A005810 -id:A005810 - cp's OEIS frontend

A082368 a(n) = (4n-1)! / (n! n! * n! * (n-1)! * 3!).

1, 105, 15400, 2627625, 488864376, 96197645544, 19688264481600, 4148378852099625, 893864677761055000, 196056702961398759480, 43627992869961630486720, 9825387560922608865863400, 2235197406895366368301560000, 512889830640524227455318600000

Offset: 1

John A. Trono (jtrono(AT)smcvt.edu), May 10 2003

Number of combinations that are possible when placing teams ranked #1 to #4*N in a single elimination tournament where there are four columns of N teams (as in the NCAA Men's Division-1 basketball tournament that is played in March) and each column is a separate regional tournament that produces one of the four semi-finalists. (The teams in the columns appear in sorted order and the relative positions of the four columns is irrelevant.)

Number of ways of dividing 4n labeled items into 4 unlabeled boxes with n items in each box. - Dan Parrish, Apr 09 2015

			8 ranked teams (n=2) in a four region, single elimination tournament generates 105 different possible tournament orderings, where the teams in each region are ordered from best to worst. (Teams would be matched up from top to bottom and continue towards the middle two for other matchups, when more than two teams are listed in each column.) 105 tournaments is too many to list here. As this formula applies to single elimination tournaments, this enumeration formula really only makes sense when n is even.

Vincenzo Librandi, Table of n, a(n) for n = 1..300

Row 4 of A060540.

Cf. A000984, A005809, A005810, A008977.

[Factorial(4*n-1) / (Factorial(n)*Factorial(n)* Factorial(n)*Factorial(n-1)*6): n in [1..15]]; // Vincenzo Librandi, Jun 16 2017

[seq(binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24,n=1..17)]; # Zerinvary Lajos, Jun 25 2006

Table[(4 n)! / (4! n!^4), {n, 30}] (* Vincenzo Librandi, Jun 16 2017 *)

a(n)=(4*n)!/(4!*n!^4) \\ Charles R Greathouse IV, Apr 09 2015

a(n) = binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24. - Zerinvary Lajos, Jun 25 2006
a(n) = (4n)!/(4!*n!^4). - Dan Parrish, Apr 09 2015
From Robert Israel, Apr 09 2015: (Start)
a(n) = Gamma(2*n+1/2)*Gamma(n+1/2)*64^n/(24*Pi*(n!)^3).
a(n+1) = 8*(2*n+1)*(4*n+1)*(4*n+3)*a(n)/(n+1)^3.
G.f.: g(x) = x*hypergeom([1,5/4,3/2,7/4],[2,2,2],256*x) satisfies
x^4*(256*x-1)*g''''(x) + 5*x^3*(384*x-1)*g'''(x) + 4*x^2*(780*x-1)*g''(x) + 840*x^2*g'(x) = 0. (End)
From Karol A. Penson, Dec 31 2023: (Start)
a(n) = Integral_{x=0..256} x^n*W(x) dx, n>=0, where W(x) = x^(1/4)*hypergeometric3F2([1/4, 1/4, 1/4], [1/2, 3/4], x/256)/(96*Gamma(3/4)^4) - sqrt(x)*hypergeometric3F2([1/2, 1/2, 1/2], [3/4, 5/4], x/256)/(96*Pi^2) + Gamma(3/4)^4*x^(3/4)*hypergeometric3F2([3/4, 3/4, 3/4], [5/4, 3/2], x/256)/(768*Pi^4) is positive and unimodal on x = [0, 256]. It has a single maximum at approximately x = 31, and it goes to zero with W'(x) diverging, at both x = 0 and x = 256. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 256] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) = A008977(n)/24. - Vaclav Kotesovec, Feb 14 2024

More terms from Zerinvary Lajos, Jun 25 2006

A169961 a(n) = binomial(12*n, n).

Original entry on oeis.org

1, 12, 276, 7140, 194580, 5461512, 156238908, 4529365776, 132601016340, 3911395881900, 116068178638776, 3461014728350400, 103619293824707388, 3112781199432937200, 93780365051563029360, 2832430653037446854640, 85733828145510955528212, 2600022926684976508835280

Offset: 0

N. J. A. Sloane, Aug 07 2010

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Row 12 of A060539.

Cf. A000984, A005809, A005810, A001449, A004355, A004368, A004381, A169958, A169959, A169960.

[Binomial(12*n, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Table[Binomial[12 n, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
CoefficientList[Series[HypergeometricPFQ[Range[11]/12, Range[10]/11,(12^12)/(11^11)*x], {x,0,10}],x] (* Bradley Klee, Jul 01 2018 *)

a(n) = binomial(12*n, n); \\ Michel Marcus, Jul 02 2018

a(n) = C(12*n-1,n-1)*C(144*n^2,2)/(3*n*C(12*n+1,3)), n>0. -  Gary Detlefs, Jan 02 2014
From Bradley Klee, Jul 01 2018 : (Start)
G.f. G(x) and derivatives G^(n)(x)=d^n/dx^n G(x) satisfy a Picard-Fuchs type differential equation, 0=Sum_{m=0..11}(v1_{n}*x^(n+1)-v2_{n}*x^n)*G^(n)(x), with integer coefficient vectors:
v1={479001600, 647647046323200, 99278289544896000, 1290870365178240000, 4245175263164774400, 5313701967430348800, 3083267876011868160, 918801061774295040, 147161631039160320, 12624021804810240, 539424077119488, 8916100448256}
v2={0, 39916800, 14079254112000, 1273481816745600, 11475123393888000, 27687351298068000, 25909403608075680, 11200182937408080, 2427742942653600, 268452344620350, 14265583530550, 285311670611}
G.f.: G(x) = 11F10(m/12;n/11;12^12/11^11*x), m=1..11, n=1..10. (End)
From Vaclav Kotesovec, Jul 15 2018: (Start)
Recurrence: 11*n*(11*n - 10)*(11*n - 9)*(11*n - 8)*(11*n - 7)*(11*n - 6)*(11*n - 5)*(11*n - 4)*(11*n - 3)*(11*n - 2)*(11*n - 1)*a(n) = 41472*(2*n - 1)*(3*n - 2)*(3*n - 1)*(4*n - 3)*(4*n - 1)*(6*n - 5)*(6*n - 1)*(12*n - 11)*(12*n - 7)*(12*n - 5)*(12*n - 1)*a(n-1).
a(n) ~ 2^(24*n + 1/2) * 3^(12*n + 1/2) / (sqrt(Pi*n) * 11^(11*n + 1/2)). (End)
From Peter Bala, Feb 21 2022: (Start)
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 11*A(x))^11 + (12^12)*x*A(x)^12 = 0.
Sum_{n >= 1} a(n)*( x*(11*x + 12)^11/(12^12*(1 + x)^12) )^n = x. (End)
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(12*n+1,k).
G.f.: 1/(1 - 12*x*g^11) where g = 1+x*g^12.
G.f.: g/(12-11*g) where g = 1+x*g^12. (End)

A023826 Sum of exponents in prime-power factorization of C(4n,n).

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 6, 9, 9, 10, 10, 9, 9, 13, 14, 14, 14, 15, 15, 15, 16, 19, 16, 19, 18, 21, 20, 18, 19, 21, 21, 22, 22, 23, 25, 24, 24, 27, 26, 28, 26, 30, 28, 28, 29, 28, 28, 30, 30, 31, 31, 31, 31, 35, 30, 31, 31, 32, 34, 33, 33, 38, 39, 39, 37, 39, 38, 39, 40, 43, 40, 41, 41, 44, 44, 43, 44, 46, 45

Offset: 0

Clark Kimberling

nonn

Also sum of exponents of primes in multinomial coefficient M(4n; n,n,n,n)/M(3n; n,n,n).

Ivan Neretin, Table of n, a(n) for n = 0..10000

Cf. A001222, A005810, A022559.

with(numtheory):a:=proc(n) if n=0 then 0 else bigomega(binomial(4*n,n)) fi end: seq(a(n), n=0..78); # Zerinvary Lajos, Apr 11 2008

Join[{0}, Table[Total[FactorInteger[Binomial[4 n, n]][[All, 2]]], {n, 78}]] (* Ivan Neretin, Nov 02 2017 *)

a(n) = bigomega(binomial(4*n, n)); \\ Amiram Eldar, Jun 11 2025

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A005810(n)).
a(n) = A022559(4*n) - A022559(3*n) - A022559(n). (End)

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jun 11 2025

A079589 a(n) = C(5*n+1,n).

Original entry on oeis.org

1, 6, 55, 560, 5985, 65780, 736281, 8347680, 95548245, 1101716330, 12777711870, 148902215280, 1742058970275, 20448884000160, 240719591939480, 2840671544105280, 33594090947249085, 398039194165652550, 4724081931321677925, 56151322242892212960, 668324943343021950370

Offset: 0

Benoit Cloitre, Jan 26 2003

nonn

a(n) is the number of paths from (0,0) to (5n,n) taking north and east steps while avoiding exactly 2 consecutive north steps. - Shanzhen Gao, Apr 15 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A052203.
Cf. A001449, A079678, A371753, A385632, A386812.
Cf. A005810, A183160.

[Binomial(5*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

seq(binomial(5*n+1,n),n=0..100); # Robert Israel, Aug 07 2014

Table[Binomial[5n+1,n],{n,0,20}]  (* Harvey P. Dale, Jan 23 2011 *)

a(n) is asymptotic to c*(3125/256)^n/sqrt(n) with c=0.557.... [c = 5^(3/2)/(sqrt(Pi)*2^(7/2)) = 0.55753878629774... - Vaclav Kotesovec, Feb 14 2019 and Aug 20 2025]
8*n*(4*n+1)*(2*n-1)*(4*n-1)*a(n) -5*(5*n+1)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Jul 17 2014
G.f.: hypergeom([2/5, 3/5, 4/5, 6/5], [1/2, 3/4, 5/4], (3125/256)*x). - Robert Israel, Aug 07 2014
a(n) = [x^n] 1/(1 - x)^(2*(2*n+1)). - Ilya Gutkovskiy, Oct 10 2017
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} binomial(5*n-k,n-k).
G.f.: 1/(1 - x*g^3*(5+g)) where g = 1+x*g^5 is the g.f. of A002294.
G.f.: g^2/(5-4*g) where g = 1+x*g^5 is the g.f. of A002294.
G.f.: B(x)^2/(1 + 4*(B(x)-1)/5), where B(x) is the g.f. of A001449. (End)

A226733 G.f.: 1 / (1 + 8xG(x)^2 - 10xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

Original entry on oeis.org

1, 2, 18, 142, 1186, 10152, 88414, 779508, 6936066, 62159224, 560238728, 5072970366, 46114086446, 420558296888, 3846232573236, 35261290343112, 323952686556354, 2981787128165592, 27491128592627800, 253835886034173848, 2346892194318851016, 21724880414632781472

Offset: 0

Paul D. Hanna, Jun 16 2013

nonn

			G.f.: A(x) = 1 + 2*x + 18*x^2 + 142*x^3 + 1186*x^4 + 10152*x^5 +...
A related series is G(x) = 1 + x*G(x)^4, where
G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...
G(x)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 + 17710*x^6 +...
G(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 612*x^4 + 4389*x^5 + 32890*x^6 +...
such that A(x) = 1/(1 + 8*x*G(x)^2 - 10*x*G(x)^3).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A147855, A183160, A226705, A002293.

Cf. A005810, A078995, A147855, A226761, A385605, A386811.

Table[Sum[Binomial[2*n+2*k,n-k]*Binomial[2*n-2*k,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 16 2013 *)

{a(n)=local(G=1+x); for(i=0, n,G=1+x*G^4+x*O(x^n)); polcoeff(1/(1+8*x*G^2-10*x*G^3), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n,G=1+x*G^4+x*O(x^n)); polcoeff(1/(1-2*x*G^2-10*x^2*G^6), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(2*n+2*k, n-k)*binomial(2*n-2*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(2*k, n-k)*binomial(4*n-2*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(4*n+2*k, n-k)*binomial(-2*k, k))}
for(n=0, 30, print1(a(n), ", "))

a(n) = Sum_{k=0..n} C(2*k, n-k) * C(4*n-2*k, k).
a(n) = Sum_{k=0..n} C(n+2*k, n-k) * C(3*n-2*k, k).
a(n) = Sum_{k=0..n} C(2*n+2*k, n-k) * C(2*n-2*k, k).
a(n) = Sum_{k=0..n} C(3*n+2*k, n-k) * C(n-2*k, k).
a(n) = Sum_{k=0..n} C(4*n+2*k, n-k) * C(-2*k, k).
G.f.: 1 / (1 - 2*x*G(x)^2 - 10*x^2*G(x)^6) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.
a(n) ~ 2^(8*n+3/2)/(3^(3*n+3/2)*sqrt(Pi*n)). - Vaclav Kotesovec, Jun 16 2013
From Seiichi Manyama, Aug 05 2025: (Start)
a(n) = [x^n] 1/((1+2*x) * (1-x)^(3*n+1)).
a(n) = Sum_{k=0..n} (-3)^(n-k) * binomial(4*n+1,k).
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(3*n+k,k). (End)
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-2)^k * 3^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k).
G.f.: G(x)^2/((-2+3*G(x)) * (4-3*G(x))) where G(x) = 1+x*G(x)^4 is the g.f. of A002293. (End)
G.f.: B(x)^2/(1 + 3*(B(x)-1)/2), where B(x) is the g.f. of A005810. - Seiichi Manyama, Aug 15 2025

A385605 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n+1,k).

Original entry on oeis.org

1, 7, 58, 502, 4436, 39687, 358024, 3249288, 29624796, 271080124, 2487835678, 22888216006, 211010997716, 1948830506578, 18026768864736, 166976297995452, 1548523206590364, 14376415735219572, 133599985919343400, 1242638966005222648, 11567295503871866536

Offset: 0

Seiichi Manyama, Aug 03 2025

nonn

Cf. A006256, A141223, A385632.
Cf. A005810, A078995, A147855, A386811.
Cf. A385498.

a(n) = sum(k=0, n, 2^(n-k)*binomial(4*n+1, k));

a(n) = [x^n] 1/((1-3*x) * (1-x)^(3*n+1)).
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(3*n+k,k).
a(n) = 3^(4*n+1)*2^(-3*n-1) - binomial(4*n+1, n)*(hypergeom([1, -1-3*n], [1+n], -1/2) - 1). - Stefano Spezia, Aug 05 2025
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k). - Seiichi Manyama, Aug 07 2025
G.f.: g^2/((3-2*g) * (4-3*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 14 2025
G.f.: B(x)^2/(1 + (B(x)-1)/4), where B(x) is the g.f. of A005810. - Seiichi Manyama, Aug 15 2025
G.f.: 1/(1 - x*g^2*(12-5*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 16 2025

A062344 Triangle of binomial(2*n, k) with n >= k.

Original entry on oeis.org

1, 1, 2, 1, 4, 6, 1, 6, 15, 20, 1, 8, 28, 56, 70, 1, 10, 45, 120, 210, 252, 1, 12, 66, 220, 495, 792, 924, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620

Offset: 0

Henry Bottomley, Jul 06 2001

From Wolfdieter Lang, Sep 19 2012: (Start)
The triangle a(n,k) appears in the formula F(2*l+1)^(2*n) = (sum(a(n,k)*L(2*(n-k)*(2*l+1)),k=0..n-1) + a(n,n))/5^n, n>=0, l>=0, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
The signed triangle as(n,k):=a(n,k)*(-1)^k appears in the formula F(2*l)^(2*n) = (sum(as(n,k)*L(4*(n-k)*l),k=0..n-1) + as(n,n))/5^n, n>=0, l>=0. Proof with the Binet-de Moivre formula for F and L and the binomial formula. (End)

			Rows start
  (1),
  (1,2),
  (1,4,6),
  (1,6,15,20)
  etc.
Row n=2, (1,4,6):
F(2*l+1)^4 = (1*L(4*(2*l+1)) + 4*L(2*(2*l+1)) + 6)/25,
F(2*l)^4 = (1*L(8*l) - 4*L(4*l) + 6)/25, l>=0, F=A000045, L=A000032. See a comment above. - _Wolfdieter Lang_, Sep 19 2012

G. C. Greubel, Rows n = 0..100 of triangle, flattened
E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
Index entries for triangles and arrays related to Pascal's triangle

Columns include (sometimes truncated) A000012, A005843, A000384, A002492, A053134 etc. Right hand side includes A000984, A001791, A002694, A002696 etc. Row sums are A032443. Row alternate differences (e.g., 6-4+1=3 or 20-15+6-1=10) are A001700.
Cf. A122366.
a(2*n,n) gives A005810.

[[Binomial(2*n, k): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Jun 28 2018

Flatten[Table[Binomial[2 n, k], {n, 0, 20}, {k, 0, n}]] (* G. C. Greubel, Jun 28 2018 *)

create_list(binomial(2*n,k),n,0,9,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

for(n=0, 20, for(k=0, n, print1(binomial(2*n, k), ", "))) \\ G. C. Greubel, Jun 28 2018

a(n,k) = a(n,k-1)*((2*n+1)/k-1) with a(n,0)=1.

G.f.: 1/((1-sqrt(1-4*x*y))^4/(16*x*y^2) + sqrt(1-4*x*y) - x). - Vladimir Kruchinin, Jan 26 2021

A126596 a(n) = binomial(4n,n)(2n+1)/(3n+1).

Original entry on oeis.org

1, 3, 20, 154, 1260, 10659, 92092, 807300, 7152444, 63882940, 574221648, 5188082354, 47073334100, 428634152730, 3914819231400, 35848190542920, 329007937216860, 3025582795190340, 27872496751392496, 257172019222240200, 2376196095585231920, 21983235825545286435

Offset: 0

Philippe Deléham, Mar 13 2007

Number of standard Young tableaux of shape [3n,n]. Also the number of binary words with 3n 1's and n 0's such that for every prefix the number of 1's is >= the number of 0's. The a(1) = 3 words are: 1011, 1101, 1110. - Alois P. Heinz, Aug 15 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Column k=3 of A214776.

a126596 n = a005810 n * a005408 n `div` a016777 n
-- Reinhard Zumkeller, Mar 04 2012

[Binomial(4*n,n)*(2*n+1)/(3*n+1): n in [0..20]]; // Vincenzo Librandi, Nov 18 2011

seq((2*n+1)*binomial(4*n,n)/(3*n+1),n=0..22); # Emeric Deutsch, Mar 27 2007

Table[(Binomial[4n,n](2n+1))/(3n+1),{n,0,30}] (* Harvey P. Dale, Feb 06 2016 *)

a(n) = A039599(2*n,n).
a(n) = (2*n+1)*A002293(n). - Mark van Hoeij, Nov 17 2011
a(n) = A208983(2*n+1). - Reinhard Zumkeller, Mar 04 2012
a(n) = A005810(n) * A005408(n) / A016777(n). - Reinhard Zumkeller, Mar 04 2012
a(n) = [x^n] ((1 - sqrt(1 - 4*x))/(2*x))^(2*n+1). - Ilya Gutkovskiy, Nov 01 2017
Recurrence: 3*n*(3*n-1)*(3*n+1)*a(n) = 8*(2*n+1)*(4*n-3)*(4*n-1)*a(n-1). - Vaclav Kotesovec, Feb 03 2018
a(n) ~ 2^(8*n+3/2) / (3^(3*n+3/2) * sqrt(Pi*n)). - Amiram Eldar, Aug 29 2025

More terms from Emeric Deutsch, Mar 27 2007

A262261 a(n) = Product_{k=0..n} binomial(4*k,k).

Original entry on oeis.org

1, 4, 112, 24640, 44844800, 695273779200, 93581069585203200, 110803729631663996928000, 1165466869384731418887782400000, 109720873815210197693149787062272000000, 93006053830822450607559730484293052399616000000

Offset: 0

Vaclav Kotesovec, Apr 17 2016

In general, for p > 1, Product_{k=0..n} binomial(p*k,k) ~ A^(1 + 1/(p*(p-1))) * exp(n/2 - 1/12 - 1/(12*p*(p-1))) * n^(-1/3 - n/2 - 1/(12*p*(p-1))) * (p-1)^(1/(12*(p-1)) - p*n/2 - (p-1)*n^2/2) * p^(-1/(12*p) + (p+1)*n/2 + p*n^2/2) * (2*Pi)^(-1/4 - n/2) * Product_{j=1..p-1} (Gamma(j/(p-1))^(j/(p-1)) / Gamma(j/p)^(j/p)), where A = A074962 is the Glaisher-Kinkelin constant.

Cf. A007685 (p=2), A268196 (p=3).
Cf. A000178, A098694, A268504, A268505, A268506, A271946, A271947.
Cf. A005810, A165975.

Table[Product[Binomial[4*k,k],{k,0,n}],{n,0,10}]

a(n) ~ A^(13/12) * 2^(9*n/2 + 4*n^2) * exp(n/2 - 13/144) * Gamma(1/4)^(1/2) / (Gamma(1/3)^(1/3) * 3^(11/36 + 2*n + 3*n^2/2) * Pi^(7/12 + n/2) * n^(49/144 + n/2)), where A = A074962 is the Glaisher-Kinkelin constant.

A182400 Integral factorial ratio sequence: a(n) = (2n)!(8n)!/(n!(4n)!(5*n)!).

Original entry on oeis.org

1, 28, 1716, 118864, 8684340, 653817528, 50181947376, 3903669874104, 306689672988468, 24278779897856848, 1933612147959994216, 154751222973374578656, 12435284300689518633456, 1002664938117354309314220, 81080672610600385236492840, 6573062133232532447808798864

Offset: 0

Bruno Berselli, Apr 27 2012

For any nonnegative integers m, n the ratio (2*m)!*(2*n)!/(m!*(m+n)!*n!) provides an integer (theorem attributed to Catalan, see Umberto Scarpis in References), and this sequence is the case m = 4*n.

Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third edition), page 11.

Bruno Berselli, Table of n, a(n) for n = 0..100
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [math.NT], page 3 (Theorem 1.2, formula 8: case a=1, b=4).
Alexander Borisov, Quotient singularities, integer ratios of factorials and the Riemann Hypothesis, arXiv:math/0505167 [math.NT], 2005; International Mathematics Research Notices, Vol. 2008, Article ID rnn052, page 2 (Theorem 2).

Cf. A000984 (m = n), A005810 (m = 2*n), A211419 (m = 3*n).

[Factorial(2*m)*Factorial(2*n)/(Factorial(m)*Factorial(m+n)*Factorial(n)) where m is 4*n: n in [0..15]];

Table[((2 n)! (8 n)!)/(n! (4 n)! (5 n)!), {n, 0, 15}]

makelist((-1024)^n*binomial(4*n-1/2,5*n),n,0,15);

a(n) = (-1024)^n*binomial(4*n-1/2,5*n).
From Ilya Gutkovskiy, Jan 31 2017: (Start)
G.f.: 5F4(1/8,3/8,1/2,5/8,7/8; 1/5,2/5,3/5,4/5; 262144*x/3125).
E.g.f.: 5F5(1/8,3/8,1/2,5/8,7/8; 1/5,2/5,3/5,4/5,1; 262144*x/3125).
a(n) ~ 2^(18*n+1/2)/(sqrt(Pi*n)*5^(5*n+1/2)). (End)
a(n) = a(n-1)*32*(2*n - 1)*(8*n - 1)*(8*n - 3)*(8*n - 5)*(8*n - 7)/(5*n*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)). - Neven Sajko, Jul 21 2023

cp's OEIS Frontend

A082368 a(n) = (4*n-1)! / (n! * n! * n! * (n-1)! * 3!).

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A169961 a(n) = binomial(12*n, n).

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A023826 Sum of exponents in prime-power factorization of C(4n,n).

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A079589 a(n) = C(5*n+1,n).

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A226733 G.f.: 1 / (1 + 8*x*G(x)^2 - 10*x*G(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

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A385605 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n+1,k).

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A062344 Triangle of binomial(2*n, k) with n >= k.

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A082368 a(n) = (4n-1)! / (n! n! * n! * (n-1)! * 3!).

A226733 G.f.: 1 / (1 + 8xG(x)^2 - 10xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

A126596 a(n) = binomial(4n,n)(2n+1)/(3n+1).

A182400 Integral factorial ratio sequence: a(n) = (2n)!(8n)!/(n!(4n)!(5*n)!).