A100327 -id:A100327 - cp's OEIS frontend

A003168 Number of blobs with 2n+1 edges.

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

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).

Original entry on oeis.org

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)

A219538 G.f. satisfies A(x) = 1 + xA(x)^2(1 + A(x))^2/2.

Original entry on oeis.org

1, 2, 12, 98, 924, 9468, 102432, 1151410, 13315692, 157406876, 1893480264, 23103024084, 285233168760, 3556744196000, 44730062281800, 566683825859730, 7225564521956940, 92653105887920556, 1194068058333608136, 15457771628663418748, 200916876963088849992

Offset: 0

Paul D. Hanna, Nov 22 2012

nonn

			G.f.: A(x) = 1 + 2*x + 12*x^2 + 98*x^3 + 924*x^4 + 9468*x^5 + 102432*x^6 +...
Related expansions:
A(x)^2 = 1 + 4*x + 28*x^2 + 244*x^3 + 2384*x^4 + 24984*x^5 +...
A(2)^3 = 1 + 6*x + 48*x^2 + 446*x^3 + 4524*x^4 + 48588*x^5 +...
A(2)^4 = 1 + 8*x + 72*x^2 + 712*x^3 + 7504*x^4 + 82704*x^5 +...
where A(x) = 1 + x*(A(x)^2 + 2*A(x)^3 + A(x)^4)/2.
The g.f. satisfies A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2) where
F(x) = 1 + 2*x + 4*x^2 + 10*x^3 + 28*x^4 + 84*x^5 + 264*x^6 +...+ 2*A000108(n)*x^n +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 2*x + 8*x^2 + 42*x^3 + 252*x^4 + 1636*x^5 +...+ A100327(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..870

Cf. A068875, A100327.

Cf. A219534, A219537, A371658.

CoefficientList[Sqrt[1/x*InverseSeries[Series[x^3/(1-x-Sqrt[1-4*x])^2, {x, 0, 20}], x]],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*A(x)^2*(1 + A(x))^2/2: */
{a(n)=local(A=1);for(i=1,n,A=1+x*A^2*(1+A +x*O(x^n))^2/2);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion involving Catalan numbers: */
{a(n)=local(A=1);A=(1-x-sqrt(1-4*x +x^3*O(x^n)))/x; polcoeff(sqrt(1/x*serreverse(x/A^2)), n)}
for(n=0,25,print1(a(n),", "))

Let F(x) = (1-x - sqrt(1 - 4*x)) / x, then g.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion(x/F(x)^2) ),
(2) A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2),
where F(x) = 2*C(x) - 1 such that C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
Let G(x) be the g.f. of A100327, then g.f. A(x) satisfies:
(3) A(x) = (1/x)*Series_Reversion(x/G(x)),
(4) A(x) = G(x*A(x)) and G(x) = A(x/G(x)).
Recurrence: 3*n*(3*n-1)*(3*n+1)*(11*n-14)*a(n) = 3*(2*n-1)*(693*n^3 - 1575*n^2 + 1026*n - 176)*a(n-1) + 2*(n-2)*(2*n-3)*(2*n-1)*(11*n-3)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ sqrt(242+66*sqrt(33)) * (7+11/9*sqrt(33))^n / (66*sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} 2^(n-2*k) * binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 02 2024

A100328 Column 1 of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324, with leading zero omitted.

Original entry on oeis.org

1, 4, 20, 116, 736, 4952, 34716, 250868, 1855520, 13979192, 106901032, 827644424, 6474611984, 51100656544, 406400018092, 3253636464756, 26201323746880, 212093247874904, 1724793778005528, 14084738953391768, 115447965121881856

Offset: 0

Paul D. Hanna, Nov 17 2004

Self-convolution of A100327, which equals the row sums of triangle A100326.

Vaclav Kotesovec, Table of n, a(n) for n = 0..650

Cf. A003168, A003169, A100324, A100326, A100327.

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

G.f.: (1+G003169(x))*G003169(x)/x, where G003169(x) is the g.f. of A003169.
Recurrence: 4*(n+1)*(2*n+1)*(17*n^2 - 28*n + 12)*a(n) = (1207*n^4 - 1988*n^3 + 1013*n^2 - 124*n - 12)*a(n-1) - 2*(n-2)*(2*n-3)*(17*n^2 + 6*n + 1)*a(n-2). - Vaclav Kotesovec, Jul 05 2014
a(n) ~ sqrt(95+393/sqrt(17)) * ((71+17*sqrt(17))/16)^n / (4*sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Jul 05 2014
From Peter Bala, Sep 08 2024: (Start)
a(n) = (2/n) * Sum_{k = 0..n} binomial(n+1, n-k-1)*binomial(2*n, k)*2^(n-k) for n >= 1.
a(n) = 4*Jacobi_P(n-1, 2, n+1, 3)/n for n >= 1. Cf. A003168. (End)

cp's OEIS Frontend

A003168 Number of blobs with 2n+1 edges.

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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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A219538 G.f. satisfies A(x) = 1 + x*A(x)^2*(1 + A(x))^2/2.

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A100328 Column 1 of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324, with leading zero omitted.

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A371655 G.f. satisfies A(x) = 1 + x * A(x) * (1 + A(x))^2.

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A371669 G.f. satisfies A(x) = 1 + x * A(x)^3 * (1 + A(x))^2/2.

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A219538 G.f. satisfies A(x) = 1 + xA(x)^2(1 + A(x))^2/2.