Exploring the A004730 Sequence: Uncovering Its Meaning

[icystrike]

Homework Statement

Can someone explain to me what is this sequence referred in the page?
http://oeis.org/A004730"

 

[Dick]

It says what it is on the page, right? '!' is 'double factorial'. If n is even n! is the product of all of the even numbers up to n and if n is odd then it's the product of all of the odd numbers. I.e. 5!/6!=(1*3*5)/(2*4*6)=5/16. Then take the numerator. That's the 5 entry.

 

[icystrike]

Thanks! =D I didnt know there is a double factorial

 

[Dick]

Now I'm confused. 4!/5! should be 2*4/(3*5)=8/15. I don't see a 15 in the series.

 

[Dick]

Now I'm confused. 4!/5! should be 2*4/(3*5)=8/15. I don't see a 15 in the series.

Never mind. I was looking at the denominator instead of the numerator.

 

[icystrike]

The series has included both numerator and denominator... So its Ok! =D

 

[HallsofIvy]

These "double factorials" can always be written in terms of the usual factorial:

If n is even, say n= 2k, then
[itex]n!= 2(4)(6)...(2k-2)(2k)= (2(1))(2(2))(2(3))...(2(k-1))(2k)= 2^k k![/tex]

If n is odd, say n= 2k+ 1, then
$$n!= 3(5)(7)...(2k-1)(2k+1)= \frac{2(3)(4)(5)(6)(7)...(2k-1)(2k)(2k+1)}{2(4)(6)...(2k)}$$
$$= \frac{(2k+1)!}{2^k k!}$$