A051133 -id:A051133 - cp's OEIS frontend

A002802 a(n) = (2n+3)!/(6n!*(n+1)!).

1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600

Offset: 0

N. J. A. Sloane

For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001
a(n) is half the number of (n+2) X 2 Young tableaux with a three horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, and A002457 for two. - Michael Wallner, Jan 31 2022
From Robert Coquereaux, Feb 12 2024: (Start)
Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.
When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).
Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.
The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)

			G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...

C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See pp. 4, 12.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023. See pp. 9, 19.
R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv preprint arXiv:1306.4628 [math.CO], 2013.
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
Alain Goupil and Gilles Schaeffer, Factoring N-Cycles and Counting Maps of Given Genus, Europ. J. Combinatorics (1998) 19 819-834.
T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. I, J. Comb. Theory B 13 (1972), 192-218 (Tab.1).
Liang Zhao and Fengyao Yan, Note on Total Positivity for a Class of Recursive Matrices, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5.
Notes.

Cf. A035309, A000108 (for genus 0 maps), A046521 (third column).
Cf. A000984, A000911, A002457, A002697, A051133.
Column g=1 of A370235.

F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # G. C. Greubel, Jul 20 2019

F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019

seq(simplify(4^n*hypergeom([-n,-3/2], [1], 1)),n=0..25); # Peter Luschny, Apr 26 2016

Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 25}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

{a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */

{a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */

f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019

G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).
Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).
a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004
a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004
a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007
From Rui Duarte, Oct 08 2011: (Start)
Also convolution of A000984 with A002697, also convolution of A000302 with A002457.
a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).
a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).
a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).
a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)
D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014
a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018
Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - Amiram Eldar, Oct 13 2020
E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - Robert Coquereaux, Feb 12 2024

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 3, 12, 10, 0, 4, 30, 60, 35, 0, 5, 60, 210, 280, 126, 0, 6, 105, 560, 1260, 1260, 462, 0, 7, 168, 1260, 4200, 6930, 5544, 1716, 0, 8, 252, 2520, 11550, 27720, 36036, 24024, 6435, 0, 9, 360, 4620, 27720, 90090, 168168, 180180, 102960, 24310

Offset: 0

Peter Luschny, Mar 20 2015

The rows give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1) / (n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022

This is related to the cluster fans of type B (see Fomin and Zelevinsky reference) - F. Chapoton, Nov 17 2022.

			[1]
[0, 1]
[0, 2,   3]
[0, 3,  12,   10]
[0, 4,  30,   60,   35]
[0, 5,  60,  210,  280,  126]
[0, 6, 105,  560, 1260, 1260,  462]
[0, 7, 168, 1260, 4200, 6930, 5544, 1716]
.
R_0(x) = 1/(x-1)^0.
R_1(x) = 0/(x-1)^1 + 1/(x-1)^2.
R_2(x) = 0/(x-1)^2 + 2/(x-1)^3 + 3/(x-1)^4.
R_3(x) = 0/(x-1)^3 + 3/(x-1)^4 + 12/(x-1)^5 + 10/(x-1)^6.
Then k!*[x^k] R_n(x) is A001286(k+2) and A001754(k+3) for n = 2, 3 respectively.
.
Seen as an array A(n, k) = binomial(n + k, k)*binomial(n + 2*k - 1, n + k):
[0] 1, 1,   3,   10,    35,    126,     462, ...
[1] 0, 2,  12,   60,   280,   1260,    5544, ...
[2] 0, 3,  30,  210,  1260,   6930,   36036, ...
[3] 0, 4,  60,  560,  4200,  27720,  168168, ...
[4] 0, 5, 105, 1260, 11550,  90090,  630630, ...
[5] 0, 6, 168, 2520, 27720, 252252, 2018016, ...
[6] 0, 7, 252, 4620, 60060, 630630, 5717712, ...

Seiichi Manyama, Rows n = 0..139, flattened
F. Chapoton Enumerative Properties of Generalized Associahedra, Sém. Loth. Comb. B51b (2004).
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
Mark Dukes and Chris D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, Electronic Journal Of Combinatorics, 23(1) (2016), #P1.45.
S. Fomin and A. Zelevinsky Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3.
Yunlong Shen and Lixin Shen, Orthogonal Fourier-Mellin moments for invariant pattern recognition, J. Opt. Soc. Am. A 11 (6) (1994) p 1748-1757, eq. (6).

T(n, n) = C(2*n-1, n) = A001700(n-1).
T(n, n-1) = A005430(n-1) for n >= 1.
T(n, n-2) = A051133(n-2) for n >= 2.
T(n, 2) = A027480(n-1) for n >= 2.
T(2*n, n) = A208881(n) for n >= 0.
A002002 (row sums).
Cf. A000142, A001286, A001754, A001755, A001777, A182626, A266732 (row k=3), A008316, A033282, A063007, A345013, A269944.

T_row := proc(n) local egf, k, F, t;
if n=0 then RETURN(1) fi;
egf := (x/(1-x))^n/n!; t := diff(egf,[x$n]);
F := convert(t,parfrac,x);
# print(seq(k!*coeff(series(F,x,20),x,k),k=0..7));
# gives A000142, A001286, A001754, A001755, A001777, ...
seq(coeff(F,(x-1)^(-k)),k=n..2*n) end:
seq(print(T_row(n)),n=0..7);
# 2nd version by R. J. Mathar, Dec 18 2016:
A253283 := proc(n,k)
    binomial(n,k)*binomial(n+k-1,k-1) ;
end proc:

Table[Binomial[n, k] Binomial[n + k - 1, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)

T(n,k) = binomial(n,k)*binomial(n+k-1,k-1);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Apr 29 2018

The exponential generating functions for the rows of the square array L(n,k) = ((n+k)!/n!)*C(n+k-1,n-1) (associated to the unsigned Lah numbers) are given by R_n(x) = Sum_{k=0..n} T(n,k)/(x-1)^(n+k).
T(n,k) = C(n,k)*C(n+k-1,k-1).
Sum_{k=0..n} T(n,k) = (-1)^n*hypergeom([-n,n],[1],2) = (-1)^n*A182626(n).
Row generating function: Sum_{k>=1} T(n,k)*z^k = z*n* 2F1(1-n,n+1 ; 2; -z). - R. J. Mathar, Dec 18 2016
From Peter Bala, Feb 22 2017: (Start)
G.f.: (1/2)*( 1 + (1 - t)/sqrt(1 - 2*(2*x + 1)*t + t^2) ) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n,x) = (1/2)*(LegendreP(n, 2*x + 1) - LegendreP(n-1, 2*x + 1)) for n >= 1.
The row polynomials are the black diamond product of the polynomials x^n and x^(n+1) (see Dukes and White 2016 for the definition of this product).
exp(Sum_{n >= 1} R(n,x)*t^n/n) = 1 + x*t + x*(1 + 2*x)*t^2 + x*(1 + 5*x + 5*x^2)*t^3 + ... is a g.f. for A033282, but with a different offset.
The polynomials P(n,x) := (-1)^n/n!*x^(2*n)*(d/dx)^n(1 + 1/x)^n begin 1, 3 + 2*x , 10 + 12*x + 3*x^2, ... and are the row polynomials for the row reverse of this triangle. (End)
Let Q(n, x) = Sum_{j=0..n} (-1)^(n - j)*A269944(n, j)*x^(2*j - 1) and P(x, y) = (LegendreP(x, 2*y + 1) - LegendreP(x-1, 2*y + 1)) / 2 (see Peter Bala above). Then n!*(n - 1)!*[y^n] P(x, y) = Q(n, x) for n >= 1. - Peter Luschny, Oct 31 2022
From Peter Bala, Apr 18 2024: (Start)
G.f.: Sum_{n >= 0} binomial(2*n-1, n)*(x*t)^n/(1 - t)^(2*n) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n, x) = [t^n] ( (1 - t)/(1 - (1 + x)*t) )^n.
It follows that for integer x, the sequence {R(n, x) : n >= 0} satisfies the Gauss congruences: R(n*p^r, x) == R(n*p^(r-1), x) (mod p^r) for all primes p and positive integers n and r.
R(n, -2) = (-1)^n * A002003(n) for n >= 1.
R(n, 3) = A299507(n). (End)

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).

Original entry on oeis.org

3, 6, 15, 60, 105, 21, 126, 360, 495, 330, 429, 1092, 1365, 420, 1020, 2448, 2907, 1710, 1995, 4620, 5313, 759, 3450, 7800, 8775, 4914, 5481, 12180, 13485, 3720, 8184, 17952, 19635, 10710, 11655, 25308, 27417, 3705, 15990, 34440, 37023, 19866, 21285, 45540

Offset: 0

Paul Curtz, Apr 18 2014

All terms are multiples of 3.
Difference table of c(n):
   1/3,     1/6,    2/15,    7/60,    2/21,...
  -1/6,    -1/30,  -1/60,   -1/84,   -1/105,...
   2/15,    1/60,   1/210,   1/420,   1/630,...
  -7/60,   -1/84,  -1/420,  -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
c(n) is a companion to A026741(n)/A045896(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
No: a(5) = 21 but a(45) = 12972. # Robert Israel, Jul 17 2023
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
  1,    1/2,  1/3,    1/4,     1/5,     1/6,   ...
  0,    1/6,  1/6,    3/20,    2/15,    5/42,  ...
  1/3,  1/6,  2/15,   7/60,   11/105,   2/21,  ...
  0,    1/10, 1/10,  13/140,   3/35,    5/63,  ...
  1/5,  1/10, 3/35,  11/140,  23/315,  43/630, ...
  0,    1/14, 1/14,  17/252,   4/63,   ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
Second row: A026741/A045896.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .

Robert Israel, Table of n, a(n) for n = 0..10000

seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))),n=0..1000);

Denominator[Table[(n^2+n+2)/Times@@(n+{1,2,3}),{n,0,50}]] (* Harvey P. Dale, Mar 27 2015 *)

for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014

c(n) = A014206(n)/A007531(n+3).
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10*A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023

More terms from Colin Barker, Apr 18 2014

A000911 a(n) = (2n+3)! /( n! * (n+1)! ).

Original entry on oeis.org

6, 60, 420, 2520, 13860, 72072, 360360, 1750320, 8314020, 38798760, 178474296, 811246800, 3650610600, 16287339600, 72129646800, 317370445920, 1388495700900, 6044040109800, 26190840475800, 113034153632400, 486046860619320, 2083057974082800, 8900338616535600

Offset: 0

N. J. A. Sloane

nonn

			6 + 60*x + 420*x^2 + 2520*x^3 + 13860*x^4 + 72072*x^5 + 360360*x^6 + ...

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 99.

T. D. Noe, Table of n, a(n) for n = 0..200

Cf. A000217, A000984, A001801, A002802, A051133, A086466, A093602.

seq(binomial(2*n,n)*binomial(n,(n-2)), n=2..21); # Zerinvary Lajos, May 10 2007

Table[(2 n + 3)!/(n!*(n + 1)!), {n, 0, 20}] (* T. D. Noe, Jun 20 2012 *)

a(n) = 2^(n+4)*polcoeff(pollegendre(n+4),n) /* Ralf Stephan */

a(n) = 2 * A051133(n+1).
a(n) = A000984(n+1)*A000217(n). - Zerinvary Lajos, May 10 2007
a(n) = 6 * A002802(n). - Zerinvary Lajos, Jun 02 2007
n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jun 07 2013
G.f.: 6*(1+10*x/( G(0)- 10*x)), where G(k)= 2*x*(2*k+5) + k + 1 - 2*x*(k+1)*(2*k+7)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 14 2013
Sum_{n>=0} (-1)^n/a(n) = 5*A086466-2 = 2*log(phi)*sqrt(5)-2 = 0.1520447... - Jean-François Alcover, Apr 22 2014
From Ilya Gutkovskiy, Jan 31 2017: (Start)
G.f.: 6/(1 - 4*x)^(5/2).
a(n) ~ 2^(2*n+3)*n^(3/2)/sqrt(Pi). (End)
Sum_{n>=0} 1/a(n) = 2 - Pi/sqrt(3) = 2 - A093602. - Amiram Eldar, Oct 13 2020

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A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!).

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A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

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A241269 Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).

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A000911 a(n) = (2n+3)! /( n! * (n+1)! ).

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A171075 Total number of balls on lawn in variant of the tennis ball problem (cf. A171074).

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A171076 Total number of balls in room in variant of the tennis ball problem (cf. A171074).

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A002802 a(n) = (2n+3)!/(6n!*(n+1)!).

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).