A103885 -id:A103885 - cp's OEIS frontend

A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).

1, 3, 19, 138, 1059, 8378, 67582, 552576, 4563235, 37972290, 317894394, 2674398268, 22590697614, 191475925332, 1627653567916, 13870754053388, 118464647799075, 1013709715774130, 8689197042438274, 74594573994750972, 641252293546113434, 5519339268476249676, 47558930664216470628

Offset: 0

Paul Barry, Feb 17 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A156894

Cf. A002003, A003169, A103885, A114496, A123164.

A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >;
[A156894(n): n in [0..30]]; // G. C. Greubel, Jan 06 2022

a := n -> hypergeom([-n, 2*n], [1], -1);
seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Aug 02 2014

Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Nov 12 2014 *)

a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k)));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015

[round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # G. C. Greubel, Jan 06 2022

a(n) = [x^n] ((1+x)/(1-x)^2)^n.
a(n) = (4*(n+1)*(2*n+1)*A003169(n+1) - (5*n+1)*(2*n-1)*A003169(n))/(17*n + 5) for n>0. - Mark van Hoeij, Jul 14 2010
a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - Peter Luschny, Aug 02 2014
Conjecture: 64*n*(2*n-1)*a(n) -16*(89*n^2 -134*n +63)*a(n-1) +4*(661*n^2 -2619*n +2576)*a(n-2) -3*(119*n^2 -713*n +1092)*a(n-3) +6*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 16*n*(782*n +5365)*(2*n-1)*a(n) +8*(3128*n^3 -362053*n^2 +593930*n -290328)*a(n-1) -3*(726869*n^3 -5105981*n^2 +11667946*n -8715544)*a(n-2) +158*(2*n-5)*(n-3)*(391*n -764)*a(n-3) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 4*n*(2*n-1)*(17*n^2 -52*n +39)*a(n) -(1207*n^4 -4899*n^3 +6692*n^2 -3504*n +576)*a(n-1) +2*(n-2)*(2*n-3)*(17*n^2 -18*n +4)*a(n-2) = 0. - R. J. Mathar, Feb 05 2015 [the Maple command sumrecursion (binomial(n,k) * binomial(2*n+k-1,k), k, a(n)) verifies this recurrence. - Peter Bala, Oct 05 2015 ]
a(n) ~ sqrt(578 + 306*sqrt(17)) * (71 + 17*sqrt(17))^n / (17 * sqrt(Pi*n) * 2^(4*n+2)). - Vaclav Kotesovec, Feb 05 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 14*x^2 + 79*x^3 + ... is the o.g.f. of A003169 (taken with offset 0). - Peter Bala, Oct 05 2015
From Peter Bala, Mar 20 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A119259.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)

A370102 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(5*n-k-1,n-k).

Original entry on oeis.org

1, 8, 128, 2312, 44032, 864008, 17282432, 350353928, 7172939776, 147972367880, 3070951360128, 64044689834760, 1341056098444288, 28176478479561992, 593725756425591680, 12542160174109922312, 265525958014053580800, 5632170795392966388744

Offset: 0

Seiichi Manyama, Feb 10 2024

Cf. A006318, A103885, A365847, A370098, A370099.

Cf. A001448, A370100, A370101.

seq(simplify(binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1)), n = 0..20); # Peter Bala, Jul 29 2024

a(n) = sum(k=0, n, binomial(4*n, k)*binomial(5*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^4/(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)^4/(1+x)^4 ). See A365847.
From Peter Bala, Jul 20 2024: (Start)
a(n) = binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1).
For n >=1, a(n) = (4/3) * [x^n] S(x)^(3*n) = (4/5) * [x^n] (1/S(-x))^(5*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
n*(4*n - 3)*(2*n - 1)*(4*n - 1)*(85*n^4 - 510*n^3 + 1138*n^2 - 1119*n + 409)*a(n) = 2*(29665*n^8 - 237320*n^7 + 794282*n^6 - 1443212*n^5 + 1544750*n^4 - 987560*n^3 + 363568*n^2 - 69168*n + 5040)*a(n-1) + (n - 2)*(4*n - 7)*(2*n - 3)*(4*n - 5)*(85*n^4 - 170*n^3 + 118*n^2 - 33*n + 3)*a(n-2) with a(0) = 1 and a(1) = 8.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. (End)
a(n) ~ (349 + 85*sqrt(17))^n / (17^(1/4) * sqrt(Pi*n) * 2^(5*n - 1/2)). - Vaclav Kotesovec, Aug 08 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(4*n).
a(n) = Sum_{k=0..n} 2^k * binomial(4*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n+k-1,k) * binomial(n-1,n-k). (End)

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

A333481 a(n) = [x^n] S(x)^(2n), where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 4, 48, 652, 9344, 138004, 2077968, 31712412, 488793088, 7591462564, 118615816048, 1862444310060, 29361743698304, 464472032918196, 7368841675386960, 117200150284494652, 1868129273410953216, 29834667873867329348, 477283242733227391152, 7647021589988643092428

Offset: 0

Peter Bala, Mar 24 2020

Compare with the sequence A103885(n) = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318. See also A333482.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for prime p and positive integers n and k.
We conjecture that the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 5 and positive integers n and k.
More generally, we conjecture that for any positive integer a and any integer b the sequence u(a,b;n) := [x^(a*n)] S(x)^(b*n) also satisfies the above congruences.

			Examples of congruences:
a(17) - a(1) = 29834667873867329348 - 4 = (2^6)*(17^3)*401*236619262717 == 0 ( mod 17^3 ).
a(2*7) - a(2) = 7368841675386960 - 48 = (2^5)*3*(7^4)*577*3229*17159 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 132585158051380247023537388004 - 138004 = (2^4)*(5^6)* 39461*13439614612035199009 == 0 ( mod 5^6 )

Cf. A006318, A103885, A333482.

[1, seq((2/3)*add(binomial(3*n,k)*binomial(4*n-k-1,3*n-1), k = 0..n), n = 1..25)];
# alternative program
S := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x, n) -> series(S(x)^(2*n), x, 76):
seq(coeff(G(x, n), x, n), n = 0..25);

Join[{1}, Table[2*Binomial[4*n-1, 3*n-1] * Hypergeometric2F1[-3*n, -n, 1 - 4*n, -1]/3, {n,1,20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = (2/3) * Sum_{k = 0..n} C(3*n,k)*C(4*n-k-1,3*n-1) for n >= 1.
P-recursive: P(n)*a(n + 1) = 4*(7805*n^6 - 7132*n^4 + 1559*n^2 - 72)*a(n) -
P(-n)*a(n - 1), where P(n) =(2*n - 1)*(3*n + 1)*(3*n + 2)*(3*n + 3)*(35*n^2 - 35*n + 6).
a(n) ~ 2^(1/4) * (223 + 70*sqrt(10))^n / (5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = (1/2) * [x^n] 1/S(-x)^(4*n) for n >= 1. - Peter Bala, Sep 16 2024

A336521 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is the coefficient of x^(k*n) in expansion of ( (1 + x)/(1 - x) )^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 8, 1, 1, 2, 16, 38, 1, 1, 2, 24, 146, 192, 1, 1, 2, 32, 326, 1408, 1002, 1, 1, 2, 40, 578, 4672, 14002, 5336, 1, 1, 2, 48, 902, 11008, 69002, 142000, 28814, 1, 1, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 1, 1, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146, 1

Offset: 0

Seiichi Manyama, Jul 24 2020

			Square array begins:
  1,    1,     1,     1,      1,      1, ...
  1,    2,     2,     2,      2,      2, ...
  1,    8,    16,    24,     32,     40, ...
  1,   38,   146,   326,    578,    902, ...
  1,  192,  1408,  4672,  11008,  21440, ...
  1, 1002, 14002, 69002, 216002, 525002, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Column k=0-3 give A000012, A123164, A103885, A333715.
Main diagonal gives A336522.
Cf. A122542, A266213, A336534.

T[n_, 0] := 1; T[n_, k_] := Sum[Binomial[n, j] * Binomial[k*n + j - 1, n - 1], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 24 2020 *)

T(n,k) = (1/k) * [x^n] ( (1 + x)/(1 - x) )^(k*n) for k > 0 and n > 0.
T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(k*n+j-1,n-1).
T(n,k) = (1/k) * Sum_{j=0..n} binomial(k*n,n-j) * binomial(k*n+j-1,j) for k > 0 and n > 0.
T(n,k) = Sum_{j=1..n} 2^j * binomial(n,j) * binomial(k*n-1,j-1) for n > 0.
T(n,k) = binomial(k*n-1, n-1)*hypergeom([-n, k*n], [1+(k-1)*n], -1) for k > 0. - Stefano Spezia, Aug 09 2025

A122538 Riordan array (1, x*f(x)) where f(x)is the g.f. of A006318.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 4, 1, 0, 22, 16, 6, 1, 0, 90, 68, 30, 8, 1, 0, 394, 304, 146, 48, 10, 1, 0, 1806, 1412, 714, 264, 70, 12, 1, 0, 8558, 6752, 3534, 1408, 430, 96, 14, 1, 0, 41586, 33028, 17718, 7432, 2490, 652, 126, 16, 1, 0, 206098, 164512, 89898, 39152, 14002, 4080, 938, 160, 18, 1

Offset: 0

Philippe Deléham, Sep 18 2006

Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 2, 1, 2, 1, 2, 1, ...] DELTA [1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 . Inverse is Riordan array (1, x*(1-x)/(1+x)).

T(n, r) gives the number of [0,r]-covering hierarchies with n segments terminating at r (see Kreweras work). - Michel Marcus, Nov 22 2014

			Triangle begins:
  1;
  0,    1:
  0,    2,    1;
  0,    6,    4,    1;
  0,   22,   16,    6,    1;
  0,   90,   68,   30,    8,   1;
  0,  394,  304,  146,   48,  10,  1;
  0, 1806, 1412,  714,  264,  70, 12,  1;
  0, 8558, 6752, 3534, 1408, 430, 96, 14, 1;
Production matrix is:
  0...1
  0...2...1
  0...2...2...1
  0...2...2...2...1
  0...2...2...2...2...1
  0...2...2...2...2...2...1
  0...2...2...2...2...2...2...1
  0...2...2...2...2...2...2...2...1
  0...2...2...2...2...2...2...2...2...1
  ... - _Philippe Deléham_, Feb 09 2014

G. C. Greubel, Rows n = 0..100 of the triangle, flattened
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973), see page 15.

Another version : A080247, A080245, A033877.
Columns: A000007, A006318, A006319, A006320, A006321.
Diagonals: A000012, A005843, A054000.
Sums include: A001003 (row and alternating sign), A006603 (diagonal).
Cf. A103885.

function T(n,k) // T = A122538
  if k eq 0 then return 0^n;
  elif k eq n then return 1;
  else return T(n-1,k-1) + T(n-1,k) + T(n,k+1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 27 2024

T[n_, n_]= 1; T[, 0]= 0; T[n, k_]:= T[n, k]= T[n-1, k-1] + T[n-1, k] + T[n, k+1];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 13 2019 *)

def A122538_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)-2*sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k)*prec(n, k) for k in (0..n)]
for n in (0..12): print(A122538_row(n)) # Peter Luschny, Mar 16 2016

T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n,k+1) if k > 0, with T(n, 0) = 0^n, and T(n, n) = 1.
Sum_{k=0..n} T(n, k) = A001003(n).
From G. C. Greubel, Oct 27 2024: (Start)
T(2*n, n) = A103885(n).
Sum_{k=0..n} (-1)^k*T(n, k) = -A001003(n-1).
Sum_{k=0..floor(n/2)} T(n-k, k) = [n=0] + 0*[n=1] + A006603(n-2)*[n>1]. (End)

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).

A363418 Square array read by ascending antidiagonals: T(n,k) = [x^(n*k)] ((1 + x)/(1 - x))^k for n, k >= 1.

Original entry on oeis.org

2, 2, 8, 2, 16, 38, 2, 24, 146, 192, 2, 32, 326, 1408, 1002, 2, 40, 578, 4672, 14002, 5336, 2, 48, 902, 11008, 69002, 142000, 28814, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146

Offset: 1

Peter Bala, Jun 12 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.

			Square array begins
 n\k |  1   2     3      4        5          6           7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  2   8    38    192     1002       5336       28814   ...   (A002003)
  2  |  2  16   146   1408    14002     142000     1459810   ...   (A103885)
  3  |  2  24   326   4672    69002    1038984    15856206   ...   (A333715)
  4  |  2  32   578  11008   216002    4320608    87588482   ...
  5  |  2  40   902  21440   525002   13104184   331482062   ...
  6  |  2  48  1298  36992  1086002   32497680   985524066   ...
  7  |  2  56  1766  58688  2009002   70097384  2478629134   ...
  8  |  2  64  2306  87552  3424002  136485568  5513464322   ...

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

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 antidiagonals)

A002003 (row 1), A103885 (row 2), A333715 (row 3). Cf. A035607, A362724 - A362733, A363419.

# display as a square array
T := (n,k) -> add( binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j) , j = 0..k): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
#alternative program
seq(print(seq(simplify(2*k*hypergeom([1 - k, 1 - n*k], [2], 2)), k = 1..10)), n = 1..10);
# display as a sequence
seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

T(n,k) = sum(j=0, k, binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j)) \\ Andrew Howroyd, Jan 05 2024

T(n,k) = Sum_{j = 0..k} binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j).
T(n,k) = 1/n * [x^k] ((1 + x)/(1 - x))*(n*k).
T(n,k) = (1/n)*Sum_{j = 0..k} binomial(n*k, j)*binomial((n + 1)*k - j - 1, k - j).
T(2*n,k) = [x^(n*k)] Chebyshev_T(k,(1 + x)/(1 - x)), where Chebyshev_T(n,x) denotes the n-th Chebyshev polynomial of the first kind. See A053120.
T(n,k) = Sum_{j = 1..k} (2^j)*binomial(k, j)*binomial(n*k - 1, n*k - j).
T(n,k) = (2*k) * hypergeom([1 - k, 1 - n*k], [2], 2).
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))^n.
E(m,x) appears to be the g.f. of the (m + 1)-Schroeder numbers. See A027307 (m = 2) and the cross references there.
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 + x)/(1 - x), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

A370099 a(n) = Sum_{k=0..n} binomial(2n,k) binomial(3*n-k-1,n-k).

Original entry on oeis.org

1, 4, 32, 292, 2816, 28004, 284000, 2919620, 30316544, 317222212, 3339504032, 35329425124, 375282559232, 4000059761572, 42760427177696, 458259268924292, 4921911787962368, 52965710906750084, 570951048018417440, 6164049197776406180, 66639047280436354816

Offset: 0

Seiichi Manyama, Feb 10 2024

Cf. A370098, A370102.

Cf. A032349, A103885.

a(n) = sum(k=0, n, binomial(2*n, k)*binomial(3*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^2/(1-x)^2 )^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)^2 ).
a(n) = 2 * A103885(n) for n >= 1. - Peter Bala, Sep 16 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(2*n).
a(n) = Sum_{k=0..n} 2^k * binomial(2*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n+k-1,k) * binomial(n-1,n-k). (End)

A333482 a(n) = [x^(2n)] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 6, 304, 17718, 1093760, 69690006, 4530426640, 298634382374, 19886739416064, 1334658881073894, 90125657301992304, 6116315760393531094, 416791616968522726784, 28500344434239455360758, 1954614576511349850157392, 134392738169746273774331718, 9260873342398000417556078592

Offset: 0

Peter Bala, Mar 24 2020

Compare with the sequence A103885(n) = [x^n] S(x)^n. See also A333481.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
We conjecture that the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 5 and positive integers n and k.
More generally, we conjecture that for any positive integer a and any integer b, the sequence u(a,b;n) := [x^(a*n)] S(x)^(b*n) also satisfies the above congruences.

			Examples of congruences:
a(17) - a(1) = 639400846289617183203551941830 - 6 = (2^6)*3*(17^3)* 677836910460361523003969 == 0 ( mod 17^3 ).
a(2*7) - a(2) = 1954614576511349850157392 - 304 = (2^5)*(7^4)*12408377* 2050236754217 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 346904370487885277935868635823142219775940006 - 69690006 = (2^4)*(5^8)*911*39097145981*1558354721574649484551883 == 0 ( mod 5^6 )

Cf. A006318, A103885, A333481.

[1, seq((1/3)*add(binomial(3*n,k)*binomial(5*n-k-1,3*n-1), k = 0..3*n), n = 1..25)];
# alternative program
S := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x, n) -> series(S(x)^n, x, 82):
seq(coeff(G(x, n), x, 2*n), n = 0..25);

Join[{1}, Table[Binomial[5*n-1, 3*n-1] * Hypergeometric2F1[-3*n, -2*n, 1 - 5*n, -1]/3, {n,1,20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = (1/3) * Sum_{k = 0..2*n} C(3*n,k)*C(5*n-k-1,3*n-1) for n >= 1.
P-recursive: P(n)*s(n+1) = 2*(1023893*n^8 - 1278327*n^6 + 474507*n^4 - 57533*n^2 + 1620)*s(n) + P(-n)*s(n-1), where P(n) = 3*(n + 1)*(2*n + 1)*(3*n + 1)*(3*n + 2)*(533*n^4 - 1066*n^3 + 736*n^2 - 203*n + 18).
a(n) ~ (1921 + 533*sqrt(13))^n / (13^(1/4) * sqrt(Pi*n) * 2^(n+1) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

cp's OEIS Frontend

A156894 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k-1,k).

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A370102 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(5*n-k-1,n-k).

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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.

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A333481 a(n) = [x^n] S(x)^(2*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.

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A336521 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is the coefficient of x^(k*n) in expansion of ( (1 + x)/(1 - x) )^n.

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A122538 Riordan array (1, x*f(x)) where f(x)is the g.f. of A006318.

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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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A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).

A370102 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(5*n-k-1,n-k).

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.

A333481 a(n) = [x^n] S(x)^(2n), where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

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

A370099 a(n) = Sum_{k=0..n} binomial(2n,k) binomial(3*n-k-1,n-k).

A333482 a(n) = [x^(2n)] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.