Missing Integer in a Sequence

[a1call]

What is the single missing integer in the sequence below?


2848011316845460139375001
5732846780213531494140625
9640421297171656741590841
14813097570041941504134801
21536438558058637251039001
30145238514364280374308241
?
54645469404718204483995601
71516631514692508175312281
92249999834554578632336721
117542503824321100231988281
148192710813027984619140625
185112909229287000187387161
229342360883365061854482961
282061813379963068278467641

[chalsall]

Originally Posted by a1call

What is the single missing integer in the sequence below?

Just wondering... Is there a reason you consistently ask asinine questions?

[Brian-E]

Those integers are obviously special when you express them as a product of prime factors, all of them being extremely smooth, none of them having a prime factor higher than 359.

After taking that obvious step, I can only see several other properties, not nearly enough to narrow down to a single answer to the puzzle. They are: all are divisible by 7 and 11, none by 2, and most of them have their highest four or more prime factors all ending with the same digit, the only two exceptions to that being those which are divisible by 5.

[cuBerBruce]

Originally Posted by Brian-E

Those integers are obviously special

,,,

Yes, from the prime factorization, I guessed that the sequence is a sequence of multifactorials. Consider in particular the numbers ending in 5 and its not too hard to find the multifactorials that produce them, and then determining what the sequence is becomes easy.

[a1call]

Originally Posted by Brian-E

all are divisible by 7 and 11[/SPOILER]

That might be a bigger discovery than you might expect. But all that in good time.

[a1call]

Originally Posted by cuBerBruce

,,,

Yes, from the prime factorization, I guessed that the sequence is a sequence of multifactorials. Consider in particular the numbers ending in 5 and its not too hard to find the multifactorials that produce them, and then determining what the sequence is becomes easy.

Bingo, half way there they are all multifactorials of the form:
n!^(k) for consecutive odd ns.
But there is something special about the k and its relation with the ns.

At this stage brute force can make light work of the puzzle.

[cuBerBruce]

Originally Posted by a1call

Bingo, half way there they are all multifactorials of the form:
n!^(k) for consecutive odd ns.

Of course, I already knew the answer when I made my previous post. I'll reveal it now.

The answer is:
41030221999423150634765625

The number 5732846780213531494140625 has many factors of 5, suggesting a multifactorial with n an odd multiple of 5, and k a multiple of 10. It has 67 as the largest prime factor, suggesting n might be 335 (equal to 67*5). After a little trial and error, I found n=335 and k=30 worked. After this, it wasn't too hard to figure out the sequence used k=30, and n ranged over the odd values from 333 to 361.

[a1call]

Originally Posted by cuBerBruce

Of course, I already knew the answer when I made my previous post. I'll reveal it now.

The answer is:
41030221999423150634765625

The number 5732846780213531494140625 has many factors of 5, suggesting a multifactorial with n an odd multiple of 5, and k a multiple of 10. It has 67 as the largest prime factor, suggesting n might be 335 (equal to 67*5). After a little trial and error, I found n=335 and k=30 worked. After this, it wasn't too hard to figure out the sequence used k=30, and n ranged over the odd values from 333 to 361.

Right on the nose. Thank you for taking the time.
Now the big reveal:
* 30 is a primorial which has prime factors less than 7. This seems to have the general effect (for any primorial) that at some point with n large enough the consecutive multufactorials will all be divisible by any given prime larger than 5. But never divisable by 3 and 5 together.
Not sure If I made any sense there, but there is a pattern there somewhere.

[a1call]

The next question is a follow up and not a puzzle. I posted this puzzle to see if someone could find the answer by some shortcut from the pattern rather than multiple multiplication. Sort of by finding a point on a curve.
There is the progressively more complicating polynomials of very high powers but that is going to get more complicated and time consuming than the multiplication.