec 04 2013
Recurrence (an alternative): (n+1)*a(n) = 8*(n-2)*a(n-3) + 4*(n-2)*a(n-2) + 2*(-n-1)*a(n-1), n>=3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) ~ (-1)^n*2^(n+1/2)/sqrt(n*Pi). - Fung Lam, Mar 22 2014
E.g.f.: BesselI(0,2*x) - BesselI(1,2*x). - Peter Luschny, Dec 17 2014
a(n) = 2^n*hypergeom([3/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
G.f. A(x) satisfies: A(x) = 1/(1 + 2*x) + x*A(x)^2. - Ilya Gutkovskiy, Jul 10 2020

A099251 Bisection of Motzkin sums (A005043).

Original entry on oeis.org

1, 1, 3, 15, 91, 603, 4213, 30537, 227475, 1730787, 13393689, 105089229, 834086421, 6684761125, 54022715451, 439742222071, 3602118427251, 29671013856627, 245613376802185, 2042162142208813, 17047255430494497, 142816973618414817

Offset: 0

N. J. A. Sloane, Nov 16 2004

The Kn4 triangle sums of A175136 lead to the sequence given above (n >= 1). For the definition of the Kn4 and other triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Equals the expected value of trace(O)^(2n), where O is a 3 X 3 orthogonal matrix randomly selected according to Haar measure (see MathOverflow link). - Nathaniel Johnston, Sep 05 2014
From Petros Hadjicostas, Jul 23 2020: (Start)
In Smith (1985), we apparently have a(n) = P(2*n), where P(n) is the number of linearly independent three-dimensional n-th order isotropic tensors. In the paper, he refers to Smith (1968) for more details. It is not clear why he does not list the values of P(2*n+1). See also the 1978 letter of D. L. Andrews to N. J. A. Sloane.
Eric Weisstein gives some details on how the material in Smith (1968) about isotropic tensors is related to Motzkin sums. (End)

G. F. Smith, On isotropic tensors and rotation tensors of dimension m and order n, Tensor (N.S.), Vol. 19 (1968), 79-88 (MR0224008).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. L. Andrews, Letter to N. J. A. Sloane, Apr 10 1978.
Georgia Benkart and A. Elduque, Cross products, invariants, and centralizers, arXiv:1606.07588 [math.RT], 2016.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
MathOverflow, Moments of the trace of orthogonal matrices.
G. F. Smith, Lectures on constitutive expressions, Mathematical models and methods in mechanics, pp. 645-678, Banach Center Publ., 15, PWN, Warsaw, 1985 (MR0874855). See p. 653.
Eric Weisstein's World of Mathematics, Isotropic tensor.

Cf. A005043, A099252, A246860, A247304.

G := (1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x)): Gser := series(G,x=0,60):
1, seq(coeff(Gser, x^(2*n)), n=1..25); # Emeric Deutsch
a := n -> hypergeom([1/2, -2*n], [2], 4):
seq(simplify(a(n)), n=0..21); # Peter Luschny, Jul 25 2020

Take[CoefficientList[Series[(1 + x - Sqrt[1 - 2 * x - 3 * x^2])/(2 * x * (1 + x)), {x, 0, 60}], x], {1, -1, 2}] (* Vaclav Kotesovec, Oct 17 2012 *)

a(n):=sum(binomial(2*j,j)*(-1)^(j)*binomial(2*n+1,j+1),j,0,2*n+1)/(2*n+1); /*Vladimir Kruchinin, Apr 02 2017*/

x='x+O('x^66); v=Vec((1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x))); vector(#v\2,n,v[2*n-1]) \\ Joerg Arndt, May 12 2013

Recurrence: n*(2*n + 1)*a(n) = (2*n - 1)*(13*n - 10)*a(n-1) - 3*(26*n^2 - 87*n + 76)*a(n-2) + 27*(n - 2)*(2*n - 5)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ 3^(2*n + 3/2)/(16*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
Conjecture: a(n) = (2/Pi)*Integral_{t=0..1} sqrt((1 - t)/t)*(1 - 8*t + 16*t^2)^n. - Benedict W. J. Irwin, Oct 05 2016
a(n) = Sum_{j=0..2*n+1} (C(2*j,j)*(-1)^(j)*C(2*n+1,j+1))/(2*n+1). - Vladimir Kruchinin, Apr 02 2017
a(n) = hypergeom([1/2, -2*n], [2], 4). - Peter Luschny, Jul 25 2020

More terms from Emeric Deutsch, Nov 18 2004

A358376 Numbers k such that the k-th standard ordered rooted tree is lone-child-avoiding (counted by A005043).

Original entry on oeis.org

1, 4, 8, 16, 18, 25, 32, 36, 50, 57, 64, 72, 100, 114, 121, 128, 137, 144, 200, 228, 242, 249, 256, 258, 274, 281, 288, 385, 393, 400, 456, 484, 498, 505, 512, 516, 548, 562, 569, 576, 770, 786, 793, 800, 897, 905, 912, 968, 996, 1010, 1017, 1024, 1032, 1096

Offset: 1

Gus Wiseman, Nov 14 2022

nonn

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

			The initial terms and their corresponding trees:
    1: o
    4: (oo)
    8: (ooo)
   16: (oooo)
   18: ((oo)o)
   25: (o(oo))
   32: (ooooo)
   36: ((oo)oo)
   50: (o(oo)o)
   57: (oo(oo))
   64: (oooooo)
   72: ((oo)ooo)
  100: (o(oo)oo)
  114: (oo(oo)o)
  121: (ooo(oo))
  128: (ooooooo)
  137: ((oo)(oo))
  144: ((oo)oooo)
  200: (o(oo)ooo)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

These trees are counted by A005043.
The series-reduced case appears to be counted by A284778.
The unordered version is A291636, counted by A001678.
A000081 counts unlabeled rooted trees, ranked by A358378.
A358371 and A358372 count leaves and nodes in standard ordered rooted trees.
A358374 ranks ordered identity trees, counted by A032027.
A358375 ranks ordered binary trees, counted by A126120.
Cf. A000014, A001263, A001679, A004249, A061775, A063895, A187306, A331489, A331490, A331934, A358373, A358377.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
srt[n_]:=If[n==1,{},srt/@stc[n-1]];
Select[Range[100],FreeQ[srt[#],[_]?(Length[#]==1&)]&]

A132081 Triangle (read by rows) with row sums = Motzkin sums (also called Riordan numbers) (A005043): T(n,s) = (1/n)C(n,s)(C(n-s,s+1) - C(n-s-2,s-1)).

Original entry on oeis.org

1, 1, 2, 1, 5, 1, 9, 5, 1, 14, 21, 1, 20, 56, 14, 1, 27, 120, 84, 1, 35, 225, 300, 42, 1, 44, 385, 825, 330, 1, 54, 616, 1925, 1485, 132, 1, 65, 936, 4004, 5005, 1287, 1, 77, 1365, 7644, 14014, 7007, 429

Offset: 3

Frank R. Bernhart (farb45(AT)gmail.com), Oct 30 2007

Whereas A005043 counts certain trees, or noncrossed partitions, this subdivides the counts according to the number of leaves, or the lattice rank. Analogous to the Narayana triangle (A001263), where rows sum to the Catalan numbers.
Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012
Related to the number of certain non-crossing partitions for the root system A_n. Cf. p. 12, Athanasiadis and Savvidou. See also A108263 and A100754. - Tom Copeland, Oct 19 2014

			A005043(6) = 15 = 1+9+5 since NC (noncrossed, planar) partitions of 6-point cycle without singletons have 1,9,5 items with 1,2,3 blocks.
Triangle begins:
  1;
  1,   2;
  1,   5;
  1,   9,   5;
  1,  14,  21;
  1,  20,  56,  14;
  1,  27, 120,  84;
  1,  35, 225, 300,  42;
  1,  44, 385, 825, 330;
  ...

C. Athanasiadis and C. Savvidou, The local h-vector of the cluster subdivision of a simplex, arXiv preprint arXiv:1204.0362 [math.CO], 2012.
F. R. Bernhart, Catalan, Motzkin and Riordan numbers, Discr. Math., 204 (1999), 73-112.
F. R. Bernhart & N. J. A. Sloane, Emails, April-May 1994
T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015.

Row sums are A007404.

Cf. A001263, A005043, A033282, A100754, A108263, A108767, A132081.

/* triangle excluding 0 */ [[Binomial(n,k)*Binomial(n-2-k,k)/(k+1): k in [0..n-3]]: n in [3..15]]; // Vincenzo Librandi, Oct 19 2014

Map[Most, Table[(1/n) Binomial[n, s] (Binomial[n - s, s + 1] - Binomial[n - s - 2, s - 1]), {n, 3, 14}, {s, 0, n}] /. k_ /; k <= 0 -> Nothing] // Flatten (* Michael De Vlieger, Jan 09 2016 *)

a(n,k) = binomial(n,k)*binomial(n-2-k,k)/(k+1). - David Callan, Jul 22 2008
From Peter Bala, Oct 22 2008: (Start)
O.g.f.: (1 + x - sqrt(1 - 2*x + x^2*(1 - 4*a)))/(2*x*(1 + a*x)) = 1 + a*x^2 + a*x^3 + (a + 2*a^2)*x^4 + (a + 5*a^2)*x^5 + (a + 9*a^2 + 5*a^3)*x^6 + ... . [corrected by Jason Yuen, Sep 22 2024]
Define a functional I on formal power series of the form f(x) = 1 + a*x + b*x^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim n -> infinity f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
Let now f(x) = 1 + a*x^2 + a*x^3 + a*x^4 + ... . Then the o.g.f. for this table is I(f(x)) = 1 + a*x^2 + a*x^3 + (a + 2*a^2)*x^4 + (a + 5*a^2)*x^5 + (a + 9*a^2 + 5*a^3)*x^6 + ... . Cf. A001263 and A108767. (End)

Edited by N. J. A. Sloane, Jul 01 2008 at the suggestion of R. J. Mathar

Name corrected by Emeric Deutsch, Dec 20 2014

A099252 Bisection of A005043.

Original entry on oeis.org

0, 1, 6, 36, 232, 1585, 11298, 83097, 625992, 4805595, 37458330, 295673994, 2358641376, 18985057351, 154000562758, 1257643249140, 10331450919456, 85317692667643, 707854577312178, 5897493615536452, 49320944483427000, 413887836110423787, 3484084625456932134, 29412628894558563849

Offset: 0

N. J. A. Sloane, Nov 16 2004

G. F. Smith, On isotropic tensors and rotation tensors of dimension m and order n, Tensor (N.S.), Vol. 19 (1968), 79-88 (MR0224008).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. L. Andrews, Letter to N. J. A. Sloane, Apr 10 1978.

Cf. A005043, A099251.

G := (1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x)): Gser := series(G,x=0,60):
seq(coeff(Gser, x^(2*n-1)), n=1..25); # Emeric Deutsch
a := n -> -hypergeom([-2*n-1, 1/2], [2], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Jul 26 2020

Take[CoefficientList[Series[(1+x-Sqrt[1-2*x-3*x^2])/(2*x*(1+x)), {x, 0, 60}], x], {2, -1, 2}] (* Vaclav Kotesovec, Oct 17 2012 *)

x='x+O('x^66); v=Vec((1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x))); vector(#v\2,n,v[2*n]) \\ Joerg Arndt, May 12 2013

def A():
    a, b, c, d, n = 0, 1, 1, -1, 1
    yield 0
    while True:
        n += 1
        a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
        c, d = d, (3*(n-1)*c-(2*n-1)*d)//n
        if n%2: yield -(d + b)*(1-(-1)^n)//2
A099252  = A()
print([next(A099252) for  in range(24)]) # _Peter Luschny, May 16 2016

Recurrence: (n+1)*(2*n+1)*a(n) = n*(26*n-7)*a(n-1) - 3*(26*n^2 - 61*n + 39)*a(n-2) + 27*(n-2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ 3^(2*n+5/2)/(16*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
a(n) = -hypergeom([-2*n - 1, 1/2], [2], 4). - Peter Luschny, Jul 26 2020

More terms from Emeric Deutsch, Nov 18 2004

A185812 Riordan array ( 1/(1-x), x*A005043(x) ).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 6, 5, 4, 1, 1, 1, 12, 12, 7, 5, 1, 1, 1, 27, 26, 19, 9, 6, 1, 1, 1, 63, 63, 43, 27, 11, 7, 1, 1, 1, 154, 153, 110, 63, 36, 13, 8, 1, 1, 1, 386, 386, 275, 169, 86, 46, 15, 9, 1, 1

Offset: 0

Vladimir Kruchinin, Feb 05 2011

			Array begins:
  1;
  1,  1;
  1,  1,  1;
  1,  2,  1,  1;
  1,  3,  3,  1,  1;
  1,  6,  5,  4,  1,  1;
  1, 12, 12,  7,  5,  1,  1;
  1, 27, 26, 19,  9,  6,  1,  1;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A082395, apparently R(n,1), A097332 (row sums). - R. J. Mathar, Feb 10 2011

A185812 := proc(n,k) if n = k  or k =0 then 1; else k*add(1/(n-i)*add(binomial(2*j-k-1,j-1) *(-1)^(n-j-i) *binomial(n-i,j),j=k..n-i),i=0..n-k) ; end if; end proc:
seq(seq(A185812(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Feb 10 2011

r[n_, k_] := k*Sum[Binomial[2*j - k - 1, j - 1]*(-1)^(n - j - i)*Binomial[n - i, j]/(n - i), {i, 0, n - k}, {j, k, n - i}]; r[n_, 0] = 1; Table[r[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 21 2013 *)

R(n,k) = k*Sum_{i=0..(n-k)} (Sum_{j=k..(n-i)} binomial(2*j-k-1,j-1) *(-1)^(n-j-i) *binomial(n-i,j))/(n-i), k>0.

R(n,0)=1.

A194589 a(n) = A194588(n) - A005043(n); complementary Riordan numbers.

Original entry on oeis.org

0, 0, 1, 1, 5, 11, 34, 92, 265, 751, 2156, 6194, 17874, 51702, 149941, 435749, 1268761, 3700391, 10808548, 31613474, 92577784, 271407896, 796484503, 2339561795, 6877992334, 20236257626, 59581937299, 175546527727, 517538571125, 1526679067331, 4505996000730

Offset: 0

Peter Luschny, Aug 30 2011

nonn

The inverse binomial transform of a(n) is A194590(n).

Peter Luschny, The lost Catalan numbers.

Cf. A005043, A189912, A194588, A100071, A005717.

# First method, describes the derivation:
A056040 := n -> n!/iquo(n,2)!^2:
A057977 := n -> A056040(n)/(iquo(n,2)+1);
A001006 := n -> add(binomial(n,k)*A057977(k)*irem(k+1,2),k=0..n):
A005043 := n -> `if`(n=0,1,A001006(n-1)-A005043(n-1)):
A189912 := n -> add(binomial(n,k)*A057977(k),k=0..n):
A194588 := n -> `if`(n=0,1,A189912(n-1)-A194588(n-1)):
A194589 := n -> A194588(n)-A005043(n):
# Second method, more efficient:
A100071 := n -> A056040(n)*(n/2)^(n-1 mod 2):
A194589 := proc(n) local k;
(n mod 2)+(1/2)*add((-1)^k*binomial(n,k)*A100071(k+1),k=1..n) end:
# Alternatively:
a := n -> `if`(n<3,iquo(n,2),hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4)): seq(simplify(a(n)), n=0..30); # Peter Luschny, Mar 07 2017

sf[n_] := With[{f = Floor[n/2]}, Pochhammer[f+1, n-f]/f!]; a[n_] := Mod[n, 2] + (1/2)*Sum[(-1)^k*Binomial[n, k]*2^-Mod[k, 2]*(k+1)^Mod[k, 2]*sf[k+1], {k, 1, n}]; Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jul 30 2013, from 2nd method *)
Table[If[n < 3, Quotient[n, 2], HypergeometricPFQ[{1 - n/2, -n, 3/2 - n/2}, {1, 2-n}, 4]], {n,0,30}] (* Peter Luschny, Mar 07 2017 *)

a(n):=sum(binomial(n+2,k)*binomial(n-k,k),k,0,(n)/2); /* Vladimir Kruchinin, Sep 28 2015 */

a(n) = sum(k=0, n/2, binomial(n+2,k)*binomial(n-k,k));
vector(30, n, a(n-3)) \\ Altug Alkan, Sep 28 2015

a(n) = sum_{k=0..n} C(n,k)*A194590(k).
a(n) = (n mod 2)+(1/2)*sum_{k=1..n} (-1)^k*C(n,k)*(k+1)$*((k+1)/2)^(k mod 2). Here n$ denotes the swinging factorial A056040(n).
a(n) = PSUMSIGN([0,0,1,2,6,16,45,..] = PSUMSIGN([0,0,A005717]) where PSUMSIGN is from Sloane's "Transformations of integer sequences". - Peter Luschny, Jan 17 2012
A(x) = B'(x)*(1/x^2-1/(B(x)*x)), where B(x)/x is g.f. of A005043. - Vladimir Kruchinin, Sep 28 2015
a(n) = Sum_{k=0..n/2} C(n+2,k)*C(n-k,k). - Vladimir Kruchinin, Sep 28 2015
a(n) = hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4) for n>=3. - Peter Luschny, Mar 07 2017
a(n) ~ 3^(n + 1/2) / (8*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 17 2024

A124926 Triangle read by rows: T(n,k) = binomial(n,k)*r(k), where r(k) are the Riordan numbers (r(k) = A005043(k); 0 <= k <= n).

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 3, 1, 1, 0, 6, 4, 3, 1, 0, 10, 10, 15, 6, 1, 0, 15, 20, 45, 36, 15, 1, 0, 21, 35, 105, 126, 105, 36, 1, 0, 28, 56, 210, 336, 420, 288, 91, 1, 0, 36, 84, 378, 756, 1260, 1296, 819, 232, 1, 0, 45, 120, 630, 1512, 3150, 4320, 4095, 2320, 603

Offset: 0

Gary W. Adamson, Nov 12 2006

Row sums = Catalan numbers, A000108: (1, 1, 2, 5, 14, 42...); e.g., sum of row 4 terms = A000108(4) = 14 = (1 + 0 + 6 + 4 + 3). A005043 is the inverse binomial transform of the Catalan numbers.

			First few rows of the triangle:
  1;
  1,  0;
  1,  0,  1;
  1,  0,  3,  1;
  1,  0,  6,  4,  3;
  1,  0, 10, 10, 15,  6;
  1,  0, 15, 20, 45, 36, 15;
  ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Chao-Jen Wang, Applications of the Goulden-Jackson cluster method to counting Dyck paths by occurrences of subwords, Dissertation, Brandeis University, 2011.

Cf. A000108, A005043.

B:=Binomial;;  Flat(List([0..12], n-> List([0..n], k-> B(n,k)* Sum([0..k], j-> (-1)^j*B(k+1,j)*B(2*(k-j), k-j))/(k+1) ))); # G. C. Greubel, Nov 19 2019

B:=Binomial; [B(n,k)*(&+[(-1)^j*B(k+1,j)*B(2*(k-j), k-j): j in [0..k]])/(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 19 2019

r:=n->(1/(n+1))*sum((-1)^i*binomial(n+1,i)*binomial(2*n-2*i,n-i),i=0..n): T:=(n,k)->r(k)*binomial(n,k): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_]:= T[n, k]= Binomial[n, k]*Sum[(-1)^j*Binomial[k+1, j]* Binomial[2*(k-j), k-j], {j,0,k}]/(k+1); Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 19 2019 *)

T(n,k) = b=binomial; b(n,k)*sum(j=0,k, (-1)^j*b(k+1,j)*b(2*(k-j), k-j))/(k+1); \\ G. C. Greubel, Nov 19 2019

b=binomial; [[b(n,k)*sum((-1)^j*b(k+1,j)*b(2*(k-j), k-j) for j in (0..k))/(k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 19 2019

Edited by N. J. A. Sloane, Nov 29 2006

A185813 Riordan array (A000045(x), x*A005043(x)).

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 3, 3, 1, 1, 0, 5, 5, 4, 1, 1, 0, 8, 11, 7, 5, 1, 1, 0, 13, 22, 18, 9, 6, 1, 1, 0, 21, 48, 39, 26, 11, 7, 1, 1, 0, 34, 106, 94, 59, 35, 13, 8, 1, 1, 0, 55, 245, 223, 152, 82, 45, 15, 9, 1, 1, 0

Offset: 0

Vladimir Kruchinin, Feb 05 2011

			Array begins:
   0;
   1,   0;
   1,   1,   0;
   2,   1,   1,   0;
   3,   3,   1,   1,  0;
   5,   5,   4,   1,  1,  0;
   8,  11,   7,   5,  1,  1,  0;
  13,  22,  18,   9,  6,  1,  1, 0;
  21,  48,  39,  26, 11,  7,  1, 1, 0;
  34, 106,  94,  59, 35, 13,  8, 1, 1, 0;
  55, 245, 223, 152, 82, 45, 15, 9, 1, 1, 0;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.

Cf. A000045 (Fibonacci).

A185813 := proc(n,k) if n = k then 0; elif k = 0 then combinat[fibonacci](n) ; else k*add(1/(n-i)*combinat[fibonacci](i)*add(binomial(2*j-k-1,j-1) *(-1)^(n-j-i) *binomial(n-i,j),j=k..n-i),i=0..n-k) ; end if; end proc:
seq(seq(A185813(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Feb 10 2011

r[n_, k_] := k*Sum[((-1)^(n+k-i)*Fibonacci[i]*(n-i)!*HypergeometricPFQ[{k/2 + 1/2, k/2, i+k-n}, {k, k+1}, 4])/((n-i)*k!*(n-i-k)!), {i, 0, n-k}]; r[n_, 0] := Fibonacci[n]; Table[r[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 21 2013 *)

R(n,k) = k*Sum_{i=0..(n-k)} Fibonacci(i)*(Sum_{j=k..(n-i)} binomial(2*j-k-1,j-1)*(-1)^(n-j-i)*binomial(n-i,j))/(n-i), k>1.

R(n,0) = Fibonacci(n).

cp's OEIS Frontend

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A132081 Triangle (read by rows) with row sums = Motzkin sums (also called Riordan numbers) (A005043): T(n,s) = (1/n)*C(n,s)*(C(n-s,s+1) - C(n-s-2,s-1)).

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A132081 Triangle (read by rows) with row sums = Motzkin sums (also called Riordan numbers) (A005043): T(n,s) = (1/n)C(n,s)(C(n-s,s+1) - C(n-s-2,s-1)).