x->x*ceiling(x)

[T. D. Noe]

>I've carried out 12/11 and 20/19 to 25 terms, without the denominator
>going to 1.  I looked at the Modulus of the numerators last night, trying
>to see patterns that might allow for a reduction in size.
>
>So, you can view the sequence as
>{1, 2, 3, 4, 19, 3, 4, 5, 7, 8, ?, 5, 10, 4, 5, 9, 7, 5, ?, 12}
>or
>{0, 1, 2, 3, 18, 2, 3, 4, 6, 7, ?, 4, 9, 3, 4, 8, 6, 4, ?, 11}
>depending on whether you count the starting condition or not.


12/11 takes 26 terms (not counting the starting condition).  The final
integer is over 10^(10^7).

Tony