Dividing n consecutive integers by (1,2,3,...,n)

[Jack Brennen]

Leroy Quet wrote:
> 
> It seems likely to me that there is a more direct
> and simpler way, simpler and more direct than
> just checking stretches of integers, to calculate
> the terms of this sequence. Is there an easy way
> to calculate the sequence's terms without not
> just checking all stretches of n consecutive
> integers for their divisibility?
> 

Well, consider that the n consecutive integers
beginning at a(n) must contain at most 1 prime.


So finding a(n) at least to some extent involves
finding prime gaps, or something of similar
difficulty.  So, no, I don't think there is any
easy way to calculate the sequence's terms.

Note that a(20) is already as high as 60458,
assuming my program is correct.  The unique
way to fit the divisors for that solution:

  2  3 20  1 18
13 16 15 14 11
12 17 10  9  8
  7  6  5  4 19


One thing that is interesting about that solution
is that 13 of the 20 numbers are divisible by
the difference between themselves and 60480, which
is a very smooth number with many small divisors.

I wonder if it can be shown that any solution below
a certain bound must be affiliated with such a
"super-smooth" number like 60480.