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A247701
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Numbers whose digits are nondecreasing and which have exactly d digits "d" whenever there is at least one digit "d".
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0
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1, 22, 122, 333, 1333, 4444, 14444, 22333, 55555, 122333, 155555, 224444, 666666, 1224444, 1666666, 2255555, 3334444, 7777777, 12255555, 13334444, 17777777, 22666666, 33355555, 88888888, 122666666, 133355555, 188888888, 223334444, 227777777, 333666666, 999999999
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OFFSET
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1,2
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COMMENTS
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Subsequence of terms with (weakly) increasing digits (when read from left to right) in A108571.
It happens that the last term displayed in the usual three lines of data is 999999999, but there is of course not any reason to think that this would be the last term of the sequence, it is only the largest term with 9 digits but many more follow up to the (maybe final) 122333...999, or infinitely many further terms if a convention is fixed to extend A108571 and the present sequence to digits beyond "9". (Clarification added in view of e-mail received privately.) - M. F. Hasler, Oct 04 2014
One possibility of encoding digits d > 9 in a sequence like the present one where no digit 0 occurs, is to write them as (d-9k)*10^k for 9*k < d < 9*k+10, i.e., d=10 as "10", d=11 as "20",..., d=18 as "90", d=19 as "100", etc. See the link for other variants which are more compact for larger digits. - M. F. Hasler, Oct 05 2014
Observation by R. J. Cano: The subset of terms with no digit larger than b has 2^b-1 elements. Proof: They can be coded as b-digit (nonzero, whence the -1) binary word, where the k-th bit is 1 iff digit k is present in the term. - M. F. Hasler, Oct 08 2014
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LINKS
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PROG
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(PARI) a=[]; N=9; for(m=1, min(N, 9), a=concat(a, n=10^m\9*m); for(i=1, #a-1, #Str(a[i])>N-m && break; a=concat(a, a[i]*10^m+n))); Set(a)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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