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A330430
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Consider integers k=0,1,2,... as strings. Compare each digit in the string, in left-to-right order, with the contents of a bag (multiset) of digits, which is initially empty. If digit d is in the bag delete d from both the string and the bag. What remains of any nonempty string is appended to the sequence and its digits are added to the bag.
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3
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 1, 1, 1, 20, 1, 22, 3, 4, 25, 6, 27, 8, 29, 3, 33, 3, 3, 40, 1, 42, 3, 44, 5, 6, 47, 8, 49, 5, 5, 55, 5, 60, 1, 62, 3, 64, 5, 66, 7, 8, 69, 7, 7, 7, 77, 80, 1, 82, 3, 84, 5, 86, 7, 88, 9, 9, 9, 9, 9, 99, 100, 1, 2, 103, 4, 105
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OFFSET
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1,3
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LINKS
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EXAMPLE
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Beginning with k=0, the bag is initially empty and 0 matches nothing in the bag, so the sequence is extended and 0 is added to the bag. Likewise, for k=1-9, none of these digits match anything in the bag, resulting in the sequence [0,1,2,3,4,5,6,7,8,9] and a bag that contains {0,1,2,3,4,5,6,7,8,9}.
With k=10, Both the 1 and the 0 in k are matched with the 1 and the 0 in the bag and cancel out. This leaves an empty string, so the sequence is not extended, and the bag now contains {2,3,4,5,6,7,8,9}.
With k=11, neither digit cancels out, since there are no ones in the bag, so the sequence is extended, [0,1,2,3,4,5,6,7,8,9,11], and the bag contains {1,1,2,3,4,5,6,7,8,9}.
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PROG
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(Python)
seq = []
bag = ""
for k in range(1000):
m = str(k)
for digit in m:
if digit in bag:
mndx = m.index(digit)
m = m[:mndx] + m[mndx+1:]
bndx = bag.index(digit)
bag = bag[:bndx] + bag[bndx+1:]
if m:
seq.append(int(m))
bag = bag + m
print(seq)
(PARI) { my (b=vector(base=10), n=0); for (k=0, 105, d = if (k, digits(k, base), [0]); t = []; for (i=1, #d, if (b[1+d[i]], b[1+d[i]]--, t=concat(t, d[i]); ); ); if (#t, print1 (fromdigits(t, base)", "); for (j=1, #t, b[1+t[j]]++))) } \\ Rémy Sigrist, Dec 15 2019
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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STATUS
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approved
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