A007054 -id:A007054 - cp's OEIS frontend

A000917 a(n) = (2n+3)!/(n!*(n+2)!).

3, 20, 105, 504, 2310, 10296, 45045, 194480, 831402, 3527160, 14872858, 62403600, 260757900, 1085822640, 4508102925, 18668849760, 77138650050, 318107374200, 1309542023790, 5382578744400, 22093039119060, 90567738003600, 370847442355650, 1516927277253024

Offset: 0

N. J. A. Sloane

G.f.: c(x)*(4-c(x))/(1-4*x)^(3/2), c(x) = g.f. for Catalan numbers A000108 (agrees with Hansen, 1975, p. 99, (5.27.9)). Convolution of A038679 with A000984 (central binomial coefficients); also convolution of A038665 with A000302 (powers of 4). - Wolfdieter Lang, Dec 11 1999
Appears as diagonal in A003506. - Zerinvary Lajos, Apr 12 2006
a(n) is the number of double rises in all Grand Dyck paths of semilength n+2. Example: a(0)=3 because in the 6 (=A000984(2)) Grand Dyck paths of semilength 2, namely udud, (uu)dd, uddu, d(uu)d, dudu, dd(uu), we have a total of 3 uu's (shown between parentheses). - Emeric Deutsch, Nov 29 2008

Eldon R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 99, (5.27.9).

T. D. Noe, Table of n, a(n) for n = 0..200
Jian Zhou, On Some Mathematics Related to the Interpolating Statistics, arXiv:2108.10514 [math-ph], 2021.

Cf. A001622, A001791, A007054, A038665, A038679, A000108, A000984, A000302, A003506, A073010.

1/beta(n, n+2) in A061928.

[(n+1)*Binomial(2*n+3, n+1): n in [0..25]]; // Vincenzo Librandi, Jun 01 2016

a := proc(n) (n+1)*binomial(2*n+3, n+2) end: seq(a(n), n=0..23); # Zerinvary Lajos, Nov 26 2006
seq((n+1)*binomial(2*n+4, n+2)/2, n=0..23); # Zerinvary Lajos, Feb 28 2007

Table[(2*n + 3)!/(n!*(n + 2)!), {n, 0, 25}] (* T. D. Noe, Jun 20 2012 *)

a(n) = (n+1)*binomial(2*n+3, n+1) = (n+1)*A001700(n+1). - Vincenzo Librandi, Jun 01 2016
a(n) = (2*n+3)*A001791(n+1). - R. J. Mathar, Nov 09 2021
D-finite with recurrence +(n+2)*a(n) +10*(-n-1)*a(n-1) +12*(2*n+1)*a(n-2)=0. - R. J. Mathar, Nov 09 2021
D-finite with recurrence n*(n+2)*a(n) -2*(2*n+3)*(n+1)*a(n-1)=0. - R. J. Mathar, Nov 09 2021
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 - Pi/(3*sqrt(3)) = 1 - A073010.
Sum_{n>=0} (-1)^n/a(n) = 6*log(phi)/sqrt(5) - 1, where phi is the golden ratio (A001622). (End)

A110331 Row sums of a number triangle related to the Pell numbers.

Original entry on oeis.org

1, -1, -5, -11, -19, -29, -41, -55, -71, -89, -109, -131, -155, -181, -209, -239, -271, -305, -341, -379, -419, -461, -505, -551, -599, -649, -701, -755, -811, -869, -929, -991, -1055, -1121, -1189, -1259, -1331, -1405, -1481, -1559, -1639, -1721, -1805, -1891, -1979, -2069, -2161, -2255, -2351, -2449

Offset: 0

Paul Barry, Jul 20 2005

Row sums of A110330. Results from a general construction: the row sums of the inverse of the number triangle whose columns have e.g.f. (x^k/k!)/(1-a*x-b*x^2) have g.f. (1-(a+2)x-(2b-a-1)x^2)/(1-x)^3 and general term 1+(b-a)*n-b*n^2. This is the binomial transform of (1,-a,-2b,0,0,0,...).

Hankel transform of A007054(n)-2*0^n. - Paul Barry, Jul 20 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A028387 (absolute values). A165900 is another version.

[1-n-n^2: n in [0..50]]; // Vincenzo Librandi, Jul 08 2012

CoefficientList[Series[(1-4x+x^2)/(1-x)^3,{x,0,50}],x]  (* Vincenzo Librandi, Jul 08 2012 *)
LinearRecurrence[{3,-3,1},{1,-1,-5},60] (* Harvey P. Dale, Mar 22 2022 *)

a(n)=1-n-n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 1-n-n^2.
G.f.: (1-4*x+x^2)/(1-x)^3.
a(n) = binomial(n+2, 2) - 4*binomial(n+1, 2) + binomial(n, 2).
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Jul 08 2012
E.g.f.: exp(x)*(1-2*x-x^2). - Tom Copeland, Dec 02 2013
a(n) = -A165900(n+1) (= -A028387(n-1) for n > 0). - M. F. Hasler, Mar 01 2014

A348893 a(n) = 840(2n)!/((n + 4)!*n!).

Original entry on oeis.org

35, 14, 14, 20, 35, 70, 154, 364, 910, 2380, 6460, 18088, 52003, 152950, 458850, 1400700, 4342170, 13646820, 43421700, 139704600, 454039950, 1489251036, 4925984196, 16419947320, 55124108860, 186281471320, 633357002488, 2165672331088, 7444498638115, 25717358931670, 89254363351090

Offset: 0

Karol A. Penson, Nov 02 2021

Row 3 of array A135573.

Cf. A000108, A001622, A007054, A007272, A348898, A348899.

Maple
```
seq(840*(2*n)!/((n + 4)!*n!),n=0..30)
```

a[n_] := 4^(n + 4) Hypergeometric2F1[9/2, 1/2 - n, 11/2, 1] / (9 Pi);
Table[a[n], {n, 0, 30}] (* Peter Luschny, Nov 03 2021 *)

a(n)=35*binomial(2*n,n)/binomial(n+4,4) \\ Charles R Greathouse IV, Oct 23 2023

def A348893(n): return (-4)^(4 + n)*binomial(7/2, 4 + n)/2
print([A348893(n) for n in range(31)])  # Peter Luschny, Nov 04 2021

O.g.f: (140*z^3 - 70*z^2 + 14*z - 1 + (1 - 4*z)^(7/2))/(2*z^4).
E.g.f: 64*exp(2*z)*((-z^3 - 1/2*z^2 - 1/4*z - 3/32)*BesselI(1,2*z) + BesselI(0,2*z)*z*(z^2 + 1/4*z + 3/32))/z^3.
O.g.f. g(z) satisfies z^4*g(z)^2 + (-140*z^3 + 70*z^2 - 14*z + 1)*g(z) + 4096*z^3 - 2268*z^2 + 476*z - 35 = 0;
a(n) = Integral_{x=0..4} x^n*64*(1 - x/4)^(7/2)/(Pi*sqrt(x)). This is the integral representation as the n-th moment of a positive function on [0, 4]. The representation is unique.
Remark: this sequence is not monotonically growing with n, as a(0) > a(1) = a(2) < a(3) < a(4)... .
From Peter Luschny, Nov 03 2021: (Start)
a(n) = 14*A007272(n)/(n + 4).
a(n) ~ 105*4^n*(8*n - 81)/(n^(11/2)*sqrt(Pi)).
a(n) = 4^(n + 4)*hypergeom([9/2, 1/2 - n], [11/2], 1) / (9*Pi). (End)
a(n) = (-4)^(4 + n)*binomial(7/2, 4 + n)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 10 2023: (Start)
a(n) = 35*binomial(2*n, n) - 56*binomial(2*n, n + 1) + 28*binomial(2*n, n + 2) - 8*binomial(2*n, n + 3) + binomial(2*n, n + 4). Thus this sequence is integral.
7 divides a(n) except when n == 3 (mod 7).
P-recursive: (n + 4)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 35.
D-finite: the o.g.f. A(x) satisfies the differential equation (1 - 4*x)*A'(x) + (4 - 2*x)*A(x) - 140 = 0, with A(0) = 35. (End)
From Peter Bala, Mar 11 2023: (Start)
a(n) = Sum_{k = 0..3} (-1)^k*4^(3-k)*binomial(3,k)*Catalan(n+k) = 64*Catalan(n) - 48*Catalan(n+1) + 12*Catalan(n+2) - Catalan(n+3), where Catalan(n) = A000108(n).
a(n) is odd if n = 2^k - 4, k >= 2, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 13/70 + 4*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 72*log(phi)/(3125*sqrt(5)) - 103/43750, where phi is the golden ratio (A001622). (End)

A361029 a(n) = 120(3n)!/(n!*(n+2)!^2).

Original entry on oeis.org

30, 20, 75, 504, 4620, 51480, 656370, 9237800, 140229804, 2259901800, 38230005450, 673210036800, 12262039956000, 229872375708480, 4417859720647350, 86767376381987400, 1736954330906081100, 35364582637561485000, 730985923118395894950, 15315895532004485418000

Offset: 0

Peter Bala, Mar 01 2023

Row 1 of the square array A361027.

The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result 6*(2n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 120*A006480(n) is divisible by ((n + 1)*(n + 2))^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Row 1 of A361027. Cf. A006480, A007054, A361028, A361030, A361031, A361037.

a := proc(n) option remember; if n = 0 then 30 else 3*(3*n-1)*(3*n-2)/(n+2)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Table[120 (3n)!/(n!(n+2)!^2),{n,0,20}] (* Harvey P. Dale, Jul 02 2023 *)

a(n) = 120/((n+1)*(n+2))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5) * A006480(n+2)/((3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n +
5)), where A006480(n) = (3*n)!/n!^3.
a(n) = (10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2)) * (3*binomial(2*n,n) - 4*binomial(2*n,n+1) + binomial(2*n,n+2)), shows that a(n) is an integer for all n.
a(n) = (1/3)*27^(n+2)*binomial(4/3, n+2)*binomial(5/3, n+2).
a(n) ~ sqrt(3)*60*27^n/(Pi*n^5).
P-recursive: (n + 2)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 30.
The o.g.f. A(x) satisfies the  differential equation x^2*(1 - 27*x)*A''(x) + x*(5 - 54*x)*A'(x) + (4 - 6*x)*A(x) - 120 = 0, with A(0) = 30 and A'(0) = 20.

A184881 a(n) = A184879(2n, n) - A184879(2n, n+1) where A184879(n, k) = Hypergeometric2F1(-2k, 2k-2*n, 1, -1) if 0<=k<=n.

Original entry on oeis.org

1, -3, 2, -3, 6, -14, 36, -99, 286, -858, 2652, -8398, 27132, -89148, 297160, -1002915, 3421710, -11785890, 40940460, -143291610, 504932340, -1790214660, 6382504440, -22870640910, 82334307276, -297670187844, 1080432533656, -3935861372604, 14386251913656

Offset: 0

Paul Barry, Jan 24 2011

sign

Hankel transform is A184882.

Signed version of A007054. - Philippe Deléham, Mar 19 2014

			a(0) = 1;
a(1) = 1 - 4*1 = -3;
a(2) = 4*1 - 2 = 2;
a(3) = 5 - 4*2 = -3;
a(4) = 4*5 - 14 = 6;
a(5) = 42 - 4*14 = -14;
a(6) = 4*42 - 132 = 36;
a(7) = 429 - 4*132 = -99;
a(8) = 4*429 - 1430 = 286, etc; with A000108 = 1,1,2,5,14,42,132,429,1430, ... - _Philippe Deléham_, Mar 19 2014
G.f. = 1 - 3*x + 2*x^2 - 3*x^3 + 6*x^4 - 14*x^5 + 36*x^6 - 99*x^7 + ... - _Michael Somos_, Mar 13 2023

J. W. Layman, The Hankel Transform and Some of Its Properties, J. Integer Sequences 4, No. 01.1.5, 2001
Fumitaka Yura, Hankel Determinant Solution for Elliptic Sequence, arXiv:1411.6972 [nlin.SI], (25-November-2014); see p. 7

Cf. A000108, A002421, A007054, A184879, A184882.

A184879 := proc(n,k) if k<0 or k >n then 0; else hypergeom([-2*k,2*k-2*n],[1],-1) ; simplify(%) ; end if; end proc:
A184881 := proc(n) A184879(2*n,n)-A184879(2*n,n+1) ; end proc:
seq(A184881(n),n=0..40) ; # R. J. Mathar, Feb 05 2011

h[n_, k_] := HypergeometricPFQ[{-2k, 2k - 2n}, {1}, -1];
a[0] = 1; a[n_] := h[2n, n] - h[2n, n + 1];
Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Nov 24 2017 *)

a(n) = 0^n + Sum_{k=0..2n} (C(2n,k)^2-C(2n+2,k)*C(2n-2,k))*(-1)^k.
G.f.: (8*x+1-sqrt(1+4*x)^3)/(2*x). - Philippe Deléham, Mar 19 2014
a(0) = 1, a(n) = (-1)^n*A007054(n-1) for n>0. - Philippe Deléham, Mar 19 2014
(n+1)*a(n) +2*(2*n-3)*a(n-1)=0. - R. J. Mathar, Nov 19 2014
a(n) = (-1)^n*A002421(n+1)/2 and 0 = a(n)*(+16*a(n+1) + 14*a(n+2)) + a(n+1)*(-6*a(n+1) + a(n+2)) for all n>0. - Michael Somos, Mar 13 2023

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

Original entry on oeis.org

315, 280, 3675, 116424, 5885880, 399072960, 33129291195, 3190228041000, 344161801063080, 40616781150254400, 5155510596280207800, 695029472211496161600, 98570579229528369624000, 14597207555235045670540800, 2243893009052293495117018875, 356344642367340570239409729000

Offset: 0

Peter Bala, Mar 01 2023

Row 1 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result (2*n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 2520*A008977(n) is divisible by ((n + 1)*(n + 2))^3, leading to the present sequence. Cf. A361029.
Conjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

Cf. A007054, A008977, A361029, A361032, A361033, A361035.

seq(2520*(4*n)!/(n!*(n+2)!^3), n = 0..20);

a(n) = 2520*A008977(n)/((n+1)*(n+2))^3.
a(n) = (315/2)*A008977(n+2)/((4*n+1)*(4*n+2)*(4*n+3)*(4*n+5)*(4*n+6)*(4*n+7)).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+2)^3 * a(n-1) with a(0) = 315.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(9 - 1152*x)*A(x)'' + x*(19 - 816*x)*A(x)' + (8 - 24*x)*A(x) - 2520 = 0 with A(0) = 315, A'(0) = 280 and A''(0) = 7350.
a(n) ~ 630*sqrt(8/Pi^3) * 2^(8*n)/n^(15/2).

A348899 a(n) = 3326404^nGamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.

Original entry on oeis.org

462, 132, 99, 110, 154, 252, 462, 924, 1980, 4488, 10659, 26334, 67298, 177100, 478170, 1320660, 3721860, 10680120, 31150350, 92205036, 276615108, 840090328, 2580277436, 8007757560, 25090973688, 79319852304, 252832029219, 812127124158, 2627470107570, 8558045493228

Offset: 0

Peter Luschny, Nov 02 2021

nonn

Row 5 of array A135573.

Cf. A000108, A000984, A001622, A007054, A348893.

a := n -> 332640*4^n*GAMMA(n + 1/2)/(sqrt(Pi)*GAMMA(n + 7));
seq(a(n), n = 0..29);

a[n_] := 4^(n + 6) Hypergeometric2F1[13/2, 1/2 - n, 15/2, 1] / (13 Pi);
Table[a[n], {n, 0, 29}]
Array[332640*4^#*Gamma[# + 1/2]/(Sqrt[Pi]*Gamma[# + 7]) &, 30, 0] (* Michael De Vlieger, Nov 02 2021 *)

Let A[c, k](n) = c*4^n*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n + k)). Then
  A[1, 1](n) = A000984(n).
  A[3!, 3](n) = A007054(n).
  A[5!*7, 5](n) = A348893(n).
  A[7!*66, 7](n) = a(n).
  A[c, k](n) ~ -c*2^(2*n - 1)*(k^2 - k - 2*n + 1/4)/(n^(k + 1/2)*sqrt(Pi)).
O.g.f.: ((2048*x^5 - 1686*x^4 + 765*x^3 - 178*x^2 + 21*x - 1)*sqrt(1 - 4*x) - 3496*x^5 + 2934*x^4 - 1083*x^3 + 218*x^2 - 23*x + 1)/(sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))*x^5).
E.g.f.: 1024*exp(2*x)*((-x^5 - 3/4*x^4 - 41/64*x^3 - 123/256*x^2 - 9/32*x - 15/128)*BesselI(1, 2*x) + BesselI(0, 2*x)*x*(x^4 + 1/2*x^3 + 27/64*x^2 + 9/32*x + 15/128))/x^5.
a(n) = Integral_{x=0..4} x^n*(4-x)^(11/2)/(2*Pi*sqrt(x)). This is the integral representation as the n-th moment of a positive function on [0, 4]. The representation is unique.
a(n) = 4^(n + 6)*hypergeom([13/2, 1/2 - n], [15/2], 1) / (13*Pi).
D-finite with recurrence (n+6)*a(n) +2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jul 27 2022
From Peter Bala, Mar 11 2023: (Start)
a(n) = 332640*(2*n)!/(n!*(n + 6)!).
a(n) = Sum_{k = 0..5} (-1)^k*4^(5-k)*binomial(n,k)*Catalan(n+k), where Catalan(n) = A000108(n). Thus a(n) is an integer for all n.
a(n) is odd if n = 2^k - 6, k >= 3, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 101/3465 + 52*Pi/(6561*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 8573/54140625 + 104*log(phi)/(78125*sqrt(5)), where phi is the golden ratio (A001622). (End)

A387248 a(n) = 3/(n + 1) * Catalan(2*n).

Original entry on oeis.org

3, 3, 14, 99, 858, 8398, 89148, 1002915, 11785890, 143291610, 1790214660, 22870640910, 297670187844, 3935861372604, 52749590350072, 715309969142307, 9800129095949682, 135490673691621794, 1888389218820071604, 26510079418051005210, 374589577468070301260, 5324240442532424176260, 76082624294738699098440

Offset: 0

Peter Bala, Aug 24 2025

Bisection of A007054.
Compare with Catalan(n) = 1/(n + 1) * binomial(2*n, n).
For r >= 2, there is a constant K_r such that K_r/(n + 1) * Catalan(r*n) is integral for all n.

Paolo Xausa, Table of n, a(n) for n = 0..800

Cf. A000108, A000984, A001448, A007054, A387249, A387250.

seq( 6/((2*n+1)*(2*n+2)) * binomial(4*n, 2*n), n = 0..22);

A387248[n_] := 3*CatalanNumber[2*n]/(n + 1); Array[A387248, 25, 0] (* Paolo Xausa, Sep 02 2025 *)

a(n) = 6/((2*n + 1)*(2*n + 2)) * binomial(4*n, 2*n).
a(n) = 4*Catalan(2*n) - Catalan(2*n+1) (showing a(n) to be an integer)
G.f.: A(x) = ((2 - f(x))*sqrt(2 + 2*f(x)) - 2)/(4*x), where f(x) = sqrt(1 - 16*x).
a(n) = 2*(4*n - 1)*(4*n - 3)/((n + 1)*(2*n + 1)) * a(n-1) with a(0) = 3.
a(n) ~ 3/(2*sqrt(2*Pi)) * 16^n/n^(5/2).
a(n) is odd iff n = 2^k - 1 for some k, so a(n) has the same parity as Catalan(n).
E.g.f.: 3*hypergeom([1/4, 3/4], [3/2, 2], 16*x). - Stefano Spezia, Aug 27 2025

A038697 Convolution of A000917 with A000984 (central binomial coefficients).

Original entry on oeis.org

3, 26, 163, 894, 4558, 22196, 104739, 483062, 2189530, 9789900, 43295118, 189749676, 825364668, 3567219688, 15332925731, 65591312550, 279415474594, 1185903736412, 5016725589402, 21159849864964, 89012979703940

Offset: 0

Wolfdieter Lang

Also convolution of A007054 (Super ballot numbers) with A002697;

Robert Israel, Table of n, a(n) for n = 0..1652

Cf. A007054, A038665, A038679, A000917, A000108, A000984, A002697.

seq(n*4^(n+1)+binomial(2*n+3,n+1),n=0..30); # Robert Israel, May 22 2019

a(n) = n*4^(n+1)+binomial(2*n+3, n+1).
G.f.: c(x)*(4-c(x))/(1-4*x)^2, where c(x) = g.f. for Catalan numbers A000108.
(160+64*n)*a(n) - (160+48*n)*a(n+1) + (50+12*n)*a(n+2) - (5+n)*a(n+3)=0. - Robert Israel, May 22 2019

A091712 a(n)=6(2n-4)!/((n-2)!n!), if n>2. a(0)=1,a(1)=a(2)=2.

Original entry on oeis.org

1, 2, 2, 2, 3, 6, 14, 36, 99, 286, 858, 2652, 8398, 27132, 89148, 297160, 1002915, 3421710, 11785890, 40940460, 143291610, 504932340, 1790214660, 6382504440, 22870640910, 82334307276, 297670187844, 1080432533656, 3935861372604

Offset: 0

Michael Somos, Jan 31 2004

nonn

A007054(n)=a(n+2), if n>0.

Essentially the same as A007054.

a(n)=if(n<3,(n>=0)+(n>0),6*(2*n-4)!/n!/(n-2)!)

a(n)=if(n<0,0,polcoeff(((1+10*x-2*x^2)+(1-4*x)*sqrt(1-4*x+x*O(x^n)))/2,n))

a(n)=if(n<=0,n==0,polcoeff(subst((1-x)*(1+x)^3,x,serreverse(x-x^2+x*O(x^n))),n))

G.f.: ((1+10x-2x^2)+(1-4x)^(3/2))/2. a(n)=6(2n-4)!/((n-2)!n!), if n>2. a(n)=a(n-1)(4n-10)/n, if n>3.

G.f. A(x) = (2c(x)-1)^3/c(x)^4 = (1-c(x)x)(1+c(x)x)^3, where c(x) = g.f. for Catalan numbers A000108.

cp's OEIS Frontend

A000917 a(n) = (2n+3)!/(n!*(n+2)!).

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A110331 Row sums of a number triangle related to the Pell numbers.

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PARI

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A348893 a(n) = 840*(2*n)!/((n + 4)!*n!).

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A361029 a(n) = 120*(3*n)!/(n!*(n+2)!^2).

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A184881 a(n) = A184879(2*n, n) - A184879(2*n, n+1) where A184879(n, k) = Hypergeometric2F1(-2*k, 2*k-2*n, 1, -1) if 0<=k<=n.

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A361034 a(n) = 2520*(4*n)!/(n!*(n+2)!^3).

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A348899 a(n) = 332640*4^n*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.

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A387248 a(n) = 3/(n + 1) * Catalan(2*n).

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A348893 a(n) = 840(2n)!/((n + 4)!*n!).

A361029 a(n) = 120(3n)!/(n!*(n+2)!^2).

A184881 a(n) = A184879(2n, n) - A184879(2n, n+1) where A184879(n, k) = Hypergeometric2F1(-2k, 2k-2*n, 1, -1) if 0<=k<=n.

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

A348899 a(n) = 3326404^nGamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.