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No cycle of length 6 is presently known!
It is also known that a(7) = 420876, a(8) = 7509843, a(14) = 753098643.
One does not need to consider every integer of n digits, only the sorted sequences of n digits (of which there are binomial(n+9, 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_k-1 9...9 9-d_k ... 9-d_2 9-d_1+1
where the values d_i are in the range 1 to 9 and the sequence of 9's in the middle may be empty.
From this form it follows that for any member of a cycle,
abs(number of 8's - number of 1's) + abs(number of 7's - number of 2's) +
abs(number of 6's - number of 3's) + abs(number of 5's - number of 4's) +
max(0, number of 0's - number of 9's) <= 4,
so given the numbers of 0's, 1's, 2's, 3's and 4's 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 4-cycles up to there are the ones containing 61974 and 62964, the only 8-cycles up to there are the ones containing 7509843 and 76320987633, the only 14-cycle up to there is the one containing 753098643. All the 7-cycles so far follow the pattern
7-cycle: 420876
7-cycle: 43208766
7-cycle: 4332087666
7-cycle: 433320876666
7-cycle: 43333208766666
7-cycle: 4333332087666666 ... (End)
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