A165817 -id:A165817 - cp's OEIS frontend

A025174 a(n) = binomial(3n-1, n-1).

0, 1, 5, 28, 165, 1001, 6188, 38760, 245157, 1562275, 10015005, 64512240, 417225900, 2707475148, 17620076360, 114955808528, 751616304549, 4923689695575, 32308782859535, 212327989773900, 1397281501935165, 9206478467454345, 60727722660586800, 400978991944396320

Offset: 0

Wouter Meeussen, Emeric Deutsch

Number of standard tableaux of shape (2n-1,n). Example: a(2)=5 because in the top row we can have 123, 124, 125, 134, or 135. - Emeric Deutsch, May 23 2004
Number of peaks in all generalized {(1,2),(1,-1)}-Dyck paths of length 3n.
Positive terms in this sequence are the numbers k such that k and 2k are consecutive terms in a row of Pascal's triangle. 1001 is the only k such that k, 2k, and 3k are consecutive terms in a row of Pascal's triangle. - J. Lowell, Mar 11 2023

			L.g.f.: L(x) = x + 5*x^2/2 + 28*x^3/3 + 165*x^4/4 + 1001*x^5/5 + 6188*x^6/6 + ...
where G(x) = exp(L(x)) satisfies G(x) = 1 + x*G(x)^3, and begins:
exp(L(x)) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... + A001764(n)*x^n + ...

B. C. Berndt, Ramanujan's Notebooks Part I, Springer-Verlag, see Entry 14, Corollary 1, p. 71.

Robert Israel, Table of n, a(n) for n = 0..1190
Paul Barry, On the Central Antecedents of Integer (and Other) Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.3.
D. Kruchinin and V. Kruchinin, A Generating Function for the Diagonal T2n,n in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
Stepan Orevkov, Asymptotics of the number of lattice triangulations of rectangles of width 4 and 5, arXiv:2412.17065 [math.CO], 2024. See p. 16.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A001764 (binomial(3n,n)/(2n+1)), A117671 (C(3n+1,n+1)), A004319, A005809, A006013, A013698, A045721, A117671, A165817, A224274, A236194.

Cf. A000108, A001045, A005156, A224274.

[Binomial(3*n-1,n-1): n in [0..30]]; // Vincenzo Librandi, Nov 12 2014

with(combinat):seq(numbcomp(3*i,i), i=0..20); # Zerinvary Lajos, Jun 16 2007

Table[ GegenbauerC[ n, n, 1 ]/2, {n, 0, 24} ]
Join[{0},Table[Binomial[3n-1,n-1],{n,20}]] (* Harvey P. Dale, Oct 19 2022 *)
nmax=23; CoefficientList[Series[(2+HypergeometricPFQ[{1/3,2/3},{1/2,1},27x/4])/3-1,{x,0,nmax}],x]Range[0,nmax]! (* Stefano Spezia, Dec 31 2024 *)

vector(30, n, n--; binomial(3*n-1, n-1)) \\ Altug Alkan, Nov 04 2015

G.f.: z*g^2/(1-3*z*g^2), where g=g(z) is given by g=1+z*g^3, g(0)=1, that is, (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) = Sum_{k=0..n} ((3k+1)/(2n+k+1))C(3n, 2n+k)*A001045(k). - Paul Barry, Oct 07 2005
Hankel transform of a(n+1) is A005156(n+1). - Paul Barry, Apr 14 2008
G.f.: x*B'(x)/B(x) where B(x) is the g.f. of A001764. - Vladimir Kruchinin Feb 03 2013
D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Feb 05 2013
Logarithmic derivative of A001764; g.f. of A001764 satisfies G(x) = 1 + x*G(x)^3. - Paul D. Hanna, Jul 14 2013
G.f.: (2*cos((1/3)*arcsin((3/2)*sqrt(3*x)))-sqrt(4-27*x))/(3*sqrt(4-27*x)). - Emanuele Munarini, Oct 14 2014
a(n) = Sum_{k=1..n} binomial(n-1,n-k)*binomial(2*n,n-k). - Vladimir Kruchinin, Nov 12 2014
a(n) = [x^n] C(x)^n for n >= 1, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function for A000108 (Ramanujan). - Peter Bala, Jun 24 2015
From Peter Bala, Nov 04 2015: (Start)
Without the initial term 0, the o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. g(x)^2 is the o.g..f for A006013. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
G.f.: ( 2F1(1/3,2/3;1/2;27*x/4)-1)/3. - R. J. Mathar, Jan 27 2020
O.g.f. without the initial term 0, in the form g(x)=(2*cos(arcsin((3*sqrt(3)*sqrt(x))/2)/3)/sqrt(4-27*x)-1)/(3*x), satisfies the following algebraic equation: 1+(9*x-1)*g(x)+x*(27*x-4)*g(x)^2+x^2*(27*x-4)*g(x)^3=0. - Karol A. Penson, Oct 11 2021
O.g.f. equals f(x)/(1 - 2*f(x)), where f(x) = series reversion (x/(1 + x)^3) = x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the o.g.f. of A001764 with the initial term omitted. Cf. A224274. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/2)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+2)*n+k-1,k) = C(3*n-1,n-1) and (1/3)*Sum_{k = 0..n} (-1)^k* C(x*n,n-k)*C((x-3)*n+k-1,k) = C(3*n-1,n-1), both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
a(n) ~ 2^(-2*n)*3^(3*n)/(2*sqrt(3*n*Pi)). - Stefano Spezia, Apr 25 2024
a(n) = Sum_{k = 0..n-1} binomial(2*n+k-1, k) = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(3*n, k). - Peter Bala, Jul 21 2024
E.g.f.: (2 + hypergeom([1/3, 2/3], [1/2, 1], 27*x/4))/3 - 1. - Stefano Spezia, Dec 31 2024

A045721 a(n) = binomial(3*n+1,n).

Original entry on oeis.org

1, 4, 21, 120, 715, 4368, 27132, 170544, 1081575, 6906900, 44352165, 286097760, 1852482996, 12033222880, 78378960360, 511738760544, 3348108992991, 21945588357420, 144079707346575, 947309492837400, 6236646703759395

Offset: 0

Emeric Deutsch

Number of leaves in all noncrossing rooted trees on n nodes on a circle.
Number of standard tableaux of shape (n-1,1^(2n-3)). - Emeric Deutsch, May 25 2004
a(n) = number of Dyck (2n-3)-paths with exactly one descent of odd length. For example, a(3) counts all 5 Dyck 3-paths except UDUDUD. - David Callan, Jul 25 2005
a(n+2) gives row sums of A119301. - Paul Barry, May 13 2006
a(n) is the number of paths avoiding UU from (0,0) to (3n,n) and taking steps from {U,H}. - Shanzhen Gao, Apr 15 2010
Central coefficients of triangle A078812. - Vladimir Kruchinin, May 10 2012
Row sums of A252501. - L. Edson Jeffery, Dec 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjic, Two Enumerative Functions
D. Kruchinin and V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequence, Vol. 15 (2012), article 12.9.3.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
Index entries for sequences related to rooted trees

Cf. A252501, A263134 (partial sums), A001764, A004319, A005809, A006013, A013698, A025174, A117671, A165817, A236194.

[Binomial(3*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

[seq( binomial(3*n+1,n),n=0..40)]; # N. J. A. Sloane, Jun 09 2007

Table[Binomial[3 n + 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

a(n)=binomial(3*n+1,n) \\ Charles R Greathouse IV, Mar 18 2014

a(n) is asymptotic to c/sqrt(n)*(27/4)^n with c=0.73... - Benoit Cloitre, Jan 27 2003
G.f.: gz^2/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n+2) = C(3n+1,n) = Sum_{k=0..n} C(3n-k,n-k). - Paul Barry, May 13 2006
a(n+2) = C(3n+1,2n+1) = A078812(2n,n). - Paul Barry, Nov 09 2006
G.f.: A(x)=(2*cos(asin((3^(3/2)*sqrt(x))/2)/3)* sin(asin((3^(3/2)* sqrt(x))/2)/3))/(sqrt(3)*sqrt(1-(27*x)/4)*sqrt(x)). - Vladimir Kruchinin, Jun 10 2012
From Peter Luschny, Sep 04 2012: (Start)
O.g.f.: hypergeometric2F1([2/3, 4/3], [3/2], x*27/4).
a(n) = (n+1)*hypergeometric2F1([-2*n, -n], [2], 1).  (End)
D-finite with recurrence 2*n*(2*n+1)*a(n) - 3*(3*n-1)*(3*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = Sum_{r=0..n} C(n,r) * C(2*n+1,r). - J. M. Bergot, Mar 18 2014
From Peter Bala, Nov 04 2015: (Start)
a(n) = Sum_{k = 0..n} 1/(2*k + 1)*binomial(3*n - 3*k,n - k)*binomial(3*k, k).
O.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*(n+1)). - Ilya Gutkovskiy, Oct 10 2017
O.g.f.: (i/24)*((4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(2/3) - (4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(2/3))*sqrt(3)*8^(1/3)*sqrt(4 - 27*z)/(sqrt(z)*(-4 + 27*z)), where i = sqrt(-1). - Karol A. Penson, Dec 13 2023
a(n-1) = (1/(4*n))*binomial(2*n, n)^2 * (1 - 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 - 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + - ...) for n >= 1. Cf. A112029. - Peter Bala, Aug 08 2024

Simpler definition from Ira M. Gessel, May 26 2007. This change means that most of the offsets in the comments will now need to be changed too.

A262977 a(n) = binomial(4*n-1,n).

Original entry on oeis.org

1, 3, 21, 165, 1365, 11628, 100947, 888030, 7888725, 70607460, 635745396, 5752004349, 52251400851, 476260169700, 4353548972850, 39895566894540, 366395202809685, 3371363686069236, 31074067324187580, 286845713747883300, 2651487106659130740, 24539426037817994160

Offset: 0

Vladimir Kruchinin, Oct 06 2015

From Gus Wiseman, Sep 28 2022: (Start)
Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:
  (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)
         (2,2,4)  (2,1,4,1)  (1,1,2,1,3)
         (3,2,3)  (3,1,3,1)  (1,1,3,1,2)
         (4,2,2)  (4,1,2,1)  (1,1,4,1,1)
         (5,2,1)  (5,1,1,1)  (2,1,1,1,3)
                             (2,1,2,1,2)
                             (2,1,3,1,1)
                             (3,1,1,1,2)
                             (3,1,2,1,1)
                             (4,1,1,1,1)
The following pertain to this interpretation:
- The case of partitions is A000712, reverse A006330.
- Allowing any alternating sum gives A013777 (compositions of 4n).
- A011782 counts compositions of n.
- A034871 counts compositions of 2n with alternating sum 2k.
- A097805 counts compositions by alternating (or reverse-alternating) sum.
- A103919 counts partitions by sum and alternating sum (reverse: A344612).
- A345197 counts compositions by length and alternating sum.
(End)

V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.
Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.
Cf. A088218, A163455, A165817.

[Binomial(4*n-1,n): n in [0..20]]; // Vincenzo Librandi, Oct 06 2015

Table[Binomial[4 n - 1, n], {n, 0, 40}] (* Vincenzo Librandi, Oct 06 2015 *)

B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,20);

a(n) = binomial(4*n-1,n); \\ Michel Marcus, Oct 06 2015

G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.
a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).
a(n) = 3*A224274(n), for n > 0. - Michel Marcus, Oct 12 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A005810 (k = 0), A052203 (k = 1), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - Gus Wiseman, Sep 28 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-3*n, n).
a(n) = hypergeom([1 - 3*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)
a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 16 2024
G.f.: 1/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378

Offset: 0

Fabian Rothelius, Feb 04 2001

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018

R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, (referring to A158498 before recycling)
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, and C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv:1502.06553 [math.RT], 2015.
Dennis Walsh, Notes on finite monotonic and non-monotonic functions

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.
A000984, A001700, A001791, A002054, A002694, A004319, A005809, A025174, A045721, A088218, A165817, A054977, A165257, A000027, A000217, A006134 (trace of the symmetric Pascal matrix).

Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018

a059481 n k = a059481_tabl !! n !! n
a059481_row n = a059481_tabl !! n
a059481_tabl = map reverse a100100_tabl
-- Reinhard Zumkeller, Jan 15 2014

&cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011

for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009

t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
(* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */

{T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017

{T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */

[[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013

Edited by M. F. Hasler, Mar 05 2017

A004319 a(n) = binomial(3*n, n - 1).

Original entry on oeis.org

1, 6, 36, 220, 1365, 8568, 54264, 346104, 2220075, 14307150, 92561040, 600805296, 3910797436, 25518731280, 166871334960, 1093260079344, 7174519270695, 47153358767970, 310325523515700, 2044802197953900, 13488561475572645, 89067326568860640, 588671286046028640

Offset: 1

N. J. A. Sloane

Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

Seiichi Manyama, Table of n, a(n) for n = 1..1000
Milan Janjic, Two Enumerative Functions.
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.

Cf. A045721 (k=1), A025174 (k=2), A236194 (k=4), A013698 (k=5), A165817 (k=-1), A117671 (k=-2).

Cf. A000139, A001764, A005809, A006013, A236194, A001791, A004331, A004343, A004356, A004369, A004382.

A004319 := proc(n)
binomial(3*n,n-1);
end proc: # R. J. Mathar, Aug 10 2015

Table[Binomial[3n, n - 1], {n, 20}] (* Harvey P. Dale, Sep 21 2011 *)

a(n):=sum((binomial(3*i-1,2*i-1)*binomial(3*n-3*i-3,2*n-2*i-2))/(2*n-2*i-1),i,1,n-1)/2; /* Vladimir Kruchinin, May 15 2013 */

vector(30, n, binomial(3*n, n-1)) \\ Altug Alkan, Nov 04 2015

G.f.: (g-1)/(1-3*z*g^2), where g = g(z) is given by g = 1 + z*g^3, g(0) = 1, i.e. (in Maple notation), g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) = Sum_{i=0..n-1} binomial(i+2*n, i). - Ralf Stephan, Jun 03 2005
D-finite with recurrence -2*(2*n+1)*(n-1)*a(n) + 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = (1/2) * Sum_{i=1..n-1} binomial(3*i - 1, 2*i - 1)*binomial(3*n - 3*i - 3, 2*n - 2*i - 2)/(2*n - 2*i - 1). - Vladimir Kruchinin, May 15 2013
G.f.: x*hypergeom2F1(5/3, 4/3; 5/2; 27x/4). - R. J. Mathar, Aug 10 2015
a(n) = n*A001764(n). - R. J. Mathar, Aug 10 2015
From Peter Bala, Nov 04 2015: (Start)
With offset 0, the o.g.f. equals f(x)*g(x)^3, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k, n). See the cross-references. (End)
G.f.: cos(t)/(2*sqrt(1 - (27*x)/4)) - sin(t)/(sqrt(3)*sqrt(x)), where t = arcsin((sqrt(27*x))/2)/3. - Vladimir Kruchinin, May 13 2016
a(n) = [x^(2*n+1)] 1/(1 - x)^n. - Ilya Gutkovskiy, Oct 10 2017
a(n) = binomial(n+1, 2) * A000139(n). - F. Chapoton, Feb 23 2024

A013698 a(n) = binomial(3*n+2, n-1).

Original entry on oeis.org

1, 8, 55, 364, 2380, 15504, 100947, 657800, 4292145, 28048800, 183579396, 1203322288, 7898654920, 51915526432, 341643774795, 2250829575120, 14844575908435, 97997533741800, 647520696018735, 4282083008118300

Offset: 1

Joachim.Rosenthal(AT)nd.edu (Joachim Rosenthal), Emeric Deutsch

Degree of variety K_{2,n}^1. Also number of double-rises (or odd-level peaks) in all generalized {(1,2),(1,-1)}-Dyck paths of length 3(n+1).
Number of dissections of a convex (2n+2)-gon by n-2 noncrossing diagonals into (2j+2)-gons, 1<=j<=n-1.
a(n) is the number of lattice paths from (0,0) to (3n+1,n-1) avoiding two consecutive up-steps. - Shanzhen Gao, Apr 20 2010

Robert Israel, Table of n, a(n) for n = 1..1000
M. S. Ravi et al., Dynamic Pole Assignment and Schubert Calculus, SIAM J. Control Optimization, 34 (1996), 813-832, esp. p. 825.
Daniel W. Stasiuk, An Enumeration Problem for Sequences of n-ary Trees Arising from Algebraic Operads, Master's Thesis, University of Saskatchewan-Saskatoon (2018).

Cf. A013699 (q=2), A013700 (q=3), A013701 (q=4), A013702 (q=5).

A column of triangle A102537. Cf. A001764, A004319, A005809, A006013, A025174, A045721, A117671, A165817, A236194.

List([1..25], n-> Binomial(3*n+2, n-1)) # G. C. Greubel, Mar 21 2019

[Binomial(3*n+2, n-1): n in [1..25]]; // Vincenzo Librandi, Aug 10 2015

seq(binomial(3*n+2,n-1), n=0..30); # Robert Israel, Aug 09 2015

Table[Binomial[3*n+2, n-1], {n, 25}] (* Arkadiusz Wesolowski, Apr 02 2012 *)

first(m)=vector(m,n,binomial(3*n+2, n-1)); /* Anders Hellström, Aug 09 2015 */

[binomial(3*n+2, n-1) for n in (1..25)] # G. C. Greubel, Mar 21 2019

G.f.: g/((g-1)^3*(3*g-1)) where g*(1-g)^2 = x. - Mark van Hoeij, Nov 09 2011
a(n) = Sum_{k=0..n-1} binomial(2*n+k+2,k). - Arkadiusz Wesolowski, Apr 02 2012
D-finite with recurrence 2*(2*n+3)*(n+1)*a(n) -n*(67*n+34)*a(n-1) +30*(3*n-1)*(3*n-2)*a(n-2)=0. - R. J. Mathar, Feb 05 2013
a(n+1) = (3*n+5)*(3*n+4)*(3*n+3)*a(n)/((2*n+5)*(2*n+4)*n). - Robert Israel, Aug 09 2015
With offset 0, the o.g.f. equals f(x)*g(x)^5, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A165817 (k = -1), A117671 (k = -2). - Peter Bala, Nov 04 2015

A117671 a(n) = binomial(3*n+1, n+1).

Original entry on oeis.org

1, 6, 35, 210, 1287, 8008, 50388, 319770, 2042975, 13123110, 84672315, 548354040, 3562467300, 23206929840, 151532656696, 991493848554, 6499270398159, 42671977361650, 280576272201225

Offset: 0

Zerinvary Lajos, Apr 12 2006

a(n) = A258993(2*n+1, n). - Reinhard Zumkeller, Jun 22 2015

			if n=0 then C(3*0+1,0+1) = C(1,1) = 1.
if n=10 then C(3*10+1,10+1) = C(31,11) = 84672315.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions

Cf. A025174: binomial(3n-1,n-1), A006013.

Cf. A258993, A001764, A004319, A005809, A013698, A045721, A165817, A236194.

a117671 n = a258993 (2 * n + 1) n  -- Reinhard Zumkeller, Jun 22 2015

seq(binomial(3*n+1,n+1),n=0..30); # Robert Israel, Oct 10 2017

Table[Binomial[3n+1,n+1],{n,0,20}] (* Harvey P. Dale, Jul 19 2011 *)

vector(30, n, n--; binomial(3*n+1, n+1)) \\ Altug Alkan, Nov 04 2015

G.f.: (2*(-1+Hypergeometric2F1[-(1/3),1/3,-(1/2),(27*x)/4]))/(3*x). - Harvey P. Dale, Jul 19 2011
G.f.: A(x) = B'(x)/B(x)-B'(x)-1/x, where B(x) = 4/3*sin(1/3*asin(sqrt((27*x)/4)))^2. - Vladimir Kruchinin, Nov 26 2014
From Peter Bala, Nov 04 2015: (Start)
With an extra initial term equal to 1, the o.g.f. equals f(x)/g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1). (End)
a(n) = [x^(2*n)] 1/(1 - x)^(n+2). - Ilya Gutkovskiy, Oct 10 2017
a(n+1) = 3*(3*n+2)*(3*n+4)*a(n)/(2*(n+2)*(2*n+1)). - Robert Israel, Oct 10 2017

A128908 Riordan array (1, x/(1-x)^2).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 4, 1, 0, 4, 10, 6, 1, 0, 5, 20, 21, 8, 1, 0, 6, 35, 56, 36, 10, 1, 0, 7, 56, 126, 120, 55, 12, 1, 0, 8, 84, 252, 330, 220, 78, 14, 1, 0, 9, 120, 462, 792, 715, 364, 105, 16, 1, 0, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1

Offset: 0

Philippe Deléham, Apr 22 2007

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,2,-1/2,1/2,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Row sums give A088305. - Philippe Deléham, Nov 21 2007
Column k is C(n,2k-1) for k > 0. - Philippe Deléham, Jan 20 2012
From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
T is the convolution triangle of the positive integers (see A357368). - Peter Luschny, Oct 19 2022

			The triangle T(n,k) begins:
   n\k  0    1    2    3    4    5    6    7    8    9   10
   0:   1
   1:   0    1
   2:   0    2    1
   3:   0    3    4    1
   4:   0    4   10    6    1
   5:   0    5   20   21    8    1
   6:   0    6   35   56   36   10    1
   7:   0    7   56  126  120   55   12    1
   8:   0    8   84  252  330  220   78   14    1
   9:   0    9  120  462  792  715  364  105   16    1
  10:   0   10  165  792 1716 2002 1365  560  136   18    1
  ... reformatted by _Wolfdieter Lang_, Jul 31 2017
From _Peter Luschny_, Mar 06 2022: (Start)
The sequence can also be seen as a square array read by upwards antidiagonals.
   1, 1,   1,    1,    1,     1,     1,      1,      1, ...  A000012
   0, 2,   4,    6,    8,    10,    12,     14,     16, ...  A005843
   0, 3,  10,   21,   36,    55,    78,    105,    136, ...  A014105
   0, 4,  20,   56,  120,   220,   364,    560,    816, ...  A002492
   0, 5,  35,  126,  330,   715,  1365,   2380,   3876, ... (A053126)
   0, 6,  56,  252,  792,  2002,  4368,   8568,  15504, ... (A053127)
   0, 7,  84,  462, 1716,  5005, 12376,  27132,  54264, ... (A053128)
   0, 8, 120,  792, 3432, 11440, 31824,  77520, 170544, ... (A053129)
   0, 9, 165, 1287, 6435, 24310, 75582, 203490, 490314, ... (A053130)
    A27,A292, A389, A580,  A582, A1288, A10966, A10968, A165817       (End)

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.

Cf. A002450, A007318, A034008, A053122, A078812, A084938, A088305.
Cf. Columns : A000007, A000027, A000292, A000389, A000580, A000582, A001288, A010966 ..(+2).. A011000, A017713 ..(+2).. A017763.
Cf. A000007, A001352, A008574, A054888, A084099, A084103, A163810, A357368.
Cf. A165817 (the main diagonal of the array).

# Computing the rows of the array representation:
S := proc(n,k) option remember;
if n = k then 1 elif k < 0 or k > n then 0 else
S(n-1, k-1) + 2*S(n-1, k) - S(n-2, k) fi end:
Arow := (n, len) -> seq(S(n+k-1, k-1), k = 0..len-1):
for n from 0 to 8 do Arow(n, 9) od; # Peter Luschny, Mar 06 2022
# Uses function PMatrix from A357368.
PMatrix(10, n -> n); # Peter Luschny, Oct 19 2022

With[{nmax = 10}, CoefficientList[CoefficientList[Series[(1 - x)^2/(1 - (2 + y)*x + x^2), {x, 0, nmax}, {y, 0, nmax}], x], y]] // Flatten (* G. C. Greubel, Nov 22 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, if(k==0, 0, binomial(n+k-1,2*k-1))), ", "))) \\ G. C. Greubel, Nov 22 2017

from functools import cache
@cache
def A128908(n, k):
    if n == k: return 1
    if (k <= 0 or k > n): return 0
    return A128908(n-1, k-1) + 2*A128908(n-1, k) - A128908(n-2, k)
for n in range(10):
    print([A128908(n, k) for k in range(n+1)]) # Peter Luschny, Mar 07 2022

@cached_function
def T(k,n):
    if k==n: return 1
    if k==0: return 0
    return sum(i*T(k-1,n-i) for i in (1..n-k+1))
A128908 = lambda n,k: T(k,n)
for n in (0..10): print([A128908(n,k) for k in (0..n)]) # Peter Luschny, Mar 12 2016

T(n,0) = 0^n, T(n,k) = binomial(n+k-1, 2k-1) for k >= 1.
Sum_{k=0..n} T(n,k)*2^(n-k) = A002450(n) = (4^n-1)/3 for n>=1. - Philippe Deléham, Oct 19 2008
G.f.: (1-x)^2/(1-(2+y)*x+x^2). - Philippe Deléham, Jan 20 2012
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A001352(n), (-1)^(n+1)*A054888(n+1), (-1)^n*A008574(n), (-1)^n*A084103(n), (-1)^n*A084099(n), A163810(n), A000007(n), A088305(n) for x = -6, -5, -4, -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Jan 20 2012
Riordan array (1, x/(1-x)^2). - Philippe Deléham, Jan 20 2012

A236194 a(n) = binomial(3n+1, n-1).

Original entry on oeis.org

1, 7, 45, 286, 1820, 11628, 74613, 480700, 3108105, 20160075, 131128140, 854992152, 5586853480, 36576848168, 239877544005, 1575580702584, 10363194502115, 68248282427325, 449972009097765, 2969831763694950, 19619725782651120, 129728497393775280

Offset: 1

Bruno Berselli, Jan 20 2014

This sequence is related to A006013 by a(n)/n = A006013(n)/2.

Bruno Berselli, Table of n, a(n) for n = 1..100
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.

Cf. A006013; A025174: C(3n-1, n-1); A117671: C(3n+1, n+1).
Second column of the triangle A159841.
Third column of the triangle A119301.
Cf. A001764, A004319, A005809, A013698, A045721, A165817.
Cf. A045721, A117671.

Magma
```
[Binomial(3*n+1,n-1): n in [1..30]];
```
Mathematica
```
Table[Binomial[3n+1, n-1], {n, 30}]
```

makelist(binomial(3*n+4,n),n,0,40); /* Emanuele Munarini, Oct 14 2014 */

vector(30, n, binomial(3*n+1, n-1)) \\ Altug Alkan, Nov 04 2015

[binomial(3*n+1,n-1) for n in range(1,31)] # G. C. Greubel, Nov 09 2022

G.f.: (sqrt(4-27*x)*cos((2/3)*arcsin((3/2)*sqrt(3*x))) + sqrt(3*x)*sin((2/3)*arcsin((3/2)*sqrt(3*x))) - sqrt(4-27*x))/(3*sqrt(4-27*x)*x^2). - Emanuele Munarini, Oct 14 2014
From Peter Bala, Nov 04 2015: (Start)
With offset 0, the o.g.f. equals f(x)*g(x)^4, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] x/(1 - x)^(2*n+3). - Ilya Gutkovskiy, Oct 10 2017
From Karol A. Penson, Mar 02 2024: (Start)
G.f.: ((sqrt(3)*sqrt(x)*i + sqrt(4 - 27*x))*(4*sqrt(4 - 27*x) - 12*i*sqrt(3)*sqrt(x))^(2/3) + (-sqrt(3)*sqrt(x)*i + sqrt(4 - 27*x))*(4*sqrt(4 - 27*x) + 12*i*sqrt(3)*sqrt(x))^(2/3) - 8*sqrt(4 - 27*x))/(24*sqrt(4 - 27*x)*x^2), where i is the imaginary unit, i=sqrt(-1).
G.f.: hypergeometric3F2([5/3,2,7/3],[5/2,3],27*x/4).
G.f. = G satisfies the algebraic equation: 1 + (7*z-1)*G + (27*z-4)*z^2*G^2 + (27*z-4)*z^4*G^3 = 0. (End)

A163455 a(n) = binomial(5*n-1,n).

Original entry on oeis.org

1, 4, 36, 364, 3876, 42504, 475020, 5379616, 61523748, 708930508, 8217822536, 95722852680, 1119487075980, 13136858812224, 154603005527328, 1824010149372864, 21566576904406820, 255485622301674660, 3031718514166879020, 36030431772522503316

Offset: 0

Zak Seidov, Jul 28 2009

Also, number of terms in A163142 with n zeros in binary representation.

All terms >= 4 are divisible by 4.

			a(1)=4 because there are 4 terms in A163142 with 1 zero in binary representation {23,27,29,30}_10 ={10111,11011,11101,11110}_2
a(2)=36 because there are 36 terms in A163142 with 2 zeros in binary representation: {639,703,735,751,759,763,765,766,831,863,879,887,891,893,894,927,943,951,955,957,958,975,983,987,989,990,999,1003,1005,1006,1011,1013,1014,1017,1018,1020}_10={1001111111,...,1111111100}_2
a(3)=364 terms in A163142 from 18431 to 32760 with 3 zeros in binary representation 18431_10=100011111111111_2 and 32760_10=111111111111000_2
a(4)=3876 terms in A163142 from 557055 to 1048560 with 4 zeros in binary representation, etc.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A163142, A118971.

Cf. A088218, A165817, A262977.

[Binomial(5*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

Table[(5*n-1)!/ n!/(4*n-1)!,{n,20}]
Table[Binomial[5 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

B(x):=sum(binomial(5*n-2,n-1)/(n)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,10);

a(n) = binomial(5*n-1,n); \\ Michel Marcus, Oct 06 2015

a(n) = (5n-1)!/(n!(4n-1)!).
G.f.: A(x)=x*B'(x)/B(x), where B(x)/x is g.f. for A118971. Also a(n) = Sum_{k=0..n} (binomial(n-1,n-k)*binomial(4*n,k)). - Vladimir Kruchinin, Oct 06 2015
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-4*n, n).
a(n) = hypergeom([1 - 4*n, -n], [1], 1).
A(x) satisfies A(x/(1 + x)^5) = 1/(1 - 4*x). (End)
From Peter Bala, Jun 05 2024: (Start)
Right-hand side of the identity Sum_{k = 0..n} binomial(n+k-1, k)*binomial(4*n-k-1, n-k) = binomial(5*n-1, n).
a(n) = (3/4)*binomial(4*n, 3*n)*hypergeom([n, -n], [1 - 4*n], 1) for n >= 1. (End)
From Karol A. Penson Jan 20 2025: (Start)
G.f.: 4*z*Hypergeometric5F4([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125*z)/256) + 1.
G.f. A(z) satisfies: z*(1250*A^3 - 250*A^2 + 25*A - 1) + (-3125*z + 256)*A^4 + (3125*z - 256)*A^5 = 0. (End)
G.f.: 1/(5-4*g) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 17 2025

Entry revised by N. J. A. Sloane, Dec 07 2015

cp's OEIS Frontend

A025174 a(n) = binomial(3n-1, n-1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A045721 a(n) = binomial(3*n+1,n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A262977 a(n) = binomial(4*n-1,n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

Maxima

PARI

PARI

Sage

Formula

Extensions

A004319 a(n) = binomial(3*n, n - 1).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Formula

A013698 a(n) = binomial(3*n+2, n-1).

Views

Author