A025174 -id:A025174 - cp's OEIS frontend

A163456 a(n) = binomial(5*n,n)/5.

1, 9, 91, 969, 10626, 118755, 1344904, 15380937, 177232627, 2054455634, 23930713170, 279871768995, 3284214703056, 38650751381832, 456002537343216, 5391644226101705, 63871405575418665, 757929628541719755

Offset: 1

Zak Seidov, Jul 28 2009

For prime p, a(p) == 1 (mod p). - Gary Detlefs, Aug 03 2013
In fact, a(p) == 1 (mod p^3) for prime p >= 5. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015
From Robert Israel, Jul 12 2016: (Start)
a(p+1) == 5 (mod p) for primes p >= 5.
a(p^(k+1)) == a(p^k) mod p^(3(k+1)) for primes p >= 5. (End)

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994.

Robert Israel, Table of n, a(n) for n = 1..920
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
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. A000108, A000245, A001764, A002293, A025174, A118970, A163455, A224274, A227726.

seq(binomial(5*n,n)/5, n=1..20); # Robert Israel, Jul 12 2016

Array[Binomial[5 #, #]/5 &, {18}] (* Michael De Vlieger, Oct 09 2015 *)

a(n) = binomial(5*n,n)/5 \\ Altug Alkan, Oct 09 2015

a(n) = (5*n-1)!/(4*n!*(4*n-1)!) = A001449(n)/5 = A163455(n)/4.
a(n) = binomial(5*n,n)/5. - Gary Detlefs, Aug 03 2013
From Peter Bala, Oct 08 2015: (Start)
a(n) = (1/3)*[x^n] (C(x)^3)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274.
exp( 3*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 18*x^2 + 136*x^3 + ... is the o.g.f. for A118970. (End)
From Peter Bala,Jul 12 2016: (Start)
a(n) = 1/6*[x^n] (1 + x)/(1 - x)^(4*n + 1).
a(n) = 1/6*[x^n] ( 1/C(-x)^6 )^n. Cf. A227726. (End)
a(n) ~ 2^(-8*n-3/2)*5^(5*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
From Robert Israel, Jul 12 2016: (Start)
G.f.: x*hypergeom([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125/256)*x).
a(n) = 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)/(8*n*(4*n-3)*(2*n-1)*(4*n-1)). (End)
O.g.f.: f(x)/(1 - 4*f(x)), where f(x) = series reversion (x/(1 + x)^5) = x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + ... is the o.g.f. of A002294 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/4)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+4)*n+k-1,k) = C(5*n,n)/5 and (1/5)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-5)*n+k-1,k) = C(5*n,n)/5, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/4)*Sum_{k = 0..2*n} (-1)^k*binomial(6*n-k-1,2*n-k)*binomial(4*n+k-1,k) = binomial(5*n,n)/5, for n >= 1. - Peter Bala, Mar 09 2022
a(n) = (1/2)* [x*n] F(x)^(2*n) = [x^n] G(x)^n for n >= 1, where F(x) = Sum_{k >= 0} 1/(2*k + 1)*binomial(3*k,k)*x^k is the o.g.f. of A001764 and G(x) = Sum_{k >= 0} 1/(3*k + 1)*binomial(4*k,k)*x^k is the o.g.f. of A002293 (apply Concrete Mathematics, equation 5.60, p. 201). - Peter Bala, Apr 26 2023

Renamed by Peter Bala, Oct 08 2015

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

Original entry on oeis.org

1, 5, 95, 2496, 76063, 2524161, 88534548, 3228482908, 121171012431, 4649906785719, 181614908182551, 7196014051078368, 288537887780406468, 11686156771344086156, 477379538839242423528, 19645977861506470428324

Offset: 0

Paul D. Hanna, Dec 31 2010

nonn

The logarithmic derivative of this sequence is a bisection of the logarithmic derivative of A001764 (ternary trees).
To see this, compare the g.f. of this sequence with g.f. of A001764:
exp(Sum_{n>=1} C(3n-1,n-1)*x^n/n) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 +...

			G.f.: A(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...
log(A(x)) = 5*x + 165*x^2 + 6188*x^3 + 245157*x^4 + 10015005*x^5 +...+ A025174(2n)*x^n/n +...
G.f. satisfies: A(x) = G(x*A(x)^3) where G(x) begins:
G(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 +...

Cf. A182959, A025174, A001764, A060941, A300386 - A300389.

f[n_] := Exp[ Sum[ Binomial[6 i - 1, 2 i - 1]*x^i/i, {i, n}]]; CoefficientList[ Series[f@ 15, {x, 0, 15}], x]  (* Robert G. Wilson v, Dec 31 2010 *)

{a(n)=polcoeff(exp(sum(m=1,n,binomial(6*m-1,2*m-1)*x^m/m)+x*O(x^n)),n)}

{a(n)=polcoeff((serreverse(x*(1-2*x+sqrt(1-8*x+x*O(x^n)))^3/(8*(1+x)^6))/x)^(1/3),n)}

G.f. A(x) satisfies: A(x^2) = F(x)*F(-x) where F(x) = 1 + x*F(x)^3 = Sum_{n>=0} C(3n,n)*x^n/(2n+1) is the g.f. of A001764.
Let G(x) = 2*(1+x)^2/(1-2*x+sqrt(1-8*x)), then g.f. A(x) satisfies:
* A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x);
* A(x) = [Series_Reversion( x/G(x)^3 ) / x]^(1/3).
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 215, 10463, 537287, 28435880, 1534398353, 83920389642, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 (checked up to p = 101). - Peter Bala, Sep 14 2021
From Vaclav Kotesovec, Sep 16 2021: (Start)
Recurrence: 16*n*(2*n + 1)*(4*n - 1)*(4*n + 1)*(27*n - 31)*a(n) = 18*(69984*n^5 - 220320*n^4 + 267705*n^3 - 156510*n^2 + 42021*n - 3680)*a(n-1) - 59049*(n-1)*(2*n - 3)*(3*n - 5)*(3*n - 4)*(27*n - 4)*a(n-2).
a(n) ~ sqrt((sqrt(2) - 1)^(2/3) + (sqrt(2) + 1)^(2/3) - 2) * 3^(6*n + 3/2) / (sqrt(Pi) * n^(3/2) * 2^(4*n + 7/2)). (End)

A090763 a(n) = (3n+3)!/(3n!(2n+2)!).

Original entry on oeis.org

1, 10, 84, 660, 5005, 37128, 271320, 1961256, 14060475, 100150050, 709634640, 5006710800, 35197176924, 246681069040, 1724337127920, 12025860872784, 83702724824775, 581558091471630, 4034231805704100, 27945630038703300

Offset: 0

Al Hakanson (hawkuu(AT)excite.com), Feb 15 2004

Alois P. Heinz, Table of n, a(n) for n = 0..500

Cf. A005809, A025174, A210453, A360560.

a:= n->sum(j*binomial(n+2, j)*binomial(2*(n+1), j)/6, j=0..n+2): seq(a(n), n=0..21); # Zerinvary Lajos, Jul 31 2006
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1,
      3*(3*n+1)*(3*n+2)*a(n-1)/(2*n*(2*n+1)))
    end:
seq(a(n), n=0..30); # Alois P. Heinz, Feb 01 2014

a[n_] := 1/Integrate[(x^(2/3) - x)^n, {x, 0, 1}]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Feb 18 2004 *)
a[n_] := 1/(2*Beta[2n, n]) (* Enrique Pérez Herrero, May 17 2009 *)
a[n_] : =1/2*Sum[j*Binomial[2 n, j]*Binomial[n, j], {j, 1, n}] (* Enrique Pérez Herrero, May 22 2009 *)

[binomial(3*n,n)*n/3 for n in range(1,21)] # Zerinvary Lajos, May 17 2009

a(n) = 1/(Integral_{x=0..1} (x^(2/3)-x)^n dx).
a(n) = 1/(Integral_{x=0..1} (x-x^1.5)^n dx).
a(n) = 1/(2*Beta(2n,n)). - Enrique Pérez Herrero, May 17 2009
a(1) = 1; a(n) = a(n-1)*2*binomial(3n,3)/binomial(2n,3). - Enrique Pérez Herrero, May 19 2009
a(n) = (1/2)*Sum{j=1,n}(j*binomial(2n,j)*binomial(n,j)). - Enrique Pérez Herrero, May 22 2009
a(n) = (n+1)*A025174(n+1). - R. J. Mathar, Jun 21 2009
G.f.: Hypergeometric2F1(4/3, 5/3, 3/2, 27*x/4). - Stefano Spezia, Oct 18 2019
G.f.: (-(3*sqrt(4-27*x)*csc(arcsin((3*sqrt(3*x))/2)/3)^2)/((4*(4-27*x)^(3/2)))+(sqrt(3)*cot(arcsin((3*sqrt(3*x))/2)/3))/((4-27*x)*sqrt(x)*sqrt(4-27*x))). - Vladimir Kruchinin, Feb 12 2023
From Amiram Eldar, Dec 07 2024: (Start)
a(n) = (n+1) * A005809(n+1) / 3.
Sum_{n>=0} 1/a(n) = 3 * A210453. (End)

More terms from Robert G. Wilson v, Feb 18 2004

Simpler description from Vladeta Jovovic, Feb 22 2004

A143858 Number of pairwise disjoint unions of m integer-to-integer subintervals of [0,n]; a rectangular array by antidiagonals, n>=2m-1, m>=1.

Original entry on oeis.org

1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060

Offset: 1

Clark Kimberling, Sep 03 2008

Main diagonal: A025174.

			R(2,4) counts these unions of 2 subintervals of [0,4]: [0,1]U[2,3], [0,1]U[2,4], [0,1]U[3,4], [0,2]U[3,4], [1,2]U[3,4].
   1    3    6   10   15   21   28   36   45   55   66   78
   0    0    1    5   15   35   70  126  210  330  495  715
   0    0    0    0    1    7   28   84  210  462  924 1716
   0    0    0    0    0    0    1    9   45  165  495 1287
   0    0    0    0    0    0    0    0    1   11   66  286
   0    0    0    0    0    0    0    0    0    0    1   13

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A007318, A025174, A109954, A085478, A258993.

Seen as a triangle read by rows
a143858 n k = a143858_tabl !! (n-1) !! k
a143858_row n = a143858_tabl !! (n-1)
a143858_tabl = map ((++ [1]) . tail) a258993_tabl
-- Reinhard Zumkeller, Jun 22 2015

A143858 := proc(m,n)
    binomial(n-1+2*m,2*m) ;
end proc:
seq(seq( A143858(n,d-n),n=1..d-1),d=2..8) ; # R. J. Mathar, Nov 16 2023

R(m,n) = C(n+1,2m), where n>=2m-1, m>=1. R is also given by the absolute values of terms in A109954.

A091958 Triangle read by rows: T(n,k)=number of ordered trees with n edges and k branch nodes at odd height.

Original entry on oeis.org

1, 1, 2, 4, 1, 9, 5, 21, 21, 51, 78, 3, 127, 274, 28, 323, 927, 180, 835, 3061, 954, 12, 2188, 9933, 4510, 165, 5798, 31824, 19734, 1430, 15511, 100972, 81684, 9790, 55, 41835, 317942, 324246, 57876, 1001, 113634, 995088, 1245762, 309036, 10920

Offset: 0

Emeric Deutsch, Mar 13 2004

T(3n,n) = binomial(3n,n)/(2n+1) = A001764(n); T(n,0) = A001006(n) (the Motzkin numbers); T(n,1) = A055219(n-3) (n>=3; most probably); Row sums are the Catalan numbers (A000108).
T(n,k) = number of ordered trees on n edges with k vertices of outdegree at least 3; T(n,k) = number of ordered trees on n edges with k vertices V such that V's rightmost descendant leaf is at distance exactly 3 from V. - David Callan, Oct 24 2004
T(n,k) is the number of Dyck n-paths containing k UUUDs. For example, T(6,2) = 3 because UUUDUUUDDDDD, UUUDDUUUDDDD, UUUDDDUUUDDD each contains 2 UUUDs. - David Callan, Nov 04 2004

			T(3,1) = 1 because the only tree having 3 edges and 1 branch node at an odd level is the tree having the shape of the letter Y.
Triangle begins:
1;
1;
2;
4,       1;
9,       5;
21,     21;
51,     78,    3;
127,   274,   28;
323,   927,  180;
835,  3061,  954,  12;
2188, 9933, 4510, 165;

Alois P. Heinz, Rows n = 0..250, flattened
E. Deutsch, A bijection on ordered trees and its consequences, J. Comb. Theory, A, 90, 210-215, 2000.
A. Kuznetsov, I. Pak, A. Postnikov, Trees associated with the Motzkin numbers, J. Comb. Theory, A, 76, 145-147, 1996.
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.

Cf. A001764, A001006, A055219, A243752.

Topmost entries in each column form A001764=( binomial(3n, n)/(2n+1) )A025174=(%20binomial(3n+2,%20n)%20)">(n>=0), next to topmost entries form A025174=( binomial(3n+2, n) )(n>=0), next lower entries are given by ( (n+2)binomial(3n+4, n) )_(n>=0).

T := (n,k)->binomial((n+1),k)*sum((-1)^j*binomial(n+1-k,j)*binomial(2*n-3*k-3*j,n),j=0..floor(n/3)-k)/(n+1): seq(seq(T(n,k),k=0..floor(n/3)),n=0..18);
# second Maple program:
b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, expand(b(x-1, y+1, [2, 3, 4, 4][t])
      +b(x-1, y-1, [1, 1, 1, 1][t])*`if`(t=4, z, 1))))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, 1)):
seq(T(n), n=0..15);  # Alois P. Heinz, Jun 10 2014

Clear[a]; a[n_, k_]/;k>n/3 || k<0 := 0; a[n_, 0]/;0<=n<=1 := 1; a[n_, 0]/;n>=2 := a[n, 0] = ((2*n + 1)*a[n-1, 0] + 3*(n - 1)*a[n-2, 0])/(n + 2); a[n_, k_]/;1<=k<=n/3 && n>=2 := a[n, k] = ( (12 - 9*k + 3*n)*a[n-2, k-2] - (12 - 18*k + 3*n)*a[ n-2, k-1] - 9*k*a[ n-2, k] + (4 - 6*k + 4*n)*a[n-1, k-1] + 6*k*a[n-1, k] - (2 - k + n)*a[n, k-1] )/k; Table[a[n, k], {n, 0, 16}, {k, 0, n/3}] (Callan)
T[n_, k_] := (2*n-3*k)!*HypergeometricPFQ[{k-n-1, k-n/3, 1/3+k-n/3, 2/3+k-n/3}, {k-2*n/3, 1/3+k-2*n/3, 2/3+k-2*n/3}, 1]/(k!*(n-k+1)!*(n-3*k)!); Table[T[n, k], {n, 0, 15}, {k, 0, n/3}] // Flatten (* Jean-François Alcover, Mar 31 2015 *)

T(n,k) = binomial((n+1), k)*sum((-1)^j*binomial(n+1-k,j)*binomial(2n-3k-3j, n), j=0..floor(n/3)-k)/(n+1). G.f.: G=G(t,z) satisfies (t-1)z^3 G^3 + zG^2 - G + 1 = 0.

A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

Original entry on oeis.org

1, 1, 4, 5, 21, 28, 120, 165, 715, 1001, 4368, 6188, 27132, 38760, 170544, 245157, 1081575, 1562275, 6906900, 10015005, 44352165, 64512240, 286097760, 417225900, 1852482996, 2707475148, 12033222880, 17620076360, 78378960360, 114955808528

Offset: 0

Paul Barry, Oct 23 2004

Main diagonal of triangle A099575.
With offset 2, this is the number of compositions of n-1 into floor(n/2) parts. - T. D. Noe, Jan 05 2007
From Petros Hadjicostas, Jul 19 2018: (Start)
We clarify the above comment by T. D. Noe. The number of compositions of N into K positive parts is C(N-1, K-1). This was proved by MacMahon in 1893 (and probably by others before him). The number of compositions of N into K nonnegative parts is C(N+K-1, K-1) because for every composition b_1 + ... + b_K = N with b_i >= 0 for all i, we may create another composition c_1 + ... + c_K = N+K with c_i = b_i + 1 >= 1.
The statement of T. D. Noe above means that, for n>=2, a(n-2) is the number of compositions of N = n-1 into K = floor(n/2) nonnegative parts. Thus, a(n-2) = C(N+K-1, K-1) = C(n-1+floor(n/2)-1, floor(n/2)-1) = C(floor((3(n-2)+2)/2), floor((n-2)/2)).
This interpretation is important for T. D. Noe's comments for sequence A030077, whose unknown general formula remains an unsolved problem (as of July 2018).
It should be noted, however, that for most authors "composition" means "composition into positive parts". The phrase "weak composition" is sometimes used for a "composition into nonnegative parts".
(End)

			From _Petros Hadjicostas_, Jul 19 2018: (Start)
With n=2 there are a(2-2) = a(0) = 1 compositions of 2-1 = 1 into floor(2/2) = 1 nonnegative parts, namely 1 (only).
With n=3 there are a(3-2) = a(1) = 1 compositions of 3-1 = 2 into floor(3/2) = 1 nonnegative parts, namely 2 (only).
With n=4 there are a(4-2) = a(2) = 4 compositions of 4-1 = 3 into floor(4/2) = 2 nonnegative parts, namely 0+3, 3+0, 1+2, and 2+1.
With n=5 there are a(5-2) = a(3) = 5 compositions of 5-1 = 4 into floor(5/2) = 2 nonnegative parts, namely 0+4, 4+0, 1+3, 3+1, and 2+2.
With n=6 there are a(6-2) = a(4) = 21 compositions of 6-1 = 5 into floor(6/2) = 3 nonnegative parts, namely the 3 permutations of 1+1+3, the 3 permutations of 1+2+2, the 3 permutations of 0+0+5, the 6 permutations of 0+1+4, and the 6 permutations of 0+2+3.
(End)

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

Cf. A025174 (bisection), A030077, A045721 (bisection), A099575, A127040.

[(&+[Binomial(n+j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

A099578:=n->binomial(floor((3*n+2)/2), floor(n/2)); seq(A099578(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013

Table[Binomial[Floor[(3n+2)/2], Floor[n/2]], {n, 0, 50}] (* Wesley Ivan Hurt, Nov 01 2013 *)
CoefficientList[Series[-((Sqrt[4 -27 x^2] -2(Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]] + Sqrt[3] Sin[2/3 ArcSin[(3 Sqrt[3] x)/2]]))/(3 x Sqrt[4 -27 x^2])), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
a[n_] := Binomial[2*n+1, n]*Hypergeometric2F1[-n, n+1, -2*n-1, -1]; Flatten[Table[a[n], {n, 0, 29}]] (* Detlef Meya, Dec 25 2023 *)

a(n) = binomial((3*n+2)\2, n\2); \\ Michel Marcus, Nov 02 2013

[binomial((3*n+2)//2, n//2) for n in (0..40)] # G. C. Greubel, Jul 24 2022

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, k).
40*n*(n+1)*a(n) +36*n*(n-2)*a(n-1) -6*(45*n^2-23)*a(n-2) -27*(3*n-4)*(3*n-5)*a(n-3) = 0. - R. J. Mathar, Oct 30 2014
From Benedict W. J. Irwin, Aug 15 2016: (Start)
G.f.: -( (sqrt(4-27*x^2) - 2*(cos(arcsin(3*sqrt(3)*x/2)/3) + sqrt(3)*sin(2*arcsin(3*sqrt(3)*x/2)/3)) )/( 3*x*sqrt(4-27*x^2)) ).
E.g.f.: Hypergeometric2F3(2/3,4/3;1/2,1,3/2;27*x^2/16) + x*Hypergeometric2F3(4/3,5/3;3/2,3/2,2;27*x^2/16).
(End)
Recurrence: 4*n*(n+1)*(6*n-1)*a(n) = 18*n*a(n-1) + 3*(3*n-2)*(3*n-1)*(6*n+5)*a(n-2). - Vaclav Kotesovec, Aug 15 2016
a(n) = binomial(2*n+1, n)*hypergeom([-n, n+1], [-2*n-1], -1). - Detlef Meya, Dec 25 2023

A333473 a(n) = [x^n] ( S(x/(1 + x)) )^n, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 2, 12, 92, 752, 6352, 54768, 478928, 4231424, 37680320, 337622912, 3040354176, 27492359936, 249463806464, 2270319909632, 20714443816192, 189418898063360, 1735482632719360, 15928224355854336, 146414296847992832, 1347721096376573952, 12421053168197722112

Offset: 0

Peter Bala, Mar 23 2020

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence s(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: s(n*p^k) == s(n*p^(k-1)) ( mod p^(k) ) for any prime p and positive integers n and k. For certain power series F(x) we may get stronger congruences. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = S(x/(1 + x)) = 1 + 2*x + 4*x^2 + 12*x^3 + 40*x^4 + 154*x^5 + 544*x^6 + ...(see A025227), where S(x) is the o.g.f. of the large Schröder numbers A006318. We conjecture that the associated sequence a(n) = [x^n] ( S(x/(1 + x)) )^n satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333472.
More generally, we conjecture that for a positive integer r and integer s, the sequence a(r,s;n) := [x^(r*n)] ( S(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note the sequence b(n) := [x^n] ( S(x) )^n = A103885(n) appears to satisfy the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*S(x/(1 + x)) )^n = A333090(n) also appears to satisfy the same congruences.

			Examples of congruences:
a(11) - a(1) = 3040354176 - 2 = 2*(11^2)*13*966419 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 12421053168197722112 - 92 = (2^2)*(3^7)*5*(7^2)* 5795401942927 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 90551762251592215396352 - 6352 = (2^4)*(5^4)*293* 30905038311123623 == 0 ( mod 5^4 ).

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Main diagonal of A378317.

Cf. A006318, A025227, A103885, A333090, A333472.

Sch := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := x → Sch(x/(1+x));
H := (x, n) -> series(G(x)^n, x, 51):
seq(coeff(H(x, n), x, n), n = 0..25)

Table[SeriesCoefficient[((1 - Sqrt[1- 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = sum(k=0, n, binomial(n, k)*binomial(n+2*k-1, 2*k)); \\ Seiichi Manyama, Nov 24 2024

a(n) = [x^n] ( (1 - sqrt(1- 4*x - 4*x^2))/(2*x) )^n.
a(n) ~ sqrt(((sqrt(2) + 1)^(2/3) + (sqrt(2) - 1)^(2/3) - 1)/3) * ((3*(71 + 8*sqrt(2))^(1/3) + 3*(71 - 8*sqrt(2))^(1/3) + 13))^n / (sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+2*k-1,2*k). - Seiichi Manyama, Nov 24 2024

A036909 a(n) = (2/3) * 4^n * binomial(3*n, n).

Original entry on oeis.org

8, 160, 3584, 84480, 2050048, 50692096, 1270087680, 32133218304, 819082035200, 21002987765760, 541167892561920, 13999778090188800, 363391162981023744, 9459706464902840320, 246865719056498950144, 6456334894356662059008, 169176689745174567321600, 4440485304168581976555520

Offset: 1

N. J. A. Sloane

nonn

Identity (3.116) in H. W. Gould, Combinatorial Identities, Morgantown, 1972, page 35.

G. C. Greubel, Table of n, a(n) for n = 1..695

Cf. A005809, A025174.

[(2/3)*4^n*Binomial(3*n,n): n in [1..30]]; // G. C. Greubel, Jun 22 2022

Table[2/3 4^n Binomial[3n,n],{n,20}](* Harvey P. Dale, Mar 26 2012 *)

[(2/3)*4^n*binomial(3*n, n) for n in (0..30)] # G. C. Greubel, Jun 22 2022

Sum_{k=0..n} binomial(4*n, 2*(n-k))*binomial(k+n, n) = (2/3)*4^n*binomial(3*n, n) = (2/3)*4^n*A005809(n) = 2*4^n*A025174(n).
G.f.: (2/3) * 2F1([1/3, 2/3], [1/2], 27*x) = 2*(cos((1/6)*arccos(1-54*x))/sqrt(1-27*x) - 1) /(3*x). - Harvey P. Dale, Mar 26 2012
D-finite with recurrence n*(2*n-1)*a(n) = 6*(3*n-1)*(3*n-2)*a(n-1). - R. J. Mathar, Feb 08 2021
G.f.: (2/3)*(cos((1/3)*Arcsin(3*sqrt(3*x)))/sqrt(1-27*x) - 1). - G. C. Greubel, Jun 22 2022
a(n) ~ 3^(3*n)/sqrt(3*n*Pi). - Stefano Spezia, Apr 25 2024

A127040 a(n) = binomial(floor((3n+4)/2),floor(n/2)).

Original entry on oeis.org

1, 1, 5, 6, 28, 36, 165, 220, 1001, 1365, 6188, 8568, 38760, 54264, 245157, 346104, 1562275, 2220075, 10015005, 14307150, 64512240, 92561040, 417225900, 600805296, 2707475148, 3910797436, 17620076360, 25518731280, 114955808528

Offset: 0

T. D. Noe, Jan 03 2007

nonn

With offset 2, the number of compositions of n into floor(n/2) parts, which is an upper bound for A007874.

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

Cf. A004319 (bisection), A025174 (bisection), A099578.

seq(sum(binomial(n+k, k-1), k=0..ceil((n+1)/2)), n=0..28); # Zerinvary Lajos, Apr 11 2007

CoefficientList[Series[(-1 + (2 Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]])/Sqrt[4 - 27 x^2] + 3 x^3 Hypergeometric2F1[4/3, 5/3, 5/2, (27 x^2)/4])/(3 x^2), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 16 2016 *)
Table[Binomial[Floor[(3 n + 4)/2], Floor[n/2]], {n, 0, 28}] (* Michael De Vlieger, Aug 18 2016 *)

a(n) = binomial((3*n+4)\2, n\2); \\ Michel Marcus, Sep 09 2016

From Benedict W. J. Irwin, Aug 16 2016: (Start)
G.f.: (-1 + (2*cos(arcsin(3*sqrt(3)*x/2)/3))/sqrt(4-27*x^2) + 3*x^3*2F1(4/3,5/3;5/2;27*x^2/4))/(3*x^2).
E.g.f.: 2F3(4/3,5/3;1/2,3/2,2;27*x^2/16) + x*2F3(4/3,5/3;1,3/2,5/2;27*x^2/16).
(End)
D-finite with recurrence 8*(n+2)*(n+1)*a(n) -84*(n-1)*(n+1)*a(n-1) +6*(-33*n^2+54*n-8)*a(n-2) +9*(63*n^2-63*n-16)*a(n-3) +108*(3*n-5)*(3*n-7)*a(n-4)=0. - R. J. Mathar, Feb 08 2021

A227726 a(n) = [x^n] (1 + x)/(1 - x)^(2*n+1).

Original entry on oeis.org

1, 4, 20, 112, 660, 4004, 24752, 155040, 980628, 6249100, 40060020, 258048960, 1668903600, 10829900592, 70480305440, 459823234112, 3006465218196, 19694758782300, 129235131438140, 849311959095600, 5589126007740660, 36825913869817380, 242910890642347200

Offset: 0

Paul D. Hanna, Jul 22 2013

			G.f.: A(x) = 1 + 4*x + 20*x^2 + 112*x^3 + 660*x^4 + 4004*x^5 + ...
where a(n) equals the coefficient of x^n in (1+x)/(1-x)^(2*n+1)
and forms the main diagonal in the following table of coefficients:
(1+x)/(1-x)^1: [1, 2, 2, 2, 2, 2, 2, 2, 2, ...];
(1+x)/(1-x)^3: [1, 4, 9, 16, 25, 36, 49, 64, 81, ...];
(1+x)/(1-x)^5: [1, 6, 20, 50, 105, 196, 336, 540, ...];
(1+x)/(1-x)^7: [1, 8, 35, 112, 294, 672, 1386, 2640, ...];
(1+x)/(1-x)^9: [1, 10, 54, 210, 660, 1782, 4290, 9438, ...];
(1+x)/(1-x)^11:[1, 12, 77, 352, 1287, 4004, 11011, 27456, ...];
(1+x)/(1-x)^13:[1, 14, 104, 546, 2275, 8008, 24752, 68952, ...];
(1+x)/(1-x)^15:[1, 16, 135, 800, 3740, 14688, 50388, 155040, ...]; ...
Related series is G(x) = 1 + x*G(x)^3, which begins:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 +...+ A001764(n)*x^n +...

Cf. A052227, A001764, A000108, A025174, A224274.

Join[{1},Table[Binomial[3n,n]+Binomial[3n-1,n-1],{n,30}]] (* Harvey P. Dale, Jan 15 2015 *)

{a(n)=binomial(3*n,n)+binomial(3*n-1,n-1)}

PARI
```
{a(n)=2*binomial(3*n-1, n) - 0^n}
```

{a(n)=polcoeff((1+x)/(1-x+x*O(x^n))^(2*n+1),n)}
for(n=0,30,print1(a(n),", "))

a(n) = binomial(3*n, n) + binomial(3*n-1, n-1).
G.f.: (2*G(x) - 1) / (3 - 2*G(x)), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
2*n*(2*n - 1)*a(n) - 3*(3*n - 1)*(3*n - 2)*a(n-1)=0. - R. J. Mathar, Jul 28 2013
a(n) = Sum_{r = 0..n-1} C(n-1,r)*C(2*n,r) + Sum_{r = 0..n} C(n,r)*C(2*n,n + r)  - J. M. Bergot, Mar 18 2014
From Peter Bala, Jul 12 2016: (Start)
For n >= 1, a(n) = 4*binomial(3*n - 1, n - 1) = 4*A025174(n).
a(n) = [x^n]( 1/C(-x)^4 )^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274. (End)

cp's OEIS Frontend

A163456 a(n) = binomial(5*n,n)/5.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A090763 a(n) = (3*n+3)!/(3*n!*(2*n+2)!).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

Extensions

A143858 Number of pairwise disjoint unions of m integer-to-integer subintervals of [0,n]; a rectangular array by antidiagonals, n>=2m-1, m>=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Formula

A091958 Triangle read by rows: T(n,k)=number of ordered trees with n edges and k branch nodes at odd height.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

A333473 a(n) = [x^n] ( S(x/(1 + x)) )^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Views

A090763 a(n) = (3n+3)!/(3n!(2n+2)!).

A333473 a(n) = [x^n] ( S(x/(1 + x)) )^n, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2x) is the o.g.f. of the large Schröder numbers A006318.