A348410 -id:A348410 - cp's OEIS frontend

A234839 a(n) = Sum_{k = 0..n} (-1)^k * binomial(n,k) * binomial(2*n,k).

1, -1, -1, 8, -17, -1, 116, -344, 239, 1709, -7001, 9316, 22276, -138412, 268568, 189008, -2608913, 6809417, -1814851, -45852416, 159116983, -155628353, -720492928, 3481793888, -5558713852, -9029921876, 71541001076, -158672882224, -45300345128, 1370202238072

Offset: 0

Vaclav Kotesovec, Dec 31 2013

For each n > 0, a(p-n) == 2^(2 - 3*n)*A252355(n) (mod p), for all primes p >= 2*n+1 [Chamberland, et al., Thm. 2.3]. - L. Edson Jeffery, Dec 17 2014

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672.
Robert Osburn, Brundaban Sahu, and Armin Straub, Supercongruences for sporadic sequences, arXiv:1312.2195 [math.NT], 2014.

Cf. A005809, A252355, A348410.

Table[Sum[(-1)^k*Binomial[n,k]*Binomial[2*n,k],{k,0,n}],{n,0,20}]
Table[Hypergeometric2F1[-2*n, -n, 1, -1],{n,0,20}]

a(n) = sum(k=0, n, (-1)^k*binomial(n,k)*binomial(2*n,k)); \\ Michel Marcus, Jan 13 2016

Recurrence: 2*n*(2*n-1)*(7*n-10)*a(n) = -(91*n^3 - 221*n^2 + 160*n - 36)*a(n-1) - 16*(n-1)*(2*n-3)*(7*n-3)*a(n-2).
Lim sup n->infinity |a(n)|^(1/n) = 2*sqrt(2).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 - x + 3*x^3 - 7*x^4 + 4*x^5 + 24*x^6 - 85*x^7 + 99*x^8 + 215*x^9 - 1196*x^10 + ... appears to have integer coefficients. - Peter Bala, Jan 04 2016
From Peter Bala, Apr 02 2020: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n+k)*binomial(n,k)*binomial(n,2*k).
a(n) = hypergeom([-n, -n/2, 1/2 - n/2], [1, 1/2], 1). (End)
From Peter Bala, Feb 23 2022: (Start)
a(n) = [x^n] ((1 - x)*(1 - x^2))^n. This implies that exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients as suggested above.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. [added Apr 12 2022: this was proved in 2014 by Osburn et al.; see Example 3.3]
The o.g.f. A(x) is the diagonal of the bivariate rational function 1/(1 - t*(1-x)*(1-x^2) ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = 1/x*Series_Reversion( x/((1-x)*(1-x^2)) ). Then A(x) = 1 + x*d/dx(Log(F(x))). (End)
a(n) =  Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(2*n+k, k)*2^(n-k). - Peter Bala, Feb 14 2024

A063020 Reversion of y - y^2 - y^3 + y^4.

Original entry on oeis.org

0, 1, 1, 3, 9, 32, 119, 466, 1881, 7788, 32868, 140907, 611871, 2685732, 11896906, 53115412, 238767737, 1079780412, 4909067468, 22424085244, 102865595140, 473678981820, 2188774576575, 10145798119530, 47165267330415, 219839845852692, 1027183096151244, 4810235214490986

Offset: 0

Olivier Gérard, Jul 05 2001

Seems to be the inverse of A007858. Can someone prove this?

a(n+1) counts paths from (0,0) to (n,n) which do not go above the line y=x, using steps (1,0) and (2k,1), where k ranges over the nonnegative integers. For example, the 9 paths from (0,0) to (3,3) are the 5 Catalan paths, as well as DNEN, DENN, EDNN and ENDN. Here E=(1,0), N=(0,1), D=(2,1). - Brian Drake, Sep 20 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Brian Drake, Limits of areas under lattice paths, Discrete Math. 309 (2009), no. 12, 3936-3953.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
A. Mironov and A. Morozov, Algebra of quantum C-polynomials, arXiv:2009.11641 [hep-th], 2020.
Hanna Mularczyk, Lattice Paths and Pattern-Avoiding Uniquely Sorted Permutations, arXiv:1908.04025 [math.CO], 2019.
Index entries for reversions of series

Cf. A007848, A052709, A064641, A348410.

A:= series(RootOf(Z-_Z^2-_Z^3+_Z^4-x), x, 21): seq(coeff(A,x,i), i=0..20); # _Brian Drake, Sep 20 2007

CoefficientList[InverseSeries[Series[y - y^2 - y^3 + y^4, {y, 0, 30}], x], x]

a(n):=sum((sum(binomial(j,n-3*k+2*j-1)*(-1)^(j-k)*binomial(k,j),j,0,k))*binomial(n+k-1,n-1),k,0,n-1)/n; /* Vladimir Kruchinin, Oct 11 2011 */

a(n):=sum((-1)^(i)*binomial(n+i-1,i)*binomial(3*n-i-2,n-i-1),i,0,n-1)/n; /* Vladimir Kruchinin, Feb 13 2014 */

x='x+O('x^66); concat([0],Vec(serreverse(x-x^2-x^3+x^4))) \\ Joerg Arndt, May 28 2013

def b(n):
    h = binomial(3*n + 1, n) * hypergeometric([-n, n + 1], [-3*n - 1], -1) / (n + 1)
    return simplify(h)
print([0] + [b(n) for n in range(27)])  # Peter Luschny, Sep 21 2023

a(n) = (1/n)*Sum_{k=0..n-1} binomial(n+k-1,n-1) * Sum_{j=0..k} binomial(j,n-3*k+2*j-1)*(-1)^(j-k)*binomial(k,j). - Vladimir Kruchinin, Oct 11 2011
a(n) = (1/n)*Sum_{i=0..n-1} (-1)^(i)*binomial(n+i-1,i)*binomial(3*n-i-2,n-i-1), n > 0. - Vladimir Kruchinin, Feb 13 2014
Recurrence: 16*(n-1)*n*(2*n-1)*(17*n-27)*a(n) = (n-1)*(1819*n^3 - 6527*n^2 + 7350*n - 2520)*a(n-1) + 8*(2*n-3)*(4*n-9)*(4*n-7)*(17*n-10)*a(n-2). - Vaclav Kotesovec, Feb 13 2014
a(n) ~ sqrt(11-3/sqrt(17))/16 * (107+51*sqrt(17))^n / (sqrt(Pi) * n^(3/2) * 2^(6*n)). - Vaclav Kotesovec, Feb 13 2014
The g.f. A(x) satisfies x*A'(x)/A(x) = 1 + x + 5*x^2 + 19*x^3 + 85*x^4 + ..., the g.f. of A348410. - Peter Bala, Feb 22 2022

A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2n) + 2y_1 + 2y_2 + ... + 2y_(2*n).

Original entry on oeis.org

2, 14, 92, 654, 4752, 35204, 264112, 2000526, 15264866, 117161264, 903533380, 6995547780, 54343476072, 423360920528, 3306313730592, 25876855432846, 202909132368942, 1593755466338030, 12537009118650016, 98753463725849904, 778825917274945408, 6149069826564738780

Offset: 1

Peter Bala, Feb 22 2022

This is a companion sequence to A348410.

Suppose n identical objects are distributed in 4*n labeled baskets, 2*n colored white and 2*n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.

			n = 2: 14 distributions of 2 identical objects in 4 white and 4 black baskets
             White             Black
   1)   (0) (0) (0) (0)   [2] [0] [0] [0]
   2)   (0) (0) (0) (0)   [0] [2] [0] [0]
   3)   (0) (0) (0) (0)   [0] [0] [2] [0]
   4)   (0) (0) (0) (0)   [0] [0] [0] [2]
   5)   (2) (0) (0) (0)   [0] [0] [0] [0]
   6)   (0) (2) (0) (0)   [0] [0] [0] [0]
   7)   (0) (0) (2) (0)   [0] [0] [0] [0]
   8)   (0) (0) (0) (2)   [0] [0] [0] [0]
   9)   (1) (1) (0) (0)   [0] [0] [0] [0]
  10)   (1) (0) (1) (0)   [0] [0] [0] [0]
  11)   (1) (0) (0) (1)   [0] [0] [0] [0]
  12)   (0) (1) (1) (0)   [0] [0] [0] [0]
  13)   (0) (1) (0) (1)   [0] [0] [0] [0]
  14)   (0) (0) (1) (1)   [0] [0] [0] [0]

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Cf. A234839, A348410, A351856, A365855.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(2*n+k-1,k), k = 0..floor(n/2) ), n = 0..20);

a(n) = sum(k = 0, n\2, binomial(3*n-2*k-1, n-2*k)*binomial(2*n+k-1, k)); \\ Michel Marcus, Feb 27 2022

a(n) = [x^n] ( 1/((1 - x)*(1 - x^2)) )^(2*n).
a(n) = Sum_{k = 0..floor(n/2)} C(3*n-2*k-1,n-2*k)*C(2*n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*C(5*n-k-1,n-k)*C(2*n+k-1,k).
1024*n*(n-1)*(2*n-1)*(2*n-3)*(4*n-1)*(4*n-3)*P(n-1)*a(n) = 8*(n-1)*(2*n-3)*Q(n)*a(n-1) + 7*(7*n-8)*(7*n-9)*(7*n-10)*(7*n-11)*(7*n-12)*(7*n-13)*P(n)*a(n-2), with a(1) = 2, a(2) = 14, P(n) = 1744*n^4-3815*n^3+ 2920*n^2-912*n+96 and Q(n) = 46599680*n^8-381534880*n^7+1306363456*n^6- 2428492279*n^5+2661904813*n^4 -1747232452*n^3+664205312*n^2- 132046848*n+10321920.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) = 2*x + 14*x^2 + 92*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)*(1 - x^2))^2 ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2)^2 ) = A365855(x). Then A(x) = x*(d/dx log(F(x))). [corrected by Jason Yuen, Mar 22 2025]

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A351856 Number of nonnegative integer solutions to 2n = x_1 + x_2 + ... + x_n + 2y_1 + 2y_2 + ... + 2y_n.

Original entry on oeis.org

2, 14, 119, 1086, 10252, 98735, 963832, 9502014, 94386908, 943206264, 9471346755, 95491466655, 966026045376, 9800968460024, 99685873633744, 1016118049037630, 10377363759903252, 106161722891946356, 1087696666197827374, 11159365823946907336, 114631982782490824420

Offset: 1

Peter Bala, Feb 22 2022

This is a companion sequence to A348410.

Suppose 2*n identical objects are distributed in 2*n labeled baskets, n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.

			n = 2: 14 distributions of 4 identical objects in 2 white and 2 black baskets
        White     Black
   1)  (0) (0)   [4] [0]
   2)  (0) (0)   [0] [4]
   3)  (0) (0)   [2] [2]
   4)  (2) (0)   [2] [0]
   5)  (0) (2)   [2] [0]
   6)  (1) (1)   [2] [0]
   7)  (2) (0)   [0] [2]
   8)  (0) (2)   [0] [2]
   9)  (1) (1)   [0] [2]
  10)  (4) (0)   [0] [0]
  11)  (0) (4)   [0] [0]
  12)  (3) (1)   [0] [0]
  13)  (1) (3)   [0] [0]
  14)  (2) (2)   [0] [0]

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Cf. A234839, A348410, A351857.

seq( add(binomial(3*n-2*k-1,2*n-2*k)*binomial(n+k-1,k), k = 0..n), n = 1..20);

a(n) = [x^(2*n)] ( 1/((1 - x)*(1 - x^2)) )^n.
a(n) = Sum_{k = 0..n} C(3*n-2*k-1,2*n-2*k)*C(n+k-1,k).
a(n) = Sum_{k = 0..2*n} (-1)^k*C(4*n-k-1,2*n-k)*C(n+k-1,k).
32*n*(n-1)*(2*n-1)*(2*n-3)*(41*n^2-126*n+93)*a(n) = 2*(n-1)*(2*n-3)*(16851*n^4-68637*n^3+93680*n^2-49024*n+7680)*a(n-1) - 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(41*n^2-44*n+8)*a(n-2) with a(1) = 2 and a(2) = 14.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) = 2*x + 14*x^2 + 119*x^3 + ... is the diagonal of the bivariate rational function x*t*(x - 1)*((x - 1)^2 - t)/((x - 1)^3 - t*(2*x + t - 2)) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*sqrt((1-x)*(1-x^2)) ) and put G(x) = x*(d/dx)(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.

A350406 a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.

Original entry on oeis.org

1, -1, 1, -1, 5, -26, 91, -246, 597, -1540, 4576, -14521, 44915, -132328, 380290, -1102076, 3268437, -9838428, 29616364, -88538500, 263489380, -785026110, 2348923875, -7053379710, 21204016275, -63716916276, 191394838116, -575200476046, 1730575897202

Offset: 0

Seiichi Manyama, Dec 29 2021

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

Cf. A005725, A348410, A350383, A350407.

a := proc (n) option remember; if n = 0 then 1 elif n = 1 then -1 elif n = 2 then 1 else -(4*n*(n-1)*(5*n-9)*(550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3+33771*n^2-20232*n+4032)*a(n-2) + (16*n-24)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3))/(16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)) end if; end:
seq(a(n), n = 0..30); # Peter Bala, Mar 18 2023
a := n -> (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1): seq(simplify(a(n)), n = 0..28); # Peter Luschny, Mar 19 2023

a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)

a(n) = (-1)^n*sum(k=0, n, binomial(n-1+k, k)*binomial(n, 4*k));

a(n) = (-1)^n * Sum_{k=0..n} binomial(n-1+k,k) * binomial(n,4*k).
From Peter Bala, Mar 17 2023: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n-k)*binomial(n+k-1,k)*binomial(2*n-2*k-1,n-2*k).
16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)*a(n) = - ( 4*n*(n-1)*(5*n-9)* (550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3 +33771*n^2-20232*n+4032)*a(n-2) + 8*(2*n-3)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3) ).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) = (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1). - Peter Luschny, Mar 19 2023

A348474 Number of compositions of n into exactly 2n nonnegative parts such that each positive i-th part is odd if i is odd.

Original entry on oeis.org

1, 2, 8, 41, 220, 1212, 6803, 38691, 222196, 1285610, 7482718, 43762754, 256972507, 1514020484, 8945944435, 52990732161, 314568593860, 1870939233546, 11146516959176, 66508200091575, 397375460647690, 2377167144881136, 14236462650026064, 85346464443885086

Offset: 0

Alois P. Heinz, Oct 19 2021

nonn

			a(2) = 8: [0,0,0,2], [0,0,1,1], [0,1,0,1], [0,1,1,0], [0,2,0,0], [1,0,0,1], [1,0,1,0], [1,1,0,0].

Alois P. Heinz, Table of n, a(n) for n = 0..1272

Cf. A133656, A348410, A348476, A348478.

b:= proc(n, t) option remember; `if`(t=0, 1-signum(n),
      add(b(n-j, t-1)*iquo(j+3, 2), j=0..n))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..25);

b[n_, t_] := b[n, t] = If[t == 0, 1 - Sign[n],
     Sum[b[n - j, t - 1]*Quotient[j + 3, 2], {j, 0, n}]];
a[n_] := b[n, n];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 07 2022, after Alois P. Heinz *)

a(n) ~ c * d^n / sqrt(Pi*n), where d = 6.12846447590595003785095345916525... is the real root of the equation 32*d^4 - 195*d^3 + 12*d^2 - 112*d - 20 = 0 and c = 0.5667463795063214394117147185755881... is positive root of the equation 182464*c^8 - 45616*c^6 - 2108*c^4 - 601*c^2 - 20 = 0. - Vaclav Kotesovec, Nov 01 2021
From Peter Bala, Feb 22 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x - x^2)/((1 + x)*(1 - x)^2) )^n.
If true, then the following hold:
a(n) = Sum_{i = 0..n} Sum_{j = 0..n} binomial(n,i+2*j)*binomial(2*i+2*j-1, i)*binomial(n+j-1,j).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 6*x^2 + 23*x^3 + 99*x^4 + ... is the g.f. of A133656.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

A370103 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n+k-1,k) binomial(4*n-k-1,n-k).

Original entry on oeis.org

1, 1, 7, 28, 151, 751, 3976, 20924, 112023, 602182, 3260257, 17724928, 96766072, 529977917, 2910984412, 16027963528, 88440034711, 488918693466, 2707393587802, 15014647096172, 83380131228401, 463593653171495, 2580426581343200, 14377474236172320

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A351857, A370104.
Cf. A352373, A370098, A370101.
Cf. A001700, A348410.
Cf. A365854.

a(n) = sum(k=0, n, (-1)^k*binomial(2*n+k-1, k)*binomial(4*n-k-1, n-k));

a(n, s=2, t=3, u=1) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial(u*n, n-s*k));

a(n, s=2, t=2, u=1) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial((u+1)*n-s*k-1, n-s*k));

a(n) = [x^n] 1/( (1+x)^2 * (1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1+x)^2*(1-x)^3 ). See A365854.
a(n) = Sum_{k=0..floor(n/2)} binomial(3*n+k-1,k) * binomial(n,n-2*k).
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n+k-1,k) * binomial(2*n-2*k-1,n-2*k).

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)(1 - x^2))^(nk) for n, k >= 1.

Original entry on oeis.org

1, 1, 5, 1, 7, 19, 1, 9, 46, 85, 1, 11, 82, 327, 376, 1, 13, 127, 793, 2376, 1715, 1, 15, 181, 1547, 7876, 17602, 7890, 1, 17, 244, 2653, 19376, 79686, 132056, 36693, 1, 19, 316, 4175, 40001, 247205, 816684, 1000263, 171820, 1, 21, 397, 6177, 73501, 614389, 3195046, 8450585, 7632433, 809380

Offset: 0

Peter Bala, Jun 13 2023

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.
We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.
The table can be extended to negative values of n, and the row sequences also appear to satisfy the above supercongruences.

			The square array begins
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  1   5   19     85     376      1715       7890   ...     (A348410)
  2  |  1   7   46    327    2376     17602     132056   ...
  3  |  1   9   82    793    7876     79686     816684   ...
  4  |  1  11  127   1547   19376    247205    3195046   ...
  5  |  1  13  181   2653   40001    614389    9560097   ...
  6  |  1  15  244   4175   73501   1318236   23952720   ...
  7  |  1  17  316   6177  124251   2546288   52867620   ...
  8  |  1  19  397   8723  197251   4544407  106076867   ...
  9  |  1  21  487  11877  298126   7624551  197571088   ...
 10  |  1  23  586  15703  433126  12172550  346618308   ...
Array extended to negative values of n:
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
 -5  |  1  -7   46   -247     626      8642    -194480   ...
 -4  |  1  -5   19     -5    -874     11569    -105300   ...
 -3  |  1  -3    1     77    -749      4641     -19893   ...
 -2  |  1  -1   -8     63    -249       440       1716   ...
 -1  |  1   1   -8     17       1      -116        344   ...     (-A234839)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Row 1 A348410. Cf. A362724 - A362733, A363418.

# display as a square array
T := (n,k) -> (1/n)*add( (-1)^(k+j) * binomial(-n*k,j)*binomial(-n*k, k-2*j) , j = 0..floor(k/2)): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
# display as a sequence
seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

T(n,k) = (1/n)*Sum_{j = 0..floor(k/2)} binomial(n*k + j - 1, j)*binomial((n+1)*k - 2*j - 1, k - 2*j).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)(1 - x^2))^n.
T(n,k) = (1/n)*binomial(n*k+k-1,k) * hypergeom([n*k, -k/2, (1 - k)/2], [(1 - (n+1)*k)/2, (2 - (n+1)*k)/2], 1) except when n = 1 and k = 1 or 2.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = 1/((1 - x)*(1 - x^2)), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

A370105 a(n) = Sum_{k=0..n} (-1)^k * binomial(n+k-1,k) * binomial(5*n-k-1,n-k).

Original entry on oeis.org

1, 3, 23, 192, 1687, 15253, 140504, 1311292, 12357015, 117318162, 1120436273, 10752242592, 103596191608, 1001494496863, 9709576926716, 94369011385192, 919175964169623, 8970063281146830, 87685232945278010, 858446087522807784, 8415669293820893937

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A348410, A352373.

Cf. A365752.

a(n) = sum(k=0, n, (-1)^k*binomial(n+k-1, k)*binomial(5*n-k-1, n-k));

a(n, s=2, t=1, u=3) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial((u+1)*n-s*k-1, n-s*k));

a(n) = [x^n] 1/( (1+x) * (1-x)^4 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1+x)*(1-x)^4 ). See A365752.
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1,k) * binomial(4*n-2*k-1,n-2*k).

cp's OEIS Frontend

A234839 a(n) = Sum_{k = 0..n} (-1)^k * binomial(n,k) * binomial(2*n,k).

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A063020 Reversion of y - y^2 - y^3 + y^4.

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A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_(2*n).

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A351856 Number of nonnegative integer solutions to 2*n = x_1 + x_2 + ... + x_n + 2*y_1 + 2*y_2 + ... + 2*y_n.

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A350406 a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.

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A348474 Number of compositions of n into exactly 2n nonnegative parts such that each positive i-th part is odd if i is odd.

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A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2n) + 2y_1 + 2y_2 + ... + 2y_(2*n).

A351856 Number of nonnegative integer solutions to 2n = x_1 + x_2 + ... + x_n + 2y_1 + 2y_2 + ... + 2y_n.

A370103 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n+k-1,k) binomial(4*n-k-1,n-k).

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)(1 - x^2))^(nk) for n, k >= 1.