Need help with finding the equation to calculate this sequence

[Gouskin]

The sequence is
1, 12, 720, 30240, 1209600, 47900160 ...

I'm almost certain that it has something to do with factorials as:

1 = 1! = 0!
12 = 2! * 3!
720 = 3! * 5! = 6!
30240 = 3! * 7!
1209600 = 2! * 5! * 7!

Anyway, I need help figuring the equation used for generating these numbers! Thank you in advance!

 

[mmm4444bot]

I don't see any pattern, in those factorials. I checked a sequence encyclopedia; the sequence in your post is not listed.

Is this a school assignment?

 

[lookagain]

The sequence is
1, 12, 720, 30240, 1209600, > > > 47900160 < < < <---- Is that the entire number?...

I'm almost certain that it has something to do with factorials as:

1 = 1! = 0!
12 = 2! * 3!
720 = 3! * 5! = 6!
30240 = 3! * 7!
1209600 = 2! * 5! * 7!

I asked that question, because if 47900160 is the entire number, then it is not the product of factorials. For instance, 11 divides it once, but 11! does not divide it.

 

[daon2]

It is 6/5*11!

 

[JeffM]

The sequence is
1, 12, 720, 30240, 1209600, 47900160 ...

I'm almost certain that it has something to do with factorials as:

1 = 1! = 0!
12 = 2! * 3!
720 = 3! * 5! = 6!
30240 = 3! * 7!
1209600 = 2! * 5! * 7!

Anyway, I need help figuring the equation used for generating these numbers! Thank you in advance!

\(\displaystyle Let\ p_i\ be\ the\ i^{th}\ prime,\ i = 1,\ ...\)

\(\displaystyle First:\ \displaystyle 1 = \prod_{i=1}^\infty p_i^0 = 1 * 12^0 = 1 * 0! = 1 * 1!\)

\(\displaystyle Second:\ \displaystyle 12 = 3^1 * 2^2 * \prod_{i=3}^\infty p_i^0= 1 * 12^1 = 2 * 3!\)

\(\displaystyle Third:\ \displaystyle 720 = 5 * 3^2 * 2^4 * \prod_{i=4}^\infty p_i^0 = 5 * 12^2 = 1 * 6!\)

\(\displaystyle Fourth:\ \displaystyle 30,240 = 7 * 5 * 3^3 * 2^5 * \prod_{i=5}^\infty p_i^0 = 210 * 12^2 = 6 * 7!\)

\(\displaystyle Fifth:\ \displaystyle 1,209,600 = 7 * 5^2 * 3^3 * 2^8 * \prod_{i=5}^\infty p_i^0 = 700 * 12^3 = 30 * 8!\)

\(\displaystyle Sixth:\ \displaystyle 47,900,160 = 11 * 7 * 5 * 3^5 * 2^9 * \prod_{i=6}^\infty p_i^0 = 2310 * 12^4 = 132 * 9!\)

That is as far as I have been able to go. There may be some pattern lurking in there, but I cannot tease it out.

 

[Gouskin]

Thank you for all of your help. I'm still working on it.