A333090 -id:A333090 - cp's OEIS frontend

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

1, 2, 8, 41, 232, 1377, 8399, 52138, 327656, 2077934, 13270633, 85226594, 549837391, 3560702069, 23132584742, 150695482041, 984021596136, 6438849555963, 42208999230224, 277144740254566, 1822379123910857, 11998811140766701, 79095365076843134

Offset: 0

Peter Bala, Mar 07 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence
  {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. For cases, see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^n:
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^1 = 1 + x + O(x^2)
  n = 2: c(x)^2 = 1 + 2*x + 5*x^2 + O(x^3)
  n = 3: c(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + O(x^4)
  n = 4: c(x)^4 = 1 + 4*x + 14*x^2 + 48*x^3 + 165*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 1 = 2, a(2) = 1 + 2 + 5 = 8, a(3) = 1 + 3 + 9 + 28 = 41 and a(4) = 1 + 4 + 14 + 48 + 165 = 232.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                        row sums
  n = 0 |   1                               1
  n = 1 |   1   1                           2
  n = 2 |   5   2    1                      8
  n = 3 |  28   9    3   1                 41
  n = 4 | 165  48   14   4   1            232
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 1, 5, 28, 165, ...] = [x^n] c(x)^n = A025174(n).
Examples of supercongruences:
a(13) - a(1) = 3560702069 - 2 = (3^2)*(13^3)*31*37*157 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11998811140766701 - 41 = (2^2)*5*(7^4)*32213*7756841 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 22794614296746579502 - 1377 = (5^6)*7*53*6491*605796421 == 0 ( mod 5^6 ).

Peter Bala, Notes on A333093

Cf. A000108, A001006, A001764, A005554, A025174, A333090 through A333097, A372214, A372215.

seq(add(n/(n+k)*binomial(n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) -> series(c(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - Sqrt[(1 - 3*x)/(1 + x)]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} n/(n+k)*binomial(n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c(x/(1 + x)) )^n = [x^n] ( (1 + x)*(1 + x*M(x)) )^n, where M(x) = ( 1 - x - sqrt(1 - 2*x - 3*x^2) ) / (2*x^2) is the o.g.f. of the Motzkin numbers A001006.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... = (1/x)*Revert( x/c(x) ) is the o.g.f. of A001764.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3^(3*n + 3/2) / (7 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(n+2*k)*binomial(n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024
D-finite with recurrence 2*n*(2*n-1)*(3991*n -21664)*a(n) +(-1329757*n^3 +9119565*n^2 -18270518*n +10657440)*a(n-1) +10*(947050*n^3 -6943257*n^2 +15944396*n -11260008)*a(n-2) +12*(-787878*n^3 +5778161*n^2 -13283386*n +9383340)*a(n-3) +9*(3*n-10)*(3*n-8)*(100503*n -141587)*a(n-4)=0, n>=5. - R. J. Mathar, Nov 22 2024

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 6, 76, 1101, 16876, 266881, 4305247, 70414133, 1163355884, 19369868385, 324486751951, 5462851474614, 92346622131103, 1566455916243068, 26649562889363259, 454528917186429226, 7769463895152496364, 133064720735632286722, 2282869928179537263601, 39225214245206751480102

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4).
In general, for m > 0 and c(x)^(m*n) is a(n) ~ m * (m+2)^((m+2)*n + 3/2) / (((m+1)*(m+2)+1) * sqrt(2*Pi*n) * (m+1)^((m+1)*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

			n-th order Taylor polynomial of c(x)^(5*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^5 = 1 + 5*x + O(x^2)
  n = 2: c(x)^10 = 1 + 10*x + 65*x^2 + O(x^3)
  n = 3: c(x)^15 = 1 + 15*x + 135*x^2 + 950*x^3 + O(x^4)
  n = 4: c(x)^20 = 1 + 20*x + 230*x^2 + 2000*x^3 + 14625*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 5 = 6, a(2) = 1 + 10 + 65 = 76, a(3) = 1 + 15 + 135 + 950 = 1101 and a(4) = 1 + 20 + 230 + 2000 + 14625 = 16876.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(5*n), n >= 0, in descending powers of x begins
                                                row sums
  n = 0 |     1                                     1
  n = 1 |     5        1                            6
  n = 2 |    65       10      1                    76
  n = 3 |   950      135     15    1             1101
  n = 4 | 14625     2000    230   20    1       16876
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 1566455916243068 - 6 = 2*(13^3)*104701*3404923 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11627033261887689372357353 - 1101 = (2^2)*(7^4)*19*29* 2197177609353575713 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 1034770243516278817426081673131 - 266881 = 2*3*(5^7)*31* 13305359*5351978496238483 == 0 ( mod 5^6 ).

Cf. A000108, A233834, A333090 through A333097.

seq(add(5*n/(5*n+k)*binomial(5*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(5*n), x, 151):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[5*Binomial[7*n-1, n] * HypergeometricPFQ[{1, -6*n, -n}, {1/2 - 7*n/2, 1 - 7*n/2}, 1/4]/6, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 5*n/(5*n+k)*binomial(5*n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^5(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 5*x + 45*x^2 + 500*x^3 + 6200*x^4 + ... = (1/x)*Revert( x/c^5(x) ) is the o.g.f. of A233834.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5 * 7^(7*n + 3/2) / (43 * sqrt(Pi*n) * 2^(6*n + 1) * 3^(6*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 5*n/(5*n+2*k)*binomial(5*n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024

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

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 3, 19, 144, 1171, 9878, 85216, 746371, 6609043, 59008563, 530279894, 4790262348, 43458522976, 395683988547, 3613641184739, 33088666355144, 303670285138067, 2792497004892302, 25724693177503987, 237350917999324431, 2193027397174233046, 20288470364637624223

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) defined as the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases A099837(m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(2*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^2 = 1 + 2*x + O(x^2)
  n = 2: c(x)^4 = 1 + 4*x + 14*x^2 + O(x^3)
  n = 3: c(x)^6 = 1 + 6*x + 27*x^2 + 110*x^3 + O(x^4)
  n = 4: c(x)^8 = 1 + 8*x + 44*x^2 + 208*x^3 + 910*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 14 = 19, a(3) = 1 + 6 + 27 + 110 = 144 and a(4) = 1 + 8 + 44 + 208 + 910 = 1171.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |   1                                 1
  n = 1 |   2    1                            3
  n = 2 |  14    4    1                      19
  n = 3 | 110   27    6   1                 144
  n = 4 | 910  208   44   8   1            1171
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 395683988547 - 3 = (2^6)*(3^2)*(13^3)*312677 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 20288470364637624223 - 144 = (7^3)*17*269*12934629208861 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150194008594715226556753 - 9878 = (5^6)*2593*5471* 677584325533 == 0 ( mod 5^6 ).

Cf. A000108, A069271, A333090 through A333097.

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

Table[SeriesCoefficient[((1 + x)^2 * (1 - x - Sqrt[(1 - 3*x)*(1 + x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 2*n/(2*n+k)*binomial(2*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^2(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + ... = (1/x)*Revert( x/c^2(x) ) is the o.g.f. of A069271.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(8*n + 7/2) / (13 * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(2*n+2*k)*binomial(2*n+2*k, k) for n >= 1. - Peter Bala, Apr 19 2024

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 34, 337, 3554, 38754, 431521, 4874377, 55639010, 640177033, 7412165034, 86256322816, 1007980394849, 11820510331777, 139032549536551, 1639506780365337, 19376785465043938, 229458302589724067, 2721958273545613513, 32339465512495259708, 384758834631081248554

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(3*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^3 = 1 + 3*x + O(x^2)
  n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
  n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
  n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                              row sums
  n = 0 |    1                                    1
  n = 1 |    3       1                            4
  n = 2 |   27       6    1                      34
  n = 3 |  273      54    9   1                 337
  n = 4 | 2907     544   90  12   1            3554
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).

Cf. A000108, A118970, A333090 through A333097.

seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 5, 53, 647, 8373, 111880, 1525511, 21093476, 294663349, 4148593604, 58770091928, 836722722951, 11961868391175, 171601856667701, 2469036254872996, 35615467194043147, 514888180699419829, 7458193213805231529, 108219144962546395364, 1572690742149983040857

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(4*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^4 = 1 + 4*x + O(x^2)
  n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)
  n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)
  n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins
                                         row sums
  n = 0 |    1                               1
  n = 1 |    4     1                         5
  n = 2 |   44     8     1                  53
  n = 3 |  544    90    12     1           647
  n = 4 | 7084  1120   152    16   1      8373
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of congruences:
a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ).

Cf. A000108, A212073, A333090 through A333097.

seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(4*n), x, 126):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

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

A372215 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 3x - sqrt(1 - 10x + 9x^2))/(2x) is the g.f. of A082298.

Original entry on oeis.org

1, 5, 65, 1013, 16897, 292005, 5157569, 92456341, 1675300865, 30604622405, 562675808065, 10398428960309, 192983418877441, 3594314403564773, 67146652988725697, 1257656071818605013, 23609209650223284225, 444081728926392461445, 8367715805572617168449

Offset: 0

Peter Bala, Apr 23 2024

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these supercongruences are given below.
More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 4*x + O(x^2)
  n = 2: G(x)^2 = 1 + 8*x + 56*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 12*x + 108*x^2 + 892 *x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 16*x + 176*x^2 + 1680*x^3 + 15024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 56 = 65, a(3) = 1 + 12 +  108 + 892 = 1013 and a(4) = 1 + 16 + 176 + 1680 + 15024 = 16897.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                            row sums
  n = 0 |     1                                 1
  n = 1 |     4      1                          5
  n = 2 |    56      8     1                   65
  n = 3 |   892    108    12    1            1013
  n = 4 | 15024   1680   176   16   1       16897
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 3594314403564773 - 5 = (2^5)*(3^3)*(13^3)*29*7643*8543 == 0 (mod 13^3).
a(2*7) - a(2) = 67146652988725697 - 65 = (2^7)*3*(7^4)*23*3943*803057 = 0 (mod 7^4).

Cf. A082298, A182959, A333090, A333093, A372214.

G := x -> (1/2)*(1 - 3*x - sqrt(1 - 10*x + 9*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 21):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);

Table[SeriesCoefficient[(2*(1 + x)^2/(1 - 2*x + Sqrt[1 - 8*x]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)

G(x) = (1 - 3*x - sqrt(1 - 10*x + 9*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 4*x + 36*x^2 + 420*x^3 + 5572*x^4 + ....
Row sums of the Riordan array ( 1 + x*F'(x)/F(x), x*F(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3^(3*n/2 + 3/4) * (1 + sqrt(3))^(2*n-1) / (sqrt(Pi*n) * 2^(n+1)). - Vaclav Kotesovec, May 02 2024
a(n) = [x^n] H(x)^n, where H(x) = 2*(1 + x)^2/(1 - 2*x + sqrt(1 - 8*x)), the g.f. of A182959, satisfies [x^(n)] H(x)^(3*n) = binomial(6*n, 2*n). - Peter Bala, Nov 07 2024

A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2x - sqrt(1 - 8x + 4x^2))/(2x).

Original entry on oeis.org

1, 4, 40, 487, 6376, 86629, 1203823, 16984678, 242274280, 3484593028, 50444222665, 734066291974, 10728052396111, 157349171819155, 2314894133906086, 34145661019248487, 504810905195542504, 7478066502444399439, 110972913533524676080, 1649407167353221551706, 24549982881130265421001

Offset: 0

Peter Bala, Apr 23 2024

x*G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/2 is the o.g.f. of A047891.
The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these supercongruences are given below.
More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 3*x + O(x^2)
  n = 2: G(x)^2 = 1 + 6*x + 33*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 9*x + 63*x^2 + 414*x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 12*x + 102*x^2 + 768*x^3 + 5493*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 33 = 40, a(3) = 1 + 9 + 63 + 414 = 487 and a(4) = 1 + 12 + 102 + 768 + 5493 = 6376.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                             row sums
  n = 0 |    1                                   1
  n = 1 |    3    1                              4
  n = 2 |   33    6     1                       40
  n = 3 |  414   63     9    1                 487
  n = 4 | 5493  768   102   12   1            6376
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 157349171819155 - 4 = (3^3)*(13^3)*269*9860941 == 0 (mod 13^3).
a(2*7) - a(2) = 2314894133906086 - 40 = 2*(3^4)*(7^3)*11*12119*312509 == 0 (mod 7^3).

Cf. A047891, A219535, A333090, A333093, A372215.

G := x -> (1/2)*(1 - 2*x - sqrt(1 - 8*x + 4*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 41):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);

Table[SeriesCoefficient[(2*(1 + x)^2/(1 - x + Sqrt[1 - 6*x - 3*x^2]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)

G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + ... is the o.g.f. of A219535.
Row sums of the Riordan array ( 1 + x*F'(x)/F(x), x*F(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ sqrt(1 + 17/sqrt(33)) * (59 + 11*sqrt(33))^n / (sqrt(3*Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, May 02 2024

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 5, 57, 761, 10817, 159005, 2386857, 36348401, 559362561, 8676917429, 135445348057, 2125030235113, 33479772021953, 529326516063181, 8393856020704841, 133449301759137761, 2126391547960594433, 33948765589280671589, 542950968447834265209, 8697032976559212532953

Offset: 0

Peter Bala, Mar 22 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n), defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases see A333090 (m = 1) and A333092 (m = 3). For similarly defined sequences see A333093 through A333097.

			n-th order Taylor polynomial of S(x)^(2*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^2 = 1 + 4*x + O(x^2)
  n = 2: S(x)^4 = 1 + 8*x + 48*x^2 + O(x^3)
  n = 3: S(x)^6 = 1 + 12*x + 96*x^2 + 652*x^3 + O(x^4)
  n = 4: S(x)^8 = 1 + 16*x + 160*x^2 + 1296*x^3 + 9344*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 48 = 57, a(3) = 1 + 12 + 96 + 652 = 761 and a(4) = 1 + 16 + 160 + 1296 + 9344 = 10817.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    4    1                           5
  n = 2 |   48    8    1                     57
  n = 3 |  652   96   12   1                761
  n = 4 | 9344 1296  160  16   1          10817
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 4, 48, 652, 9344, 138004, ...]  = [x^n] S(x)^(2*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 529326516063181 - 5 = (2^3)*(13^3)*30116438101 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 2240508640665255893197949 - 761 = (2^2)*3*(7^3)*11* 49485569411283149863 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150633078429259494145205034005 - 159005 = (2^3)*(3^3)*(5^6)*11*23*61*2663*28097*119633*323083 == 0 ( mod 5^6 ).

A006318, A333090 through A333097.

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(2*n), x, 76):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1+x) * (1 - 2*x*(1+x) - Sqrt[1 - 4*x*(1+x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = [x^n] ( (1 + x)*S^2(x/(1 + x)) )^n
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 32*x^2 + 324*x^3 + 3696*x^4 + ... = (1/x)*Revert( x/S^2(x) ).
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ sqrt(120 + 39*sqrt(10)) * (223 + 70*sqrt(10))^n / (30*sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Mar 28 2020

cp's OEIS Frontend

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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

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A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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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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A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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.

A372215 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 3x - sqrt(1 - 10x + 9x^2))/(2x) is the g.f. of A082298.

A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2x - sqrt(1 - 8x + 4x^2))/(2x).

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.