No cycle of length 6 is presently known!
It is also known that a(7) = 420876, a(8) = 7509843, a(14) = 753098643.
Comments from Joseph Myers, Aug 19 2009 (Start)
One does not to need to consider every integer of n digits, only the sorted
sequences of n digits (of which there are n+9 choose 9, so 28048800 for 23
digits). Then you only need to consider those sorted sequences of digits
whose total is a multiple of 9, as the number and so the sum of its digits
is always a multiple of 9 after the first iteration, which reduces the
work by a further factor of about 9.
As a further refinement, the result of a single subtraction, if not zero, will have digit sequence of the form
d_1 d_2 ... d_k1 9...9 9d_k ... 9d_2 9d_1+1
where the values d_i are in the range 1 to 9 and the sequence of 9s in the middle may be empty.
From this form it follows that for any member of a cycle,
abs(number of 8s  number of 1s) + abs(number of 7s  number of 2s) +
abs(number of 6s  number of 3s) + abs(number of 5s  number of 4s) +
max(0, number of 0s  number of 9s) <= 4,
so given the numbers of 0s, 1s, 2s, 3s and 4s there is little freedom left
in choosing the number of each remaining digit.
No further cycle lengths exist up to at least 140 digits. The only 4cycles up
to there are the ones containing 61974 and 62964, the only 8cycles up to there are
the ones containing 7509843 and 76320987633, the only 14cycle up to there
is the one containing 753098643. All the 7cycles so far follow the pattern
7cycle: 420876
7cycle: 43208766
7cycle: 4332087666
7cycle: 433320876666
7cycle: 43333208766666
7cycle: 4333332087666666 ... (End)
