A038455 -id:A038455 - cp's OEIS frontend

A039619 Second column of Jabotinsky-triangle A038455 related to A006963.

1, 9, 107, 1650, 31594, 725592, 19471500, 598482000, 20742534576, 800575997760, 34059828307680, 1583808130195200, 79925022369273600, 4350478314951982080, 254086498336122950400, 15849890120755311667200

Offset: 2

Wolfdieter Lang

nonn

Explicit formula using partitions of n into 2 parts: cf. Knuth's paper for f(n,2), n >= 2, formula with f(k) as given above.

			G.f. = x^2 + 9*x^3 + 107*x^4 + 1650*x^5 + 31594*x^6 + 725592*x^7 + ...

D. E. Knuth, Convolution polynomials, Mathematica J. 2.1 (1992), no. 4, 67-78.
D. E. Knuth, Problem 11832,The American Mathematical Monthly, Vol. 122, No. 4 (April 2015), page 390.
David Callan, An identity for A039619.

Cf. A006963, A038455.

Cf. A039646.

Rest[Rest[CoefficientList[Series[Log[(1-Sqrt[1-4*x])/x/2]^2/2, {x, 0, 20}], x]* Range[0, 20]!]] (* Vaclav Kotesovec, Oct 07 2013 *)
a[ n_] := If[ n < 1, 0, Sum[ 1/k, {k, n + 1, 2 n - 1}] (2 n - 1)!/n!]; (* Michael Somos, May 11 2015 *)
a[ n_] := If[ n < 1, 0, Sum[ Times @@ Drop[ Range[ n - 1] + n, {k}], {k, n - 1}]]; (* Michael Somos, May 11 2015 *)

{a(n) = if( n<1, 0, sum(k=n+1, 2*n-1, 1/k) * (2*n-1)! / n!)}; /* Michael Somos, May 11 2015 */

a(n) = sum(binomial(n-1, j-1)*f(j)*f(n-j), j=1..n-1) with f(k) := A006963(k+1) = (2*k+1)!/k!, k >= 1.
E.g.f.: log((1-sqrt(1-4*x))/x/2)^2/2. - Vladeta Jovovic, May 02 2003
a(n) ~ log(2) * n^(n-1) * 2^(2*n-1/2) / exp(n). - Vaclav Kotesovec, Oct 07 2013
a(n) = (H(2*n-1) - H(n)) * (2*n-1)! / n! where H() is Harmonic numbers [Knuth 2015]. - Michael Somos, May 11 2015
a(n) = (n-1)!*sum(binomial(2n-1-k, n-1-k)/k,k=1..n-1) [Callan 2015]. - David Callan, May 17 2015

A039646 Third column of Jabotinsky-triangle A038455 related to A006963.

Original entry on oeis.org

1, 18, 335, 7155, 176554, 4985316, 159168428, 5681708100, 224518859136, 9737714177928, 460132506980640, 23537198603711520, 1296157111841533824, 76467514565810332800, 4812260962479036076800, 321826321845522830649600

Offset: 0

Wolfdieter Lang

nonn

Explicit formula for a(n-3) using partitions of n into 3 parts: cf. Knuth's paper for f(n,3) n >= 3, formula with f(k) := A006963(k+1) = (2*k-1)!/k!, k >= 1.

G. C. Greubel, Table of n, a(n) for n = 0..362
D. E. Knuth, Convolution polynomials, arXiv:math/9207221 [math.CA], 1992; Mathematica J. 2.1 (1992), no. 4, 67-78.

Cf. A038455, A006963, A039619.

Drop[With[{nmax = 20}, CoefficientList[Series[Log[(1 - Sqrt[1 - 4*x])/x/2]^3/6, {x, 0, nmax}], x]*Range[0, nmax]!], 3] (* G. C. Greubel, Dec 14 2017 *)

my(x='x+O('x^30)); Vec(serlaplace(log((1-sqrt(1-4*x))/x/2)^3/6)) \\ G. C. Greubel, Dec 14 2017

a(n) = Sum_{j=0..n} binomial(n+2, j)*A006936(j+2)*A039619(n+2-j).

E.g.f.: log((1-sqrt(1-4*x))/x/2)^3/6. - Vladeta Jovovic, May 02 2003

A001761 a(n) = (2*n)!/(n+1)!.

Original entry on oeis.org

1, 1, 4, 30, 336, 5040, 95040, 2162160, 57657600, 1764322560, 60949324800, 2346549004800, 99638080819200, 4626053752320000, 233153109116928000, 12677700308232960000, 739781100339240960000, 46113021921146019840000

Offset: 0

N. J. A. Sloane

nonn

According to the Beineke and Pippert paper, the number of dissections of a disk is given by D(n)=R(n)/(n-2)!, where R(n)=A001761(n-2) is the number of labeled planar 2-trees having n vertices and rooted at a given exterior edge. [Clarified by M. F. Hasler, Feb 22 2012]
a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left and middle child have a larger label than their parent. - Brian Drake, Jul 28 2008
For n>0: a(n) = A173333(2*n,n+1); cf. A006963, A001813. - Reinhard Zumkeller, Feb 19 2010

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted in Math. Annalen, Vol. 191 (1971), pp. 87-98.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Peter J. Cameron, Some treelike objects, Quart. J. Math. Oxford, Vol. 38, No. 2 (1987), pp. 155-183. See p. 166. - _N. J. A. Sloane_, Apr 18 2014
Ali Chouria, Vlad-Florin Drǎgoi, and Jean-Gabriel Luque, On recursively defined combinatorial classes and labelled trees, arXiv:2004.04203 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 80.
K. A. Penson and J.-M. Sixdeniers, Integral Representations of Catalan and Related Numbers, J. Integer Sequences, Vol. 4 (2001), Article 01.2.5.
Karol A. Penson and Allan I. Solomon, Coherent states from combinatorial sequences, arXiv:quant-ph/0111151, 2001.

Cf. A000108, A173333, A006963, A001813.

Main diagonal of A255982, A256061.

seq(mul((n+k), k=2..n), n=0..17); # Zerinvary Lajos, Feb 15 2008

Table[(2*n)!/(n+1)!,{n,0,20}] (* Vincenzo Librandi, Feb 23 2012 *)

combinat::catalan(n)*n! $ n = 0..17; // Zerinvary Lajos, Feb 15 2007

A001761(n)=binomial(2*n,n+1)*(n-1)!  \\ M. F. Hasler, Feb 23 2012

{Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
{a(n)=sum(k=0,n,(-1)^(n-k)*(n+1)^(k-1)*Stirling1(n,k))} \\ Paul D. Hanna, Nov 09 2012

[binomial(2*n,n)/(1+n)*factorial(n) for n in range(0, 18)] # Zerinvary Lajos, Dec 03 2009

a(n) = n!*Catalan(n) =n!* A000108(n). - N. J. A. Sloane, Apr 18 2014
a(n+2) = sum(A038455(n, m), m=1..n), n >= 1. - Wolfdieter Lang
E.g.f. for this sequence = o.g.f. for A000108. - Len Smiley, Dec 07 2001
Integral representation as the moment of a positive function on the positive half-axis: in Maple notation, a(n)=int(x^n*(-1/2+exp(-x/4)/sqrt(Pi*x)+erf(sqrt(x)/2)/2), x=0..infinity), n=0, 1... This representation is unique. - Karol A. Penson, Aug 21 2001
G.f.: If G_N(x)=1+sum('(2*k)!*(x^k)/(k+1)!', 'k'=1..N),  G_N(x)=1+2*x/(G(0)-2*x); G(k)=4*x*(k^2)+6*k*x+k+2*x+2-2*x*(2*k+3)*((k+2)^2)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
a(n) = Sum_{k=0..n} (-1)^(n-k) * (n+1)^(k-1) * Stirling1(n,k). - Paul D. Hanna, Nov 09 2012
G.f.: Q(0) where Q(k) =  1 + x*(2*k+1)*(4*k+1)/(k+1 - 4*x*(k+1)^2*(4*k+3)/(4*x*(k+1)*(4*k+3) + (2*k+3)/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 05 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x/(x + (k+2)/(2*k+2)/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
Let A(x) = sum(k>=0, a(k)*x^k /(2*k)! ) = ( exp(x)-1)/x, then A(x) = 1/Q(0), where Q(k) = 1 - x/( 1 + (2*k+1)/(1 - x/( 1 + 2*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 24 2013
From Ilya Gutkovskiy, Jan 21 2017: (Start)
a(n) ~ sqrt(2)*4^n*n^(n-1)/exp(n).
Sum_{n>=0} 1/a(n) = (7*exp(1/4)*sqrt(Pi)*erf(1/2) + 10)/8 = 2.2865189388213215..., where erf() is the error function. (End)
D-finite with recurrence: (n+1)*a(n) -2*n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Feb 16 2020
Sum_{n>=0} (-1)^n/a(n) = 3/4 - 5*sqrt(Pi)*erfi(1/2)/(8*exp(1/4)), where erfi() is the imaginary error function. - Amiram Eldar, Apr 03 2022

A143395 Triangle read by rows: T(n,k) = number of forests of k labeled rooted trees of height at most 1, with n labels, where any root may contain >= 1 labels, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 7, 9, 1, 0, 15, 55, 18, 1, 0, 31, 285, 205, 30, 1, 0, 63, 1351, 1890, 545, 45, 1, 0, 127, 6069, 15421, 7770, 1190, 63, 1, 0, 255, 26335, 116298, 95781, 24150, 2282, 84, 1, 0, 511, 111645, 830845, 1071630, 416451, 62370, 3990, 108, 1

Offset: 0

Alois P. Heinz, Aug 12 2008

This is the Sheffer triangle (1,exp(x)*(exp(x)-1)) (Jobotinsky type). See the e.g.f. given by V. Jovovic below, and the W. Lang link under A006232 (second part) for general Sheffer remarks and the conversion to the umbral notation of S. Roman's book. - Wolfdieter Lang, Oct 08 2011
From Peter Bala, Jan 07 2015: (Start)
T(n,k) counts the ways a set of size n can be partitioned into k nonempty blocks and then a nonempty subset chosen from each block. An example is given below.
This triangle is the particular case a = 1, b = 1, c = 0 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. A008277 is the case a = 1, b = 0, c = 0.
Define a polynomial sequence x_(n) by putting x_(0) = 1, x_(1) = x and for n >= 2 setting x_(n) = x*(x - (n+1))*(x - (n+2))*...*(x - (2*n-1)), that is, x_(n) = (-1)^(n+1)*n!*(x/(2*n - x))*binomial(2*n - x,n) for n >= 0. Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis polynomials x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k), n = 0,1,2,.... Examples are given below.
Matrix factorization: Let M be the infinite lower unit triangular array with (n,k)-th entry (2^(n+1-k)-1)*binomial(n,k). 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. It follows from the recurrence equation given in the Formula section that the infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to the present triangle (but with the first row and column omitted). See the Example section. (End)
The Bell transform of 2^(n+1)-1. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			T(3,2) = 9: {1}{2}<-3, {1}{3}<-2, {1}{2,3}, {2}{1}<-3, {2}{3}<-1, {2}{1,3}, {3}{1}<-2, {3}{2}<-1, {3}{1,2}.
Triangle begins:
  1;
  0,   1;
  0,   3,    1;
  0,   7,    9,     1;
  0,  15,   55,    18,    1;
  0,  31,  285,   205,   30,    1;
  0,  63, 1351,  1890,  545,   45,  1;
  0, 127, 6069, 15421, 7770, 1190, 63,  1;
  ...
From _Peter Bala_, Jan 07 2015: (Start)
T(4,2) = 55: There are 7 partitions of the set {1,2,3,4} into 2 blocks. For the 3 set partitions of the type {a,b}{c,d} we can choose a nonempty subset from each block in one of 3*3 ways giving 3*3*3 = 27 possibilities in all. The remaining 4 set partitions of {1,2,3,4} into 2 blocks are of the form {a,b,c}{d} and we can choose a nonempty subset from each block in 7*1 ways giving 4*7*1 = 28 possible choices. Thus in total T(4,2) = 27 + 28 = 55.
Recurrence equation example:
T(4,2) = sum {j = 1..3} (2^(4-j) - 1)*binomial(3,j)*T(j,1) = 7*3*1 + 3*3*3 + 1*1*7 = 55.
Connection constants:
Row 3 = [0, 7, 9, 1]. Hence x^3 = 7*x + 9*x*(x - 3) + x*(x - 4)*(x - 5); Row 4 = [0, 15, 55, 18, 1]. Hence x^4 = 15*x + 55*x*(x - 3) + 18*x*(x - 4)*(x - 5) + x*(x - 5)*(x - 6)*(x - 7).
With the array 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      |
| 7  6  1  ||0  3  1     ||0 0 1    |... = | 7  9  1   |
|15 21 9 1 ||0  7  6  1  ||0 0 3 1  |      |15 55 18 1 |
|...       ||0 15 21  9 1||0 0 7 6 1|      |...        |
|...       ||...         ||...      |      |           |
(End)

Alois P. Heinz, Rows n = 0..140, flattened
Eli Bagno, Riccardo Biagioli, and David Garber, Some identities involving second kind Stirling numbers of types B and D, arXiv:1901.07830 [math.CO], 2019.
Peter Bala, A 3 parameter family of generalized Stirling numbers
Takao Komatsu, Eli Bagno, and David Garber, A q,r-analogue of poly-Stirling numbers of second kind with combinatorial applications, arXiv:2209.06674 [math.CO], 2022.
Index entries for sequences related to rooted trees

Columns k=0-9: A000007, A000225, A016269(n-2), A028025(n-3), A143399, A143400, A143401, A143402, A143403, A143404.
Diagonal: A000012.
T(2*n,n) gives A383869.
See also A048993, A008277, A007318, A143405 for row sums.

[[(&+[Binomial(n,j)*StirlingSecond(j,k)*k^(n-j): j in [k..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019

T:= (n, k)-> add(binomial(n,t)*Stirling2(t,k)*k^(n-t), t=k..n):
seq(seq(T(n, k), k=0..n), n=0..11);

t[0, 0]=1; t[n_, k_]:= SeriesCoefficient[Exp[y*Exp[x]*(Exp[x]-1)], {x, 0, n}, {y, 0, k}]*n!; Table[t[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Dec 05 2013, after Vladeta Jovovic *)
Table[If[n==k==0, 1, If[k==0, 0, Sum[Binomial[n, j]*StirlingS2[j, k]* k^(n-j), {j,k,n}]]], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 07 2019 *)

{T(n,k) = sum(j=k, n, binomial(n,j)*stirling(j,k,2)*k^(n-j))};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Mar 07 2019

# uses[bell_matrix from A264428]
bell_matrix(lambda n: 2^(n+1)-1, 10) # Peter Luschny, Jan 18 2016

G.f. for column k: x^k/Product_{t=k..2*k} (1-t*x).
T(n,k) = Sum_{t=k..n} C(n,t) * Stirling2(t,k) * k^(n-t).
E.g.f.: exp(y*exp(x)*(exp(x)-1)). - Vladeta Jovovic, Dec 08 2008
T(n,k) = Sum_{m=0..k} Stirling2(n,k+m)*(k+m)!/(m!*(k-m)!). - Vladimir Kruchinin, Apr 06 2011
Let P be Pascal's triangle A007318. The first column of the array exp(t*(P^2-P)) gives the row generating polynomials of this triangle.
The row polynomials R(n,t) satisfy the recurrence R(n+1,t) = t*(Sum_{k = 0..n} (2^(k+1)-1)*C(n,k)*R(n-k,t)) with R(0,t) = 1. For example, the row 4 polynomial R(4,t) = 15*t + 55*t^2 + 18*t^3 + t^4 = t*((7*t + 9*t^2 + t^3) + 3*3*(3*t+t^2) + 7*3*t + 15*1). - Peter Bala, Oct 12 2011
From Peter Bala, Jan 07 2015: (Start)
T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(j + k)^n.
Recurrence equation: T(n+1,k+1) = Sum_{j = k..n} (2^(n-j+1) - 1)*binomial(n,j)*T(j,k) with T(0,0) = 1 and T(n,0) = 0 for n >= 1. This leads to the matrix factorization noted in the Comments section.
The inverse array is a signed version of A038455. (End)

A257635 Triangle with n-th row polynomial equal to Product_{k = 1..n} (x + n + k).

Original entry on oeis.org

1, 2, 1, 12, 7, 1, 120, 74, 15, 1, 1680, 1066, 251, 26, 1, 30240, 19524, 5000, 635, 40, 1, 665280, 434568, 117454, 16815, 1345, 57, 1, 17297280, 11393808, 3197348, 495544, 45815, 2527, 77, 1, 518918400, 343976400, 99236556, 16275700, 1659889, 107800, 4354, 100, 1

Offset: 0

Peter Bala, Nov 05 2015

The row polynomials are a Sheffer sequence. For the associated polynomial sequence of binomial type see A038455.

			Triangle begins:
[0]       1;
[1]       2,      1;
[2]      12,      7,      1;
[3]     120,     74,     15,     1;
[4]    1680,   1066,    251,    26,    1;
[5]   30240,  19524,   5000,   635,   40,  1;
[6]  665280, 434568, 117454, 16815, 1345, 57, 1;
  ...

R. Sprugnoli, An Introduction to Mathematical Methods in Combinatorics, Section 5.6 CreateSpace Independent Publishing Platform 2006, ISBN-13: 978-1502925244.
Wikipedia, Sheffer sequence

Cf. A001813 (column 0), A005449 (first subdiagonal), A098118 (column 1).
Cf. A006963 (row sums), A000407 (alternating row sum).
Cf. A054649, A000108, A000984, A038455, A092932.

seq(seq(coeff(product(n + x + k, k = 1 .. n), x, i), i = 0..n), n = 0..8);
# Alternative:
p := n -> n!*hypergeom([-n, -x + n], [-n], 1):
seq(seq((-1)^k*coeff(simplify(p(n)), x, k), k=0..n), n=0..6); # Peter Luschny, Nov 27 2021

p[n_, x_] := FunctionExpand[Gamma[2*n + x + 1] / Gamma[n + x + 1]];
Table[CoefficientList[p[n, x], x], {n,0,8}] // Flatten (* Peter Luschny, Mar 21 2022 *)

E.g.f.: A(x,t) = B(t)*C(t)^x = 1 + (2 + x)*t + (3 + x)*(4 + x)*t^2/2! + (4 + x)*(5 + x)*(6 + x)*t^3/3! + ..., where B(t) = 1/sqrt(1 - 4*t) is the o.g.f. for A000984 and C(t) = (1 - sqrt(1 - 4*t))/(2*t) is the o.g.f. for A000108.
n-th row polynomial: n!*binomial(2*n + x,n).
T(n, k) = (-1)^k*n!*[x^k] hypergeom([-n, -x + n], [-n], 1). - Peter Luschny, Nov 27 2021
T(n, k) = [x^k] Gamma(2*n + x + 1) / Gamma(n + x + 1). - Peter Luschny, Mar 21 2022

cp's OEIS Frontend

A039619 Second column of Jabotinsky-triangle A038455 related to A006963.

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A039646 Third column of Jabotinsky-triangle A038455 related to A006963.

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A001761 a(n) = (2*n)!/(n+1)!.

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A143395 Triangle read by rows: T(n,k) = number of forests of k labeled rooted trees of height at most 1, with n labels, where any root may contain >= 1 labels, n >= 0, 0 <= k <= n.

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A257635 Triangle with n-th row polynomial equal to Product_{k = 1..n} (x + n + k).

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