|
|
A182400
|
|
Integral factorial ratio sequence: a(n) = (2*n)!*(8*n)!/(n!*(4*n)!*(5*n)!).
|
|
6
|
|
|
1, 28, 1716, 118864, 8684340, 653817528, 50181947376, 3903669874104, 306689672988468, 24278779897856848, 1933612147959994216, 154751222973374578656, 12435284300689518633456, 1002664938117354309314220, 81080672610600385236492840, 6573062133232532447808798864
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
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.
|
|
REFERENCES
|
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.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (-1024)^n*binomial(4*n-1/2,5*n).
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
|
|
MATHEMATICA
|
Table[((2 n)! (8 n)!)/(n! (4 n)! (5 n)!), {n, 0, 15}]
|
|
PROG
|
(Magma)
[Factorial(2*m)*Factorial(2*n)/(Factorial(m)*Factorial(m+n)*Factorial(n)) where m is 4*n: n in [0..15]];
(Maxima) makelist((-1024)^n*binomial(4*n-1/2, 5*n), n, 0, 15);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|