A003169 -id:A003169 - cp's OEIS frontend

A100326 Triangle, read by rows, where row n equals the inverse binomial of column n of square array A100324, which lists the self-convolutions of SHIFT(A003169).

1, 1, 1, 3, 4, 1, 14, 20, 7, 1, 79, 116, 46, 10, 1, 494, 736, 311, 81, 13, 1, 3294, 4952, 2174, 626, 125, 16, 1, 22952, 34716, 15634, 4798, 1088, 178, 19, 1, 165127, 250868, 115048, 36896, 9094, 1724, 240, 22, 1, 1217270, 1855520, 862607, 285689, 74687, 15629, 2561, 311, 25, 1

Offset: 0

Paul D. Hanna, Nov 17 2004

The leftmost column equals A003169 shift one place right.
Each column k > 0 equals the convolution of the prior column and A003169.
Row sums form A100327.
The elements of the matrix inverse are T^(-1)(n,k) = (-1)^(n+k) * A158687(n,k). - R. J. Mathar, Mar 15 2013

			Rows begin:
        1;
        1,       1;
        3,       4,      1;
       14,      20,      7,      1;
       79,     116,     46,     10,     1;
      494,     736,    311,     81,    13,     1;
     3294,    4952,   2174,    626,   125,    16,    1;
    22952,   34716,  15634,   4798,  1088,   178,   19,   1;
   165127,  250868, 115048,  36896,  9094,  1724,  240,  22,  1;
  1217270, 1855520, 862607, 285689, 74687, 15629, 2561, 311, 25,  1;
  ...
First column forms A003169 shift right.
Binomial transform of row 3 forms column 3 of square A100324: BINOMIAL([14,20,7,1]) = [14,34,61,96,140,194,259,...].
Binomial transform of row 4 forms column 4 of square A100324: BINOMIAL([79,116,46,10,1]) = [79,195,357,575,860,1224,...].

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened

Cf. A003169, A100324, A100327 (row sums), A158687, A264717 (central terms).

import Data.List (transpose)
a100326 n k = a100326_tabl !! n !! k
a100326_row n = a100326_tabl !! n
a100326_tabl = [1] : f [[1]] where
f xss@(xs:_) = ys : f (ys : xss) where
ys = y : map (sum . zipWith (*) (zs ++ [y])) (map reverse zss)
y = sum $ zipWith (*) [1..] xs
zss@((:zs):) = transpose $ reverse xss
-- Reinhard Zumkeller, Nov 21 2015

A100326 := proc(n,k)
    if k < 0 or k > n then
        0 ;
    elif n = 0 then
        1 ;
    elif k = 0 then
        A003169(n)
    else
        add(procname(i+1,0)*procname(n-i-1,k-1),i=0..n-k) ;
    end if;
end proc: # R. J. Mathar, Mar 15 2013

lim= 9; t[0, 0]=1; t[n_, 0]:= t[n, 0]= Sum[(k+1)*t[n-1,k], {k,0,n-1}]; t[n_, k_]:= t[n, k]= Sum[t[j+1, 0]*t[n-j-1, k-1], {j,0,n-k}]; Flatten[Table[t[n, k], {n,0,lim}, {k,0,n}]] (* Jean-François Alcover, Sep 20 2011 *)

PARI
```
T(n,k)=if(n
				
```

@CachedFunction
def T(n,k): # T = A100326
    if (k<0 or k>n): return 0
    elif (k==n): return 1
    elif (k==0): return sum((j+1)*T(n-1,j) for j in range(n))
    else: return sum(T(j+1,0)*T(n-j-1,k-1) for j in range(n-k+1))
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 30 2023

T(n, 0) = A003169(n) = Sum_{k=0..n-1} (k+1)*T(n-1, k) for n>0, with T(0, 0)=1.
T(n, k) = Sum_{i=0..n-k} T(i+1, 0)*T(n-i-1, k-1) for n > 0.
T(2*n, n) = A264717(n).
Sum_{k=0..n} T(n, k) = A100327(n).
G.f.: A(x, y) = (1 + G(x))/(1 - y*G(x)), where G(x) is the g.f. of A003169.
From G. C. Greubel, Jan 30 2023: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A033999(n). (End)

A100324 Square array, read by antidiagonals, where rows are successive self-convolutions of the top row, which equals A003169 shifted one place right.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 7, 14, 1, 4, 12, 34, 79, 1, 5, 18, 61, 195, 494, 1, 6, 25, 96, 357, 1230, 3294, 1, 7, 33, 140, 575, 2277, 8246, 22952, 1, 8, 42, 194, 860, 3716, 15372, 57668, 165127, 1, 9, 52, 259, 1224, 5641, 25298, 108018, 415995, 1217270

Offset: 0

Paul D. Hanna, Nov 16 2004

Column k forms the binomial transform of row k in triangle A100326 for k>=0.

			Array, A(n,k), begins as:
  1, 1,  3,  14,   79,   494,  3294, ...;
  1, 2,  7,  34,  195,  1230,  8246, ...;
  1, 3, 12,  61,  357,  2277, 15372, ...;
  1, 4, 18,  96,  575,  3716, 25298, ...;
  1, 5, 25, 140,  860,  5641, 38775, ...;
  1, 6, 33, 194, 1224,  8160, 56695, ...;
  1, 7, 42, 259, 1680, 11396, 80108, ...;
Antidiagonal triangle, T(n,k), begins as:
  1;
  1, 1;
  1, 2,  3;
  1, 3,  7,  14;
  1, 4, 12,  34,  79;
  1, 5, 18,  61, 195,  494;
  1, 6, 25,  96, 357, 1230, 3294;
  1, 7, 33, 140, 575, 2277, 8246, 22952;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A003169, A032349, A100325, A100326.

f[n_]:= f[n]= If[n<2, 1, If[n==2, 3, ((324*n^2-708*n+360)*f[n-1] - (371*n^2-1831*n+2250)*f[n-2] +(20*n^2-130*n+210)*f[n-3])/(16*n*(2*n -1)) ]]; (* f = A003169 *)
A[n_, k_]:= A[n, k]= If[n==0, f[k], If[k==0, 1, Sum[A[0,k-j]*A[n-1,j], {j,0,k}]]]; (* A = A100324 *)
T[n_, k_]:= A[n-k, k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 31 2023 *)

{A(n,k)=if(k==0,1,if(n>0,sum(i=0,k,A(0,k-i)*A(n-1,i)), if(k==1,1,if(k==2,3,( (324*k^2-708*k+360)*A(0,k-1)-(371*k^2-1831*k+2250)*A(0,k-2)+(20*k^2-130*k+210)*A(0,k-3))/(16*k*(2*k-1)) )));)}

def f(n): # f = A003169
    if (n<2): return 1
    elif (n==2): return 3
    else: return ((324*n^2-708*n+360)*f(n-1) - (371*n^2-1831*n+2250)*f(n-2) + (20*n^2-130*n+210)*f(n-3))/(16*n*(2*n-1))
@CachedFunction
def A(n, k): # A = 100324
    if (n==0): return f(k)
    elif (k==0): return 1
    else: return sum( A(0,k-j)*A(n-1, j) for j in range(k+1) )
def T(n,k): return A(n-k,k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 31 2023

A(n, k) = Sum_{i=0..k} A(0, k-i)*A(n-1, i) for n>0.
A(0, k) = A003169(k+1) = ( (324*k^2-708*k+360)*A(0, k-1) - (371*k^2-1831*k+2250)*A(0, k-2) +(20*k^2-130*k+210)*A(0, k-3) )/(16*k*(2*k-1)) for k>2, with A(0, 0) = A(0, 1)=1, A(0, 2)=3.
A(n, n) = (n+1)*A032349(n+1).
T(n, k) = A(n-k, k) (Antidiagonal triangle).
T(n, n) = A003169(n+1).
Sum_{k=0..n} T(n, k) = A100325(n) (Antidiagonal row sums).

A100325 Antidiagonal sums of square array A100324, which lists the self-convolutions of SHIFT(A003169).

Original entry on oeis.org

1, 2, 6, 25, 130, 774, 5009, 34231, 242988, 1773767, 13229272, 100362848, 772016385, 6007208105, 47198747457, 373929821070, 2983774582206, 23958802697161, 193448157014605, 1569625544848531, 12791865082236462

Offset: 0

Paul D. Hanna, Nov 17 2004

nonn

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

Cf. A003169, A100324.

f[n_]:= f[n]= If[n<2, 1, If[n==2, 3, ((324*n^2 -708*n +360)*f[n-1] -(371*n^2 -1831*n +2250)*f[n-2] + (20*n^2 -130*n +210)*f[n-3])/(16*n*(2*n-1)) ]]; (* f = A003169 *)
A[n_, k_]:= A[n, k]= If[n==0, f[k], If[k==0, 1, Sum[f[k-j]*A[n-1,j], {j,0,k}]]]; (* A = 100324 *)
a[n_]:= a[n]= Sum[A[n-k,k], {k,0,n}]; (* a = A100325 *)
Table[a[n], {n, 0, 40}] (* G. C. Greubel, Jan 31 2023 *)

{a(n)=local(A=1+x+x*O(x^n));if(n==0,1, for(i=1,n,A=1+x*A/(2-A)^2); sum(k=0,n,polcoeff(A^(n-k+1),k)))}

@CachedFunction
def f(n): # f = A003169
    if (n<2): return 1
    elif (n==2): return 3
    else: return ((324*n^2-708*n+360)*f(n-1) - (371*n^2-1831*n+2250)*f(n-2) + (20*n^2-130*n+210)*f(n-3))/(16*n*(2*n-1))
@CachedFunction
def A(n, k): # A = 100324
    if (n==0): return f(k)
    elif (k==0): return 1
    else: return sum( f(k-j)*A(n-1, j) for j in range(k+1) )
def T(n,k): return A(n-k, k)
def A100325(n): return sum( T(n,k) for k in range(n+1) )
[A100325(n) for n in range(41)] # G. C. Greubel, Jan 31 2023

G.f. A(x) = (1+G003169(x))/(1-x-x*G003169(x)), where G003169(x) is the g.f. of A003169.

a(n) ~ (sqrt(3056686 + 12607266/sqrt(17)) * ((71 + 17*sqrt(17))/16)^n) / (10201 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 31 2023

A003168 Number of blobs with 2n+1 edges.

Original entry on oeis.org

1, 1, 4, 21, 126, 818, 5594, 39693, 289510, 2157150, 16348960, 125642146, 976789620, 7668465964, 60708178054, 484093913917, 3884724864390, 31348290348086, 254225828706248, 2070856216759478, 16936016649259364

Offset: 0

N. J. A. Sloane

a(n) is the number of ways to dissect a convex (2n+2)-gon with non-crossing diagonals so that no (2m+1)-gons (m>0) appear. - Len Smiley
a(n) is the number of plane trees with 2n+1 leaves and all non-leaves having an odd number > 1 of children. - Jordan Tirrell, Jun 09 2017
a(n) is the number of noncrossing cacti with n+1 nodes. See A361242. - Andrew Howroyd, Mar 07 2023

			a(2)=4 because we may place exactly one diagonal in 3 ways (forming 2 quadrilaterals), or not place any (leaving 1 hexagon).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe)
Thomas H. Bertschinger, Joseph Slote, Olivia Claire Spencer, and Samuel Vinitsky, The Mathematics of Origami, Undergrad Thesis, Carleton College (2016).
D. Birmajer, J. B. Gil, and M. D. Weiner, Colored partitions of a convex polygon by noncrossing diagonals, arXiv preprint arXiv:1503.05242 [math.CO], 2015.
L. Carlitz, Enumeration of two-line arrays, Fib. Quart., Vol. 11 Number 2 (1973), 113-130.
Frédéric Chapoton and Philippe Nadeau, Combinatorics of the categories of noncrossing partitions, Séminaire Lotharingien de Combinatoire 78B (2017), Article #37.
Michael Drmota, Anna de Mier, and Marc Noy, Extremal statistics on non-crossing configurations, Discrete Math. 327 (2014), 103--117. MR3192420. See p. 116, B_b(z). - N. J. A. Sloane, May 18 2014
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 415
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Jean-Christophe Novelli and Jean-Yves Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Jean-Christophe Novelli and Jean-Yves Thibon, Noncommutative Symmetric Functions and Lagrange Inversion II: Noncrossing partitions and the Farahat-Higman algebra, arXiv:2106.08257 [math.CO], 2021-2022.
Ronald C. Read, On the enumeration of a class of plane multigraphs, Aequat. Math. 31 (1986) No. 1, 47-63.
L. Smiley, Even-gon reference
Jun Yan, Results on pattern avoidance in parking functions, arXiv:2404.07958 [math.CO], 2024. See p. 30.

Cf. A049124 (no 2m-gons).
Cf. A003169, A100327, A114496, A007318, A361242.
Row sums of A102537, A243662. Column 2 of A336573.

import Data.List (transpose)
a003168 0 = 1
a003168 n = sum (zipWith (*)
   (tail $ a007318_tabl !! n)
   ((transpose $ take (3*n+1) a007318_tabl) !! (2*n+1)))
   `div` fromIntegral n
-- Reinhard Zumkeller, Oct 27 2013

Order := 40; solve(series((A-2*A^3)/(1-A^2),A)=x,A);
A003168 := n -> `if`(n=0,1,A100327(n)/2): seq(A003168(n),n=0..20); # Peter Luschny, Jun 10 2017

a[0] = 1; a[n_] = (2^(-n-1)*(3n)!* Hypergeometric2F1[-1-2n, -2n, -3n, -1])/((2n+1)* n!*(2n)!); Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 25 2011, after Vladimir Kruchinin *)

a(n)=if(n<0,0,polcoeff(serreverse((x-2*x^3)/(1-x^2)+O(x^(2*n+2))),2*n+1))

{a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=(1+x*A)/(1-x*A)^2); sum(k=0,n,polcoeff(A^(n-k),k))} \\ Paul D. Hanna, Nov 17 2004

seq(n) = Vec( 1 + serreverse(x/((1+2*x)*(1+x)^2) + O(x*x^n)) ) \\ Andrew Howroyd, Mar 07 2023

a(n) = Sum_{k=1..n} binomial(n, k)*binomial(2*n+k, k-1)/n.
G.f.: A(x) = Sum_{n>=0} a(n)*x^(2*n+1) satisfies (A-2*A^3)/(1-A^2)=x. - Len Smiley.
D-finite with recurrence 4*n*(2*n + 1)*(17*n - 22)*a(n) = (1207*n^3 - 2769*n^2 + 1850*n - 360)*a(n - 1) - 2*(17*n - 5)*(n - 2)*(2*n - 3)*a(n - 2). - Vladeta Jovovic, Jul 16 2004
G.f.: A(x) = 1/(1-G003169(x)) where G003169(x) is the g.f. of A003169. - Paul D. Hanna, Nov 17 2004
a(n) = JacobiP(n-1,1,n+1,3)/n for n > 0. - Mark van Hoeij, Jun 02 2010
a(n) = (1/(2*n+1))*Sum_{j=0..n} (-1)^j*2^(n-j)*binomial(2*n+1,j)*binomial(3*n-j,2*n). - Vladimir Kruchinin, Dec 24 2010
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
  1, 1
  3, 3, 1
  5, 5, 3, 1
  7, 7, 5, 3, 1
  9, 9, 7, 5, 3, 1
  ... (End)
a(n) ~ sqrt(14+66/sqrt(17)) * (71+17*sqrt(17))^n / (sqrt(Pi) * n^(3/2) * 2^(4*n+4)). - Vaclav Kotesovec, Jul 01 2015
From Peter Bala, Oct 05 2015: (Start)
a(n) = (1/n) * Sum_{i = 0..n} 2^(n-i-1)*binomial(2*n,i)* binomial(n,i+1).
O.g.f. = 1 + series reversion( x/((1 + 2*x)*(1 + x)^2) ).
Logarithmically differentiating the modified g.f. 1 + 4*x + 21*x^2 + 126*x^3 + 818*x^4 + ... gives the o.g.f. for A114496, apart from the initial term. (End)
G.f.: A(x) satisfies A = 1 + x*A^3/(1-x*A^2). - Jordan Tirrell, Jun 09 2017
a(n) = A100327(n)/2 for n>=1. - Peter Luschny, Jun 10 2017

A193091 Augmentation of the triangular array A158405. See Comments.

Original entry on oeis.org

1, 1, 3, 1, 6, 14, 1, 9, 37, 79, 1, 12, 69, 242, 494, 1, 15, 110, 516, 1658, 3294, 1, 18, 160, 928, 3870, 11764, 22952, 1, 21, 219, 1505, 7589, 29307, 85741, 165127, 1, 24, 287, 2274, 13355, 61332, 224357, 638250, 1217270, 1, 27, 364, 3262, 21789, 115003

Offset: 0

Clark Kimberling, Jul 30 2011

Suppose that P is an infinite triangular array of numbers:
p(0,0)
p(1,0)...p(1,1)
p(2,0)...p(2,1)...p(2,2)
p(3,0)...p(3,1)...p(3,2)...p(3,3)...
...
Let w(0,0)=1, w(1,0)=p(1,0), w(1,1)=p(1,1), and define
    W(n)=(w(n,0), w(n,1), w(n,2),...w(n,n-1), w(n,n)) recursively by W(n)=W(n-1)*PP(n), where PP(n) is the n X (n+1) matrix given by
...
row 0 ... p(n,0) ... p(n,1) ...... p(n,n-1) ... p(n,n)
row 1 ... 0 ..... p(n-1,0) ..... p(n-1,n-2) .. p(n-1,n-1)
row 2 ... 0 ..... 0 ............ p(n-2,n-3) .. p(n-2,n-2)
...
row n-1 . 0 ..... 0 ............. p(2,1) ..... p(2,2)
row n ... 0 ..... 0 ............. p(1,0) ..... p(1,1)
...
The augmentation of P is here introduced as the triangular array whose n-th row is W(n), for n>=0. The array P may be represented as a sequence of polynomials; viz., row n is then the vector of coefficients:  p(n,0), p(n,1),...,p(n,n), from p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n). For example, (C(n,k)) is represented by ((x+1)^n); using this choice of P (that is, Pascal's triangle), the augmentation of P is calculated one row at a time, either by the above matrix products or by polynomial substitutions in the following manner:
...
row 0 of W:  1, by decree
row 1 of W:  1 augments to 1,1
...polynomial version:  1 -> x+1
row 2 of W:  1,1 augments to 1,3,2
...polynomial version:  x+1 -> (x^2+2x+1)+(x+1)=x^2+3x+2
row 3 to W:  1,3,2 augments to 1,6,11,6
...polynomial version:
    x^2+3x+2 -> (x+1)^3+3(x+1)^2+2(x+1)=(x+1)(x+2)(x+3)
...
Examples of augmented triangular arrays:
(p(n,k)=1) augments to A009766, Catalan triangle.
Catalan triangle augments to A193560.
Pascal triangle augments to A094638, Stirling triangle.
A002260=((k+1)) augments to A023531.
A154325 augments to A033878.
A158405 augments to A193091.
((k!)) augments to A193092.
A094727 augments to A193093.
A130296 augments to A193094.
A004736 augments to A193561.
...
Regarding the specific augmentation W=A193091: w(n,n)=A003169.
From Peter Bala, Aug 02 2012: (Start)
This is the table of g(n,k) in the notation of Carlitz (p. 124). The triangle enumerates two-line arrays of positive integers
............a_1 a_2 ... a_n..........
............b_1 b_2 ... b_n..........
such that
1) max(a_i, b_i) <= min(a_(i+1), b_(i+1)) for 1 <= i <= n-1
2) max(a_i, b_i) <= i for 1 <= i <= n
3) max(a_n, b_n) = k.
See A071948 and A211788 for other two-line array enumerations.
(End)

			The triangle P, at A158405, is given by rows
  1
  1...3
  1...3...5
  1...3...5...7
  1...3...5...7...9...
The augmentation of P is the array W starts with w(0,0)=1, by definition of W.
Successive polynomials (rows of W) arise from P as shown here:
  ...
  1->x+3, so that W has (row 1)=(1,3);
  ...
  x+3->(x^2+3x+5)+3*(x+3), so that W has (row 2)=(1,6,14);
  ...
  x^2+6x+14->(x^3+3x^2+5x+7)+6(x^2+3x+5)+14(x+3), so that (row 3)=(1,9,37,79).
  ...
First 7 rows of W:
  1
  1    3
  1    6    14
  1    9    37    79
  1   12    69   242    494
  1   15   110   516   1658    3294
  1   18   160   928   3870   11764   22952

L. Carlitz, Enumeration of two-line arrays, Fib. Quart., Vol. 11 Number 2 (1973), 113-130.

Cf. A003169, A193092, A193093, A193094.

Cf. A071948, A211788.

p[n_, k_] := 2 k + 1
Table[p[n, k], {n, 0, 5}, {k, 0, n}] (* A158405 *)
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]] (* A193091 *)
Flatten[Table[v[n], {n, 0, 9}]]

From Peter Bala, Aug 02 2012: (Start)
T(n,k) = (n-k+1)/n*Sum_{i=0..k} C(n+1,n-k+i+1)*C(2*n+i+1,i) for 0 <= k <= n.
Recurrence equation: T(n,k) = Sum_{i=0..k} (2*k-2*i+1)*T(n-1,i).
(End)

A100327 Row sums of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324.

Original entry on oeis.org

1, 2, 8, 42, 252, 1636, 11188, 79386, 579020, 4314300, 32697920, 251284292, 1953579240, 15336931928, 121416356108, 968187827834, 7769449728780, 62696580696172, 508451657412496, 4141712433518956, 33872033298518728, 278014853384816184, 2289376313410678312

Offset: 0

Paul D. Hanna, Nov 17 2004

nonn

Self-convolution yields A100328, which equals column 1 of triangle A100326 (omitting leading zero).

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

Cf. A003168, A003169, A100326, A219538.

A100327:= func< n | n eq 0 select 1 else (2/n)*(&+[Binomial(n, k)*Binomial(2*n+k, k-1): k in [1..n]]) >;
[A100327(n): n in [0..30]]; // G. C. Greubel, Jan 30 2023

A100327 := n -> simplify(2^n*binomial(3*n,2*n)*hypergeom([-1-2*n,-n], [-3*n], 1/2)/ (n+1/2)): seq(A100327(n), n=0..22); # Peter Luschny, Jun 10 2017

Flatten[{1,Table[Sum[2*Binomial[n,k]*Binomial[2n+k,k-1]/n,{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Oct 17 2012 *)

a(n)=if(n==0,1,sum(k=0,n,2*binomial(n,k)*binomial(2*n+k,k-1)/n))

a(n)=polcoeff((1/x)*serreverse(x*(1-x+sqrt(1-4*x +x^2*O(x^n)))/(2+x)),n)
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 22 2012

def A100327(n): return 2^n*binomial(3*n,2*n)*simplify(hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2))
[A100327(n) for n in range(31)] # G. C. Greubel, Jan 30 2023

G.f.: (1/x)*Series_Reversion( x*(1-x + sqrt(1 - 4*x)) / (2+x) ). - Paul D. Hanna, Nov 22 2012
G.f. A(x) = (1+G(x))/(1-G(x)), also A(x)^2 = (1+G(x))*G(x)/x, where G(x) = x*(1+G(x))/(1-G(x))^2 is the g.f. of A003169.
a(n) = 2*A003168(n) for n>0 with a(0)=1.
a(n) = Sum_{k=1..n} 2*binomial(n, k)*binomial(2n+k, k-1)/n for n>0 with a(0)=1.
Recurrence: 20*n*(2*n+1)*a(n) = (371*n^2 - 395*n + 96)*a(n-1) - 6*(27*n^2 - 103*n + 96)*a(n-2) + 4*(n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(4046 + 1122*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(136*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
a(n) = 2^n*binomial(3*n,2*n)*hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2). - Peter Luschny, Jun 10 2017

A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).

Original entry on oeis.org

1, 3, 19, 138, 1059, 8378, 67582, 552576, 4563235, 37972290, 317894394, 2674398268, 22590697614, 191475925332, 1627653567916, 13870754053388, 118464647799075, 1013709715774130, 8689197042438274, 74594573994750972, 641252293546113434, 5519339268476249676, 47558930664216470628

Offset: 0

Paul Barry, Feb 17 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A156894

Cf. A002003, A003169, A103885, A114496, A123164.

A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >;
[A156894(n): n in [0..30]]; // G. C. Greubel, Jan 06 2022

a := n -> hypergeom([-n, 2*n], [1], -1);
seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Aug 02 2014

Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Nov 12 2014 *)

a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k)));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015

[round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # G. C. Greubel, Jan 06 2022

a(n) = [x^n] ((1+x)/(1-x)^2)^n.
a(n) = (4*(n+1)*(2*n+1)*A003169(n+1) - (5*n+1)*(2*n-1)*A003169(n))/(17*n + 5) for n>0. - Mark van Hoeij, Jul 14 2010
a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - Peter Luschny, Aug 02 2014
Conjecture: 64*n*(2*n-1)*a(n) -16*(89*n^2 -134*n +63)*a(n-1) +4*(661*n^2 -2619*n +2576)*a(n-2) -3*(119*n^2 -713*n +1092)*a(n-3) +6*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 16*n*(782*n +5365)*(2*n-1)*a(n) +8*(3128*n^3 -362053*n^2 +593930*n -290328)*a(n-1) -3*(726869*n^3 -5105981*n^2 +11667946*n -8715544)*a(n-2) +158*(2*n-5)*(n-3)*(391*n -764)*a(n-3) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 4*n*(2*n-1)*(17*n^2 -52*n +39)*a(n) -(1207*n^4 -4899*n^3 +6692*n^2 -3504*n +576)*a(n-1) +2*(n-2)*(2*n-3)*(17*n^2 -18*n +4)*a(n-2) = 0. - R. J. Mathar, Feb 05 2015 [the Maple command sumrecursion (binomial(n,k) * binomial(2*n+k-1,k), k, a(n)) verifies this recurrence. - Peter Bala, Oct 05 2015 ]
a(n) ~ sqrt(578 + 306*sqrt(17)) * (71 + 17*sqrt(17))^n / (17 * sqrt(Pi*n) * 2^(4*n+2)). - Vaclav Kotesovec, Feb 05 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 14*x^2 + 79*x^3 + ... is the o.g.f. of A003169 (taken with offset 0). - Peter Bala, Oct 05 2015
From Peter Bala, Mar 20 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A119259.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)

A286784 Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 5, 15, 9, 1, 14, 56, 56, 16, 1, 42, 210, 300, 150, 25, 1, 132, 792, 1485, 1100, 330, 36, 1, 429, 3003, 7007, 7007, 3185, 637, 49, 1, 1430, 11440, 32032, 40768, 25480, 7840, 1120, 64, 1, 4862, 43758, 143208, 222768, 179928, 77112, 17136, 1836, 81, 1, 16796, 167960, 629850, 1162800, 1162800, 651168, 203490, 34200, 2850, 100, 1

Offset: 0

Gheorghe Coserea, May 14 2017

T(n,k) is the number of Feynman's diagrams with k fermionic loops in the order n of the perturbative expansion in dimension zero for the GW approximation of the self-energy function in a many-body theory of fermions with two-body interaction (see Molinari link).

			A(x;t) = 1 + (1 + t)*x + (2 + 4*t + t^2)*x^2 + (5 + 15*t + 9*t^2 + t^3)*x^3 + ...
Triangle starts:
n\k  [0]   [1]    [2]     [3]     [4]     [5]    [6]    [7]   [8] [9]
[0]  1;
[1]  1,    1;
[2]  2,    4,     1;
[3]  5,    15,    9,      1;
[4]  14,   56,    56,     16,     1;
[5]  42,   210,   300,    150,    25,     1;
[6]  132,  792,   1485,   1100,   330,    36,    1;
[7]  429,  3003,  7007,   7007,   3185,   637,   49,    1;
[8]  1430, 11440, 32032,  40768,  25480,  7840,  1120,  64,   1;
[9]  4862, 43758, 143208, 222768, 179928, 77112, 17136, 1836, 81, 1;
[10] ...

Gheorghe Coserea, Rows n=0..122, flattened
Luca G. Molinari, Hedin's equations and enumeration of Feynman's diagrams, arXiv:cond-mat/0401500 [cond-mat.str-el], 2005.

Cf. A286781, A286782, A286783.

/* As triangle */ [[(Binomial(2*n, n+m)*Binomial(n+1, m))/(n+1): m in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 23 2018

Flatten@Table[Binomial[2 n, n + m] Binomial[n + 1, m] / (n + 1), {n, 0, 10}, {m, 0, n}] (* Vincenzo Librandi, Sep 23 2018 *)

T(n,m):=(binomial(2*n,n+m)*binomial(n+1,m))/(n+1); /* Vladimir Kruchinin, Sep 23 2018 */

A286784_ser(N,t='t) = my(x='x+O('x^N)); serreverse(Ser(x*(1-x)^2/(1+(t-1)*x)))/x;
concat(apply(p->Vecrev(p), Vec(A286784_ser(12))))
\\ test: y=A286784_ser(50); y*(1-x*y)^2 == 1 + ('t-1)*x*y

y(x;t) = Sum_{n>=0} P_n(t)*x^n satisfies y*(1-x*y)^2 = 1 + (t-1)*x*y, where P_n(t) = Sum_{k=0..n} T(n,k)*t^k.
A000108(n) = T(n,0), A001791(n) = T(n,1), A002055(n+3) = T(n,2), A000290(n) = T(n,n-1), A006013(n) = P_n(1), A003169(n+1) = P_n(2).
T(n,m) = C(2*n,n+m)*C(n+1,m)/(n+1). - Vladimir Kruchinin, Sep 23 2018

A365764 Expansion of (1/x) * Series_Reversion( x*(1-x)^3/(1+x) ).

Original entry on oeis.org

1, 4, 25, 188, 1563, 13840, 127972, 1221260, 11938471, 118936100, 1203155633, 12325599632, 127611357300, 1333153669632, 14035828918560, 148773617605036, 1586305110768863, 17002975960876300, 183102052226442475, 1980078493171083292, 21493846031259095539

Offset: 0

Seiichi Manyama, Sep 18 2023

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..940

Cf. A003169, A006318, A365765, A365766.

Cf. A365751.

CoefficientList[(1/x) *InverseSeries[Series[x*(1-x)^3/(1+x),{x,0,21}]],x] (* Stefano Spezia, May 04 2025 *)

a(n) = sum(k=0, n, binomial(n+1, k)*binomial(4*n-k+2, n-k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+1,k) * binomial(4*n-k+2,n-k) = (1/(n+1)) * Sum_{k=0..n} binomial(3*n+k+2,k) * binomial(n+1,n-k).

A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

Original entry on oeis.org

1, 3, 1, 5, 6, 1, 7, 19, 9, 1, 9, 44, 42, 12, 1, 11, 85, 138, 74, 15, 1, 13, 146, 363, 316, 115, 18, 1, 15, 231, 819, 1059, 605, 165, 21, 1, 17, 344, 1652, 2984, 2470, 1032, 224, 24, 1, 19, 489, 3060, 7380, 8378, 4974, 1624, 292, 27, 1, 21, 670, 5301, 16488

Offset: 1

Clark Kimberling, Mar 03 2012

For a discussion and guide to related arrays, see A208510.
Riordan array ((1+x)/(1-x)^2, x(1+x)/(1-x)^2) (follows from Kruchinin formula). - Ralf Stephan, Jan 02 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A099375 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

			First five rows:
1
3...1
5...6....1
7...19...9....1
9...44...42...12...1
First five polynomials v(n,x):
1
3 + x
5 + 6x + x^2
7 + 19x + 9x^2 + x^3
9 + 44x + 42x^2 + 12x^3 + x^4
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
/1        \/1        \/1        \      /1            \
|3 1      ||0 1      ||0 1      |      |3  1         |
|5 3 1    ||0 3 1    ||0 0 1    |... = |5  6  1      |
|7 5 3 1  ||0 5 3 1  ||0 0 3 1  |      |7 19  9  1   |
|9 7 5 3 1||0 7 5 3 1||0 0 5 3 1|      |9 44 42 12 1 |
|...      ||...      ||...      |      |...
(End)

Cf. A208660, A208510. A099375.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A208660 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208904 *)

u(n,x)=u(n-1,x)+2x*v(n-1,x),
v(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Vladimir Kruchinin, Mar 11 2013: (Start)
T(n,k) = sum(i=0..n, binomial(i+k-1,2*k-1)*binomial(k,n-i))
((x+x^2)/(1-x)^2)^k = sum(n>=k, T(n,k)*x^n).
T(n,2)=A005900(n).
T(2*n-1,n) / n = A003169(n).
T(2*n,n) = A156894(n), n>1.
sum(k=1..n, T(n,k)) = A003946(n).
sum(k=1..n, T(n,k)*(-1)^(n+k)) = A078050(n).
n*sum(k=1..n, T(n,k)/k) = A058481(n). (End)
Recurrence: T(n+1,k+1) = sum {i = 0..n-k} (2*i + 1)*T(n-i,k). - Peter Bala, Jul 21 2014

cp's OEIS Frontend

A100326 Triangle, read by rows, where row n equals the inverse binomial of column n of square array A100324, which lists the self-convolutions of SHIFT(A003169).

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A100324 Square array, read by antidiagonals, where rows are successive self-convolutions of the top row, which equals A003169 shifted one place right.

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A100325 Antidiagonal sums of square array A100324, which lists the self-convolutions of SHIFT(A003169).

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A003168 Number of blobs with 2n+1 edges.

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A193091 Augmentation of the triangular array A158405. See Comments.

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A100327 Row sums of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324.

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A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).