OFFSET
1,2
COMMENTS
It appears that the difference between a(n)/a(n-1) and a(n-1)/a(n-2) approaches some factor, 5 < x < 7, as n --> oo.
It appears that 3|a(n) for n > 1.
REFERENCES
L. Euler and J. L. Lagrange, Elements of Algebra, J. Johnson and Co. 1810. See pages 332-335.
FORMULA
a(n) = ((n*(n+1))/2)/(Product_{i=0..floor((n-1)/2),n-2*i}/Product_{i=1..n}).
From Chai Wah Wu, Feb 07 2018: (Start)
a(n) = n*(n+1)!!/2.
a(n)/a(n-1) = ((n+1)!!/n!!)*(n/(n-1)) = n/b*(n-1) if n is even and n*Pi/(2*b*(n-1)) if n is odd where b = Integral_{x=0..(Pi/2)} sin^(n+1)*x dx.
Since b -> 0 as n -> oo, a(n)/a(n-1) is unbounded as n -> oo. On the other hand, a(n)/a(n-1) and a(n-1)/a(n-2) differ by a multiplicative factor of approximately Pi/2.
(End)
EXAMPLE
For n = 6, a(6) = binomial(6+1,2)/(6!!/6!) = 315.
MATHEMATICA
Table[Binomial[n+1, 2] n!/n!!, {n, 30}] (* Harvey P. Dale, Jul 27 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Justin Gaetano, Feb 06 2018
STATUS
approved