A211419 -id:A211419 - cp's OEIS frontend

A211420 a(n) = (8n)!n!/((4n)!(3n)!(2*n)!).

1, 140, 60060, 29745716, 15628090140, 8480843582640, 4697400936504900, 2638798257262351800, 1497753729733989900060, 856840435680656569701776, 493243073668546377605912560, 285369375758780754651194529300, 165789876049841088844342275759300

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 4, b = 1 of the following result (see Bober, Theorem 1.2): let a, b be nonnegative integers with a > b and GCD(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211419(a = 3, b = 2), A211421(a = 4, b = 3) and A061163 (a = 5, b = 1).

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

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2 (2009), 422-444.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061162, A061163, A211419, A211421.

a := n -> (2^(6*n)*GAMMA(4*n + 1/2))/(GAMMA(n + 1/2)*GAMMA(3*n + 1)):
seq(a(n), n = 0..12); # Peter Luschny, Jul 11 2023

Table[ 2^(6*n) * Gamma[4*n + 1/2] / (Gamma[n + 1/2] * Gamma[3*n + 1]), {n, 0, 12}] (* James C. McMahon, Feb 24 2024 *)

The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
D-finite with recurrence: 3*(3*n-1)*(2*n-1)*(3*n-2)*n*a(n) - 8*(8*n-3)*(8*n-1)*(8*n-7)*(8*n-5)*a(n-1) = 0. - Georg Fischer, Nov 30 2022
From Peter Bala, Jul 10 2023: (Start)
a(n) = Sum_{k = 0..3*n} binomial(8*n, k)*binomial(5*n-k-1, 3*n-k).
a(n) = [x^(3*n)] F(x)^n, where F(x) = (1 + x)^8/(1 - x)^2.
It follows that the o.g.f. A(x) for this sequence is the diagonal of the bivariate rational generating function 1/3*( 1/(1 - t*F(x^(1/3))) +  1/(1 - t*F(w*x^(1/3))) +  1/(1 - t*F(w^2*x^(1/3))) ), where w = exp(2*Pi*i/3), and hence A(x), as stated above, is algebraic over Q(x) by Stanley 1999, Theorem 6.33, p. 197. (End)
From Karol A. Penson, Feb 23 2024: (Start)
O.g.f.: hypergeometric4F3([1/8, 3/8, 5/8, 7/8], [1/3, 1/2, 2/3], (2^14*z)/27). (O.g.f.(z))^2 satisfies the algebraic equation of order 16, in which the powers of (o.g.f.(z))^2 are multiplied by polynomials p(n, z) with integer coefficients, in the form: Sum_{n = 0..16} p(n, z) * (o.g.f.(z))^(2*n) = 0.
Here is the list of orders, in the variable z, of all polynomials p(n, z) for n = 0..16: 8,8,8,8,9,9,9,9,10,10,10,11,11,11,11,11,12. For example p(14, z) = 6*(2^13*z + 31*27)*(2^14*z - 27)^10. (End)
a(n) ~ 2^(14*n - 1/2) / (3^(3*n + 1/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 27 2024

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

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 = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A262739 O.g.f. exp( Sum_{n >= 1} A262733(n)*x^n/n ).

Original entry on oeis.org

1, 12, 215, 4564, 106442, 2635704, 68031147, 1810302340, 49308457334, 1368019979976, 38525145673126, 1098380420669000, 31641932951483220, 919622628946689648, 26931762975278938035, 793967020231145502564, 23543663463050594677310, 701763102761640853890600, 21014048069544552257072530, 631868353403527700756671320, 19070677448561228207945931276

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) when k = 5. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A262733, A211419, A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262737 (k = 3), A262738 (k = 4), A262740 (k = 6).

A262739 := proc (n) option remember; if n = 0 then 1 else add(1/k!*(7*k)!/GAMMA(7*k/2 + 1)*GAMMA(5*k/2 + 1)/(5*k)!*A262739(n-k), k = 1 .. n)/n end if; end proc:
seq(A262739(n), n = 0..20);

a(n) = sum(k=0, n, binomial(7*(n+1),k)*binomial(6*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(7*n,i)*binomial(6*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} 1/n! * (7*n)!/(7*n/2)! * (5*n/2)!/(5*n)!*x^n/n ) = 1 + 12*x + 215*x^2 + 4564*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A262733.
O.g.f. is the series reversion of x*(1 - x)^5/(1 + x)^7,
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} 1/k! * (7*k)!/(7*k/2)! * (5*k/2)!/(5*k)!*a(n-k).

A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 2, -2, 1, 4, 6, 0, 1, 6, 30, 20, 6, 1, 8, 70, 256, 70, 0, 1, 10, 126, 924, 2310, 252, -20, 1, 12, 198, 2240, 12870, 21504, 924, 0, 1, 14, 286, 4420, 41990, 184756, 204204, 3432, 70, 1, 16, 390, 7680, 104006, 811008, 2704156, 1966080, 12870, 0

Offset: 0

Seiichi Manyama, Feb 07 2020

			Square array begins:
    1,   1,     1,      1,      1,       1, ...
    0,   2,     4,      6,      8,      10, ...
   -2,   6,    30,     70,    126,     198, ...
    0,  20,   256,    924,   2240,    4420, ...
    6,  70,  2310,  12870,  41990,  104006, ...
    0, 252, 21504, 184756, 811008, 2521260, ...

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

Columns k=1..7 give A000984, A091527, A001448, A262732, A211419, A262733, A211421.

Main diagonal is A332231.

T[n_, k_] := Sum[Binomial[(k + 1)*n, j] * Binomial[k*n - j - 1, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 05 2021 *)

T(n,k) = Sum_{j=0..n} binomial((k+1)*n,j) * binomial(k*n-j-1,n-j).

T(n,k) = 1/n! * ((k+1)*n)!/Gamma(1 + (k+1)*n/2) * Gamma(1 + (k-1)*n/2)/((k-1)*n)!.

A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 6, -10, -20, 1, 16, 70, 0, 70, 1, 30, 630, 924, 198, -252, 1, 48, 2310, 28672, 12870, 0, 924, 1, 70, 6006, 204204, 1385670, 184756, -4420, -3432, 1, 96, 12870, 860160, 19122246, 69206016, 2704156, 0, 12870, 1, 126, 24310, 2704156, 130378950, 1848483780, 3528923580, 40116600, 104006, -48620

Offset: 0

Peter Bala, Aug 07 2023

Compare with A364303 and A364519.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 1. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.
It is known that the unsigned version of row 0 (the central binomial numbers A000984) and row 2 satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that all the row sequences of the table satisfy the same supercongruences.

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252   ... see A000984
  1 |  1   0    -10        0         198             0   ... see A211419
  2 |  1   6     70      924       12870        184756   ... A001448
  3 |  1  16    630    28672     1385670      69206016   ... A091496
  4 |  1  30   2310   204204    19122246    1848483780   ... A061162
  5 |  1  48   6006   860160   130378950   20392706048   ... A276098
  6 |  1  70  12870  2704156   601080390  137846528820   ... A001448 bisected
  7 |  1  96  24310  7028736  2149374150  678057476096   ... A276099

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
Wikipedia, Dixon's identity
Wikipedia, Hypergeometric function

Cf. A000984 (row 0 unsigned), A211419 (row 1 unsigned without 0's), A001448 (row 2), A091496 (row 3), A061162 (row 4), A276098 (row 5), A001448 bisected (row 6), A276099 (row 7).

Cf. A330843, A364303, A364506, A364509, A364513, A364519.

T(n,k) = add( binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j), j = 0..2*k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

T(n,k) = sum(j = 0, 2*k, binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j));
lista(nn) = for( n=0, nn, for (k=0, n, print1(T(n-k, k), ", "))); \\ Michel Marcus, Aug 13 2023

T(n,k) = Sum_{j = 0..2*k} binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j).
T(2,k) = binomial(4*k,2*k).
For n >= 3, T(n,k) = binomial(n*k-1,2*k) * hypergeom([-(n+2)*k, -2*k], [1 - n*k], -1) except when (n,k) = (3,1).
For n >= 2, T(n,k) = ((n+2)*k)!*((n-2)*k/2)!/(((n+2)*k/2)!*((n-2)*k)!*(2*k)!) by Kummer's Theorem.
T(n,k) = [x^k] (1 - x)^(2*k) * Chebyshev_T(n*k, (1 + x)/(1 - x)).
T(n,k) = Sum_{j = 0..k} binomial(2*n*k, 2*j)*binomial((n-1)*k-j-1, k-j).
For n >= 3, T(n,k) = binomial((n-1)*k-1,k) * hypergeom([-n*k, -k, -n*k + 1/2], [1 - (n-1)*k, 1/2], 1).
The row generating functions are algebraic functions over the field of rational functions Q(x).

A173781 a(n) is the smallest entry of the n-th column of the matrix of Super Catalan numbers S(m,n).

Original entry on oeis.org

1, 2, 4, 10, 28, 72, 198, 572, 1560, 4420, 12920, 36176, 104006, 305900, 869400, 2521260, 7443720, 21360240, 62300700, 184410072, 532740208, 1560167752, 4626704368, 13432367520, 39457579590, 117177054540, 341487416088, 1005490725148, 2989296750440, 8737944347440, 25776935824948

Offset: 0

Joseph Alfano (jalfano(AT)assumption.edu), Feb 24 2010

Super Catalan number S(m,n) is [(2m)! (2n)! ] / [(m!) (n!) (m+n)! ], where m,n are nonnegative integers.
S(m,n) is a positive integer, but a combinatorial interpretation of S(m,n) is an open problem.
For each n, the sequence S(m,n) is decreasing then increasing, with minimum value at m = ceiling(n/3).
Our sequence is that list of values S( ceiling(n/3), n).
S(n,k-n) = C(2k,k) * C(k,n) / C(2k,2n). - Charlie Neder, Dec 27 2018

Seiichi Manyama, Table of n, a(n) for n = 0..2099
Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.
Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of the Numbers 6(2n)!/n!(n+2)!, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.3, 13 pp.

Cf. A211419.

nn = 30; {1}~Join~Table[Min@ Map[Function[n, ((2 m)! (2 n)!)/((m!) (n!) (m + n)!)], Range@ nn], {m, nn}] (* Michael De Vlieger, Jul 16 2016 *)

a(3*n) = A211419(n). - Peter Bala, Sep 24 2023

A364402 a(n) = (3n)!(10n)!/((2n)!(5n)!(6n)!).

Original entry on oeis.org

1, 126, 41990, 15967980, 6421422150, 2663825039876, 1127155102890908, 483537022180231320, 209536624110664757830, 91505601042318156186900, 40205863224219682380130740, 17753412284992688334256754280, 7871411119532225034145860092700, 3502017467737750755575471520717480

Offset: 0

Neven Sajko, Jul 22 2023

nonn

Member of Bober's second infinite family of integral factorial ratio sequences with a=5 and b=3 (see equation 11 at p. 16 in Bober).

Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, Journal of the London Mathematical Society, Vol. 79, No. 2 (2009), pp. 422-444; arXiv preprint, arXiv:0709.1977 [math.NT], 2007.

Bisection of A262732. Cf. A182400, A211419.

seq( (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!), n = 0..20); # Peter Bala, Sep 24 2023

a(n) = (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!); \\ Michel Marcus, Sep 20 2023

a(n) = 10*(10*n - 1)*(10*n - 3)*(10*n - 7)*(10*n - 9)/(3*n*(2*n - 1)*(6*n - 1)*(6*n - 5))*a(n-1).
a(n) ~ 2^(2*n-1) * 5^(5*n) / (sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Sep 21 2023
From Peter Bala, Sep 24 2023: (Start)
a(n) = A262732(2*n).
a(n) = [x^(2*n)] (1 + 4*x)^((10*n-1)/2) = 16^n * binomial((10*n-1)/2, 2*n).
O.g.f. A(x) = hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x).
(End)

cp's OEIS Frontend

A211420 a(n) = (8*n)!*n!/((4*n)!*(3*n)!*(2*n)!).

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A262739 O.g.f. exp( Sum_{n >= 1} A262733(n)*x^n/n ).

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A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

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A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

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A173781 a(n) is the smallest entry of the n-th column of the matrix of Super Catalan numbers S(m,n).

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A364402 a(n) = (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!).

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A211420 a(n) = (8n)!n!/((4n)!(3n)!(2*n)!).

A364402 a(n) = (3n)!(10n)!/((2n)!(5n)!(6n)!).