A002294 -id:A002294 - cp's OEIS frontend

A213227 G.f. satisfies: A(x) = 1/(1 - x/A(-x*A(x)^6)).

1, 1, 2, 8, 35, 181, 1042, 6301, 39435, 249744, 1585386, 10027385, 62696192, 385398251, 2322152120, 13727653882, 80274175978, 472701550856, 2883417403654, 18796497074750, 132728456810968, 995480740265410, 7605881152587204, 56821415293287735, 403362682583930224

Offset: 0

Paul D. Hanna, Jun 06 2012

nonn

Compare g.f. to:
(1) G(x) = 1/(1 - x/G(-x*G(x)^3)^1) when G(x) = 1/(1 - x*G(x)^1) (A000108).
(2) G(x) = 1/(1 - x/G(-x*G(x)^5)^2) when G(x) = 1/(1 - x*G(x)^2) (A001764).
(3) G(x) = 1/(1 - x/G(-x*G(x)^7)^3) when G(x) = 1/(1 - x*G(x)^3) (A002293).
(4) G(x) = 1/(1 - x/G(-x*G(x)^9)^4) when G(x) = 1/(1 - x*G(x)^4) (A002294).

			G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 35*x^4 + 181*x^5 + 1042*x^6 +...
Related expansions:
A(x)^6 = 1 + 6*x + 27*x^2 + 128*x^3 + 645*x^4 + 3462*x^5 + 19823*x^6 +...
1/A(-x*A(x)^6) = 1 + x + 5*x^2 + 20*x^3 + 108*x^4 + 638*x^5 + 3889*x^6 +...

Cf. A213225, A213226, A213228, A213229, A213230, A213231, A213232, A213233.
Cf. A213091, A213092, A213093, A213094, A213095, A213096, A213098.
Cf. A213099, A213100, A213101, A213102, A213103, A213104, A213105.
Cf. A213108, A213109, A213110, A213111, A213112, A213113.

{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1/(1-x/subst(A, x, -x*subst(A^6, x, x+x*O(x^n)))) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

A214769 G.f. satisfies: A(x) = 1/A(-x*A(x)^9).

Original entry on oeis.org

1, 2, 20, 220, 2280, 25920, 443744, 10057408, 215047552, 3841564160, 57161584256, 757459114112, 10427052678656, 166827795710208, 2728593278189568, 38108069305433088, 521570277192555520, 14195894062729323520, 594582326909611536384, 21399757674339677249536

Offset: 0

Paul D. Hanna, Jul 29 2012

nonn

Compare g.f. to: G(x) = 1/G(-x*G(x)^7) when G(x) = 1 + x*G(x)^5 (A002294).

An infinite number of functions G(x) satisfy (*) G(x) = 1/G(-x*G(x)^9); for example, (*) is satisfied by G(x) = F(m*x) = 1 + m*x*F(m*x)^5 for all m, where F(x) is the g.f. of A002294.

			G.f.: A(x) = 1 + 2*x + 20*x^2 + 220*x^3 + 2280*x^4 + 25920*x^5 + 443744*x^6 +...
A(x)^5 = 1 + 10*x + 140*x^2 + 1980*x^3 + 26680*x^4 + 362432*x^5 + 5617920*x^6 +...
A(x)^9 = 1 + 18*x + 324*x^2 + 5532*x^3 + 88776*x^4 + 1386432*x^5 + 22460832*x^6 +...

Cf. A214761, A214762, A214763, A214764, A214765, A214766, A214767, A214768.

{a(n)=local(A=1+2*x);for(i=0,n,A=(A+1/subst(A,x,-x*A^9+x*O(x^n)))/2);polcoeff(A,n)}
for(n=0,31,print1(a(n),", "))

The g.f. of this sequence is the limit of the recurrence:

(*) G_{n+1}(x) = (G_n(x) + 1/G_n(-x*G_n(x)^9))/2 starting at G_0(x) = 1+2*x.

A365181 G.f. satisfies A(x) = 1 + xA(x)^4(1 + x*A(x)^2).

Original entry on oeis.org

1, 1, 5, 32, 237, 1905, 16160, 142392, 1290613, 11955947, 112697701, 1077438356, 10422562156, 101827196684, 1003312506776, 9958506719664, 99479743121349, 999370184665407, 10090067735619023, 102330789530653912, 1041997707624103589, 10648963961114066129

Offset: 0

Seiichi Manyama, Aug 25 2023

nonn

Cf. A002294, A365178, A365180, A365182, A365183.

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

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

A371753 a(n) = Sum_{k=0..floor(n/2)} binomial(5n-2k-1,n-2*k).

Original entry on oeis.org

1, 4, 37, 376, 4013, 44064, 492871, 5585080, 63901421, 736575316, 8540549322, 99503540008, 1163910870767, 13660217796736, 160782910480936, 1897131524755896, 22433316399634669, 265775992115557076, 3154067508987675679, 37487016824453703920, 446148092364247390618

Offset: 0

Seiichi Manyama, Apr 05 2024

nonn

Cf. A026641, A147855, A183160.

Cf. A001449, A079589, A079678, A385632, A386812.

A371753 := proc(n)
    add( binomial(5*n-2*k-1,n-2*k),k=0..floor(n/2)) ;
end proc:
seq(A371753(n),n=0..50) ; # R. J. Mathar, Sep 27 2024

a(n) = sum(k=0, n\2, binomial(5*n-2*k-1, n-2*k));

a(n) = [x^n] 1/((1-x^2) * (1-x)^(4*n)).
a(n) ~ 5^(5*n + 3/2) / (3 * sqrt(Pi*n) * 2^(8*n + 5/2)). - Vaclav Kotesovec, Apr 05 2024
Conjecture D-finite with recurrence +1024*n*(796184150374453*n -1374782084855770) *(4*n-3)*(2*n-1)*(4*n-1)*a(n) +64*(-4720591427354845074*n^5 +16046598674673412696*n^4 -14164434258362644374*n^3 -6132680339747354209*n^2 +16406971563067867560*n -7312237120275595200)*a(n-1) +40*(-4968388566264801507*n^5 +51044954667717039608*n^4 -218029351288077225930*n^3 +471970442274586326109*n^2 -511707487331990011785*n +221366817798624198360)*a(n-2) -25*(5*n-11) *(719005061479699*n -1438086256867727)*(5*n-9) *(5*n-13)*(5*n-12)*a(n-3)=0. - R. J. Mathar, Sep 27 2024
From Seiichi Manyama, Aug 05 2025: (Start)
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(5*n+1,k).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(4*n+k,k). (End)
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(5*n+1,k) * binomial(5*n-k,n-k).
G.f.: g^2/((-1+2*g) * (5-4*g)) where g = 1+x*g^5 is the g.f. of A002294. (End)
G.f.: B(x)^2/(1 + 6*(B(x)-1)/5), where B(x) is the g.f. of A001449. - Seiichi Manyama, Aug 15 2025
G.f.: 1/(1 - x*g^3*(-5+9*g)) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 16 2025

A385632 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(5*n+1,k).

Original entry on oeis.org

1, 8, 81, 872, 9669, 109128, 1246419, 14359304, 166512285, 1940885504, 22717923586, 266833238328, 3143237113479, 37119019790016, 439290932937672, 5208668386199112, 61861932606093901, 735804601177846968, 8763478151940329859, 104498114621004830160, 1247410783999193335434

Offset: 0

Seiichi Manyama, Aug 03 2025

nonn

Cf. A006256, A141223, A385605.
Cf. A001449, A079589, A079678, A371753, A386812.
Cf. A386699.

a(n) = sum(k=0, n, 2^(n-k)*binomial(5*n+1, k));

a(n) = [x^n] 1/((1-3*x) * (1-x)^(4*n+1)).
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(4*n+k,k).
a(n) = 3^(5*n+1)*2^(-4*n-1) - binomial(5*n+1, n)*(hypergeom([1, -1-4*n], [1+n], -1/2) - 1). - Stefano Spezia, Aug 05 2025
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(5*n+1,k) * binomial(5*n-k,n-k). - Seiichi Manyama, Aug 07 2025
G.f.: g^2/((3-2*g) * (5-4*g)) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 14 2025
From Seiichi Manyama, Aug 16 2025: (Start)
G.f.: 1/(1 - x*g^3*(15-7*g)) where g = 1+x*g^5 is the g.f. of A002294.
G.f.: B(x)^2/(1 + 2*(B(x)-1)/5), where B(x) is the g.f. of A001449. (End)

A163455 a(n) = binomial(5*n-1,n).

Original entry on oeis.org

1, 4, 36, 364, 3876, 42504, 475020, 5379616, 61523748, 708930508, 8217822536, 95722852680, 1119487075980, 13136858812224, 154603005527328, 1824010149372864, 21566576904406820, 255485622301674660, 3031718514166879020, 36030431772522503316

Offset: 0

Zak Seidov, Jul 28 2009

Also, number of terms in A163142 with n zeros in binary representation.

All terms >= 4 are divisible by 4.

			a(1)=4 because there are 4 terms in A163142 with 1 zero in binary representation {23,27,29,30}_10 ={10111,11011,11101,11110}_2
a(2)=36 because there are 36 terms in A163142 with 2 zeros in binary representation: {639,703,735,751,759,763,765,766,831,863,879,887,891,893,894,927,943,951,955,957,958,975,983,987,989,990,999,1003,1005,1006,1011,1013,1014,1017,1018,1020}_10={1001111111,...,1111111100}_2
a(3)=364 terms in A163142 from 18431 to 32760 with 3 zeros in binary representation 18431_10=100011111111111_2 and 32760_10=111111111111000_2
a(4)=3876 terms in A163142 from 557055 to 1048560 with 4 zeros in binary representation, etc.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A163142, A118971.

Cf. A088218, A165817, A262977.

[Binomial(5*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

Table[(5*n-1)!/ n!/(4*n-1)!,{n,20}]
Table[Binomial[5 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

B(x):=sum(binomial(5*n-2,n-1)/(n)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,10);

a(n) = binomial(5*n-1,n); \\ Michel Marcus, Oct 06 2015

a(n) = (5n-1)!/(n!(4n-1)!).
G.f.: A(x)=x*B'(x)/B(x), where B(x)/x is g.f. for A118971. Also a(n) = Sum_{k=0..n} (binomial(n-1,n-k)*binomial(4*n,k)). - Vladimir Kruchinin, Oct 06 2015
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-4*n, n).
a(n) = hypergeom([1 - 4*n, -n], [1], 1).
A(x) satisfies A(x/(1 + x)^5) = 1/(1 - 4*x). (End)
From Peter Bala, Jun 05 2024: (Start)
Right-hand side of the identity Sum_{k = 0..n} binomial(n+k-1, k)*binomial(4*n-k-1, n-k) = binomial(5*n-1, n).
a(n) = (3/4)*binomial(4*n, 3*n)*hypergeom([n, -n], [1 - 4*n], 1) for n >= 1. (End)
From Karol A. Penson Jan 20 2025: (Start)
G.f.: 4*z*Hypergeometric5F4([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125*z)/256) + 1.
G.f. A(z) satisfies: z*(1250*A^3 - 250*A^2 + 25*A - 1) + (-3125*z + 256)*A^4 + (3125*z - 256)*A^5 = 0. (End)
G.f.: 1/(5-4*g) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 17 2025

Entry revised by N. J. A. Sloane, Dec 07 2015

A163456 a(n) = binomial(5*n,n)/5.

Original entry on oeis.org

1, 9, 91, 969, 10626, 118755, 1344904, 15380937, 177232627, 2054455634, 23930713170, 279871768995, 3284214703056, 38650751381832, 456002537343216, 5391644226101705, 63871405575418665, 757929628541719755

Offset: 1

Zak Seidov, Jul 28 2009

For prime p, a(p) == 1 (mod p). - Gary Detlefs, Aug 03 2013
In fact, a(p) == 1 (mod p^3) for prime p >= 5. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015
From Robert Israel, Jul 12 2016: (Start)
a(p+1) == 5 (mod p) for primes p >= 5.
a(p^(k+1)) == a(p^k) mod p^(3(k+1)) for primes p >= 5. (End)

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994.

Robert Israel, Table of n, a(n) for n = 1..920
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A000108, A000245, A001764, A002293, A025174, A118970, A163455, A224274, A227726.

seq(binomial(5*n,n)/5, n=1..20); # Robert Israel, Jul 12 2016

Array[Binomial[5 #, #]/5 &, {18}] (* Michael De Vlieger, Oct 09 2015 *)

a(n) = binomial(5*n,n)/5 \\ Altug Alkan, Oct 09 2015

a(n) = (5*n-1)!/(4*n!*(4*n-1)!) = A001449(n)/5 = A163455(n)/4.
a(n) = binomial(5*n,n)/5. - Gary Detlefs, Aug 03 2013
From Peter Bala, Oct 08 2015: (Start)
a(n) = (1/3)*[x^n] (C(x)^3)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274.
exp( 3*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 18*x^2 + 136*x^3 + ... is the o.g.f. for A118970. (End)
From Peter Bala,Jul 12 2016: (Start)
a(n) = 1/6*[x^n] (1 + x)/(1 - x)^(4*n + 1).
a(n) = 1/6*[x^n] ( 1/C(-x)^6 )^n. Cf. A227726. (End)
a(n) ~ 2^(-8*n-3/2)*5^(5*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
From Robert Israel, Jul 12 2016: (Start)
G.f.: x*hypergeom([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125/256)*x).
a(n) = 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)/(8*n*(4*n-3)*(2*n-1)*(4*n-1)). (End)
O.g.f.: f(x)/(1 - 4*f(x)), where f(x) = series reversion (x/(1 + x)^5) = x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + ... is the o.g.f. of A002294 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/4)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+4)*n+k-1,k) = C(5*n,n)/5 and (1/5)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-5)*n+k-1,k) = C(5*n,n)/5, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/4)*Sum_{k = 0..2*n} (-1)^k*binomial(6*n-k-1,2*n-k)*binomial(4*n+k-1,k) = binomial(5*n,n)/5, for n >= 1. - Peter Bala, Mar 09 2022
a(n) = (1/2)* [x*n] F(x)^(2*n) = [x^n] G(x)^n for n >= 1, where F(x) = Sum_{k >= 0} 1/(2*k + 1)*binomial(3*k,k)*x^k is the o.g.f. of A001764 and G(x) = Sum_{k >= 0} 1/(3*k + 1)*binomial(4*k,k)*x^k is the o.g.f. of A002293 (apply Concrete Mathematics, equation 5.60, p. 201). - Peter Bala, Apr 26 2023

Renamed by Peter Bala, Oct 08 2015

A227035 a(n) = Sum_{k=0..floor(n/4)} binomial(n,4k)binomial(5k,k)/(4k+1).

Original entry on oeis.org

1, 1, 1, 1, 2, 6, 16, 36, 76, 172, 436, 1156, 3006, 7606, 19202, 49466, 130156, 345356, 915196, 2421532, 6427001, 17163581, 46087911, 124133531, 334850208, 904691576, 2449891276, 6651540676, 18100561856, 49344295152, 134719523056, 368350942416, 1008680051756

Offset: 0

Vaclav Kotesovec, Jun 28 2013

Generally, Sum(binomial(n,p*k)*binomial((p+1)*k,k)/(p*k+1), k=0..floor(n/p)) is asymptotic to (p+(p+1)^(1+1/p))^(n+3/2)/(p^(n+1)*(p+1)^(1+3/(2*p))*n^(3/2)*sqrt(2*Pi)).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A002294, A007317 (p=1), A049130 (p=2), A226974 (p=3), A226910 (p=5).

Table[Sum[Binomial[n,4*k]*Binomial[5*k,k]/(4*k+1),{k,0,Floor[n/4]}],{n,0,20}]

a(n)=sum(k=0,n\4,binomial(n,4*k)*binomial(5*k,k)/(4*k+1)) \\ Charles R Greathouse IV, Jun 28 2013

Recurrence: -2869*(n-7)*(n-6)*(n-5)*(n-4)*a(n-8) + 2*(n-6)*(n-5)*(n-4)*(5226*n-17267)*a(n-7) - (n-5)*(n-4)*(11582*n^2-55156*n+50139)*a(n-6) - 3*(n-4)*(612*n^3 - 18926*n^2 + 102684*n - 155665)*a(n-5) + 5*(n-4)*(2959*n^3 - 26172*n^2 + 77408*n - 76800)*a(n-4) - 1024*(n-2)*(2*n-5)*(7*n^2-35*n+48)*a(n-3) + 1024*(n-2)*(n-1)*(7*n^2-28*n+30)*a(n-2) - 1024*(n-2)*(n-1)*n*(2*n-3)*a(n-1) + 256*(n-2)*(n-1)*n*(n+1)*a(n) = 0.
a(n) ~ (4+5^(1+1/4))^(n+3/2)/(4^(n+1)*5^(1+3/8)*n^(3/2)*sqrt(2*Pi)).
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x^4 * A(x)^5. - Ilya Gutkovskiy, Jul 25 2021
From Peter Bala, Sep 15 2021: (Start)
O.g.f.: A(x) = (1/x)*series reversion ( x*(1 - x^4)/(1 + x*(1 - x^4) )).
The g.f. of the m-th binomial transform of this sequence is equal to (1/x)*series reversion ( x*(1 - x^4)/(1 + (m + 1)*x*(1 - x^4)) ). The case m = -1 gives the sequence [1,0,0,0,1,0,0,0,5,0,0,0,35,0,0,0,285,...] - an aerated version of A002294. (End)

A345368 a(n) = Sum_{k=0..n} binomial(5k,k) / (4k + 1).

Original entry on oeis.org

1, 2, 7, 42, 327, 2857, 26608, 258488, 2588933, 26539288, 277082658, 2936050788, 31494394563, 341325970323, 3731742758203, 41108999917483, 455850863463768, 5084213586320193, 56997201842602368, 641906808539396253, 7258985455500009623, 82393287049581399283

Offset: 0

Ilya Gutkovskiy, Jul 28 2021

nonn

Partial sums of A002294.

Seiichi Manyama, Table of n, a(n) for n = 0..925

Cf. A002294, A014137, A104859, A345367, A346065, A346647, A346671, A346672.

Table[Sum[Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 21}]
nmax = 21; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^4 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

a(n) = sum(k=0, n, binomial(5*k, k)/(4*k+1)); \\ Michel Marcus, Jul 28 2021

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^4 * A(x)^5.

a(n) ~ 5^(5*n + 11/2) / (2869 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 28 2021

A365341 a(n) = (5n)!/(4n+1)!.

Original entry on oeis.org

1, 1, 10, 210, 6840, 303600, 17100720, 1168675200, 93963542400, 8691104822400, 909171781056000, 106137499051584000, 13679492361575040000, 1929327666754295808000, 295570742023171270656000, 48877281133334949335040000, 8677556868736487617966080000

Offset: 0

Seiichi Manyama, Sep 01 2023

Cf. A001761, A001763, A052795, A365340.
Cf. A004343.
Cf. A000142, A002294.

PARI
```
a(n) = (5*n)!/(4*n+1)!;
```

from sympy import ff
def A365341(n): return ff(5*n,n-1) # Chai Wah Wu, Sep 01 2023

E.g.f.: exp( 1/5 * Sum_{k>=1} binomial(5*k,k) * x^k/k ). - Seiichi Manyama, Feb 08 2024
a(n) = A000142(n)*A002294(n). - Alois P. Heinz, Feb 08 2024
From Seiichi Manyama, Aug 31 2024: (Start)
E.g.f. satisfies A(x) = 1/(1 - x*A(x)^4).
a(n) = Sum_{k=0..n} (4*n+1)^(k-1) * |Stirling1(n,k)|. (End)

cp's OEIS Frontend

A213227 G.f. satisfies: A(x) = 1/(1 - x/A(-x*A(x)^6)).

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A214769 G.f. satisfies: A(x) = 1/A(-x*A(x)^9).

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A365181 G.f. satisfies A(x) = 1 + x*A(x)^4*(1 + x*A(x)^2).

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PARI

Formula

A371753 a(n) = Sum_{k=0..floor(n/2)} binomial(5*n-2*k-1,n-2*k).

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Maple

PARI

Formula

A385632 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(5*n+1,k).

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PARI

Formula

A163455 a(n) = binomial(5*n-1,n).

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Magma

Mathematica

Maxima

PARI

Formula

Extensions

A163456 a(n) = binomial(5*n,n)/5.

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Maple

Mathematica

PARI

Formula

Extensions

A227035 a(n) = Sum_{k=0..floor(n/4)} binomial(n,4*k)*binomial(5*k,k)/(4*k+1).

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A365181 G.f. satisfies A(x) = 1 + xA(x)^4(1 + x*A(x)^2).

A371753 a(n) = Sum_{k=0..floor(n/2)} binomial(5n-2k-1,n-2*k).

A227035 a(n) = Sum_{k=0..floor(n/4)} binomial(n,4k)binomial(5k,k)/(4k+1).

A345368 a(n) = Sum_{k=0..n} binomial(5k,k) / (4k + 1).

A365341 a(n) = (5n)!/(4n+1)!.