A351857 -id:A351857 - cp's OEIS frontend

A348410 Number of nonnegative integer solutions to n = Sum_{i=1..n} (a_i + b_i), with b_i even.

1, 1, 5, 19, 85, 376, 1715, 7890, 36693, 171820, 809380, 3830619, 18201235, 86770516, 414836210, 1988138644, 9548771157, 45948159420, 221470766204, 1069091485500, 5167705849460, 25009724705460, 121171296320475, 587662804774890, 2852708925078675, 13859743127937876

Offset: 0

César Eliud Lozada, Oct 17 2021

Suppose n objects are to be distributed into 2n baskets, half of these white and half black. White baskets may contain 0 or any number of objects, while black baskets may contain 0 or an even number of objects. a(n) is the number of distinct possible distributions.

			Some examples (semicolon separates white basket from black baskets):
For n=1: {{1 ; 0}} - Total possible ways: 1.
For n=2: {{0, 0 ; 0, 2}, {0, 0 ; 2, 0}, {0, 2 ; 0, 0}, {1, 1 ; 0, 0}, {2, 0 ; 0, 0}} - Total possible ways: 5.

Alois P. Heinz, Table of n, a(n) for n = 0..1441

Cf. A033715, A033716, A034297, A063020, A088218, A234839, A294860, A348474, A351856, A351857, A352373.

b:= proc(n, t) option remember; `if`(t=0, 1-signum(n),
      add(b(n-j, t-1)*(1+iquo(j, 2)), j=0..n))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..25);  # Alois P. Heinz, Oct 17 2021

(* giveList=True produces the list of solutions *)
(* giveList=False gives the number of solutions *)
counter[objects_, giveList_: False] :=
  Module[{n = objects, nb, eq1, eqa, eqb, eqs, var, sol, var2, list},
   nb = n;
   eq1 = {Total[Map[a[#] + 2*b[#] &, Range[nb]]] - n == 0};
   eqa = {And @@ Map[0 <= a[#] <= n &, Range[nb]]};
   eqb = {And @@ Map[0 <= b[#] <= n &, Range[nb]]};
   eqs = {And @@ Join[eq1, eqa, eqb]};
   var = Flatten[Map[{a[#], b[#]} &, Range[nb]]];
   var = Join[Map[a[#] &, Range[nb]], Map[b[#] &, Range[nb]]];
   sol = Solve[eqs, var, Integers];
   var2 = Join[Map[a[#] &, Range[nb]], Map[2*b[#] &, Range[nb]]];
   list = Sort[Map[var2 /. # &, sol]];
   list = Map[StringReplace[ToString[#], {"," -> " ;"}, n] &, list];
   list = Map[StringReplace[#, {";" -> ","}, n - 1] &, list];
   Return[
    If[giveList, Print["Total: ", Length[list]]; list, Length[sol]]];
   ];
(* second program: *)
b[n_, t_] := b[n, t] = If[t == 0, 1 - Sign[n], Sum[b[n - j, t - 1]*(1 + Quotient[j, 2]), {j, 0, n}]];
a[n_] := b[n, n];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 16 2023, after Alois P. Heinz *)

Conjecture: D-finite with recurrence +7168*n*(2*n-1)*(n-1)*a(n) -64*(n-1)*(1759*n^2-5294*n+5112)*a(n-1) +12*(7561*n^3-75690*n^2+165271*n-101070)*a(n-2) +5*(110593*n^3-743946*n^2+1659971*n-1232778)*a(n-3) +2680*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4)=0. - R. J. Mathar, Oct 19 2021
From Vaclav Kotesovec, Nov 01 2021: (Start)
Recurrence (of order 2): 16*(n-1)*n*(2*n - 1)*(51*n^2 - 162*n + 127)*a(n) = (n-1)*(5457*n^4 - 22791*n^3 + 32144*n^2 - 17536*n + 3072)*a(n-1) + 8*(2*n - 3)*(4*n - 7)*(4*n - 5)*(51*n^2 - 60*n + 16)*a(n-2).
a(n) ~ sqrt(3 + 5/sqrt(17)) * (107 + 51*sqrt(17))^n / (sqrt(Pi*n) * 2^(6*n+2)). (End)
From Peter Bala, Feb 21 2022: (Start)
a(n) = [x^n] ( (1 - x)*(1 - x^2) )^(-n). Cf. A234839.
a(n) = Sum_{k = 0..floor(n/2)} binomial(2*n-2*k-1,n-2*k)*binomial(n+k-1,k).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 9*x^3 + 32*x^4 + 119*x^5 + ... is the g.f. of A063020.
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) 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)^(n+k)*binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). Cf. A352373. - Peter Bala, Jun 05 2024

More terms from Alois P. Heinz, Oct 17 2021

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

Original entry on oeis.org

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501

Offset: 0

Peter Bala, Feb 27 2022

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.
The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.
We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.
Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.
More generally, if r is a positive integer and s an integer then the sequence defined by  u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)

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

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351859, A372211, A372212, A372213.

seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);

A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).
Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).
The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022

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.

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).

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

Original entry on oeis.org

1, 1, 3, 19, 67, 251, 1137, 4803, 20035, 87013, 377753, 1634469, 7134385, 31261114, 137121113, 603206144, 2660097603, 11749336328, 51981371895, 230336544210, 1021976441817, 4539784391763, 20188837618799, 89871081815631, 400427435522737, 1785639575031501

Offset: 0

Peter Bala, Mar 01 2022

This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
a(2*7) - a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)

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 = 0..425
R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351858.

seq(add(add(add((-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)* binomial(3*n+j-1,j)*binomial(j,k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);

A351859[n_] := Sum[(-1)^j*Binomial[4*n, n-2*i-j-k]*Binomial[4*n, i]*Binomial[3*n+j-1, j]*Binomial[j, k], {i, 0, n}, {j, 0, n}, {k, 0, j}];
Array[A351859, 25, 0] (* Paolo Xausa, May 30 2025 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,(-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)*binomial(3*n+j-1,j)*binomial(j,k))));
vector(25,n,a(n-1)) \\ Paolo Xausa, May 04 2022

a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).

A370104 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n+k-1,k) binomial(6*n-k-1,n-k).

Original entry on oeis.org

1, 3, 25, 219, 2025, 19253, 186469, 1829565, 18124521, 180886260, 1815946275, 18318160358, 185518492965, 1885157971596, 19211066004995, 196258973605094, 2009302383218409, 20610411795602760, 211768072490024440, 2179156980022097775, 22454554231950998275

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A351857, A370103.

Cf. A365856.

a(n) = sum(k=0, n, (-1)^k*binomial(2*n+k-1, k)*binomial(6*n-k-1, n-k));

a(n, s=2, t=2, 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)^2 * (1-x)^5 )^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)^5 ). See A365856.
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n+k-1,k) * binomial(4*n-2*k-1,n-2*k).

cp's OEIS Frontend

A348410 Number of nonnegative integer solutions to n = Sum_{i=1..n} (a_i + b_i), with b_i even.

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A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(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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A370103 a(n) = Sum_{k=0..n} (-1)^k * binomial(2*n+k-1,k) * binomial(4*n-k-1,n-k).

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A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).

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A370104 a(n) = Sum_{k=0..n} (-1)^k * binomial(2*n+k-1,k) * binomial(6*n-k-1,n-k).

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A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

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).

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

A370104 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n+k-1,k) binomial(6*n-k-1,n-k).