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A348783
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Let c(i) be the number of times the digit i appears in n, for 0 <= i <= 9; then a(n) is the concatenation of c(9) c(8) ... c(1) c(0), with leading 0's omitted.
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4
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1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 11, 20, 110, 1010, 10010, 100010, 1000010, 10000010, 100000010, 1000000010, 101, 110, 200, 1100, 10100, 100100, 1000100, 10000100, 100000100, 1000000100, 1001, 1010, 1100, 2000, 11000
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OFFSET
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0,2
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COMMENTS
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"Leading zeros" means "zeros on the left".
21 -> 110 -> 21 is the first loop, when this operation is iterated. Another is 22 -> 200 -> 102 -> 111 -> 30 -> 1001 -> 22.
The numbers 2, 3, ..., 9 cannot appear, but 1 and all k >= 10 are terms. Specifically, if the decimal digits of k > 0 are d0^c, where d is a single nonzero digit and 0^c is a string of c > 0 concatenated 0's, then k appears as a(1^(d*10**(c-1))); otherwise, if the decimal digits of k are d0^cb where d and c are as before but c >= 0 and b is a string of digits not starting with 0, then k appears as a(1^(d*10^c) 0^b). - Michael S. Branicky, Nov 14 2021
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LINKS
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EXAMPLE
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0 -> 1
1 -> 10
2 -> 100
3 -> 1000 (one 3, zero copies of 2, 1, 0, so 1 0 0 0)
4 -> 10000
5 -> 100000
6 -> 1000000
7 -> 10000000
8 -> 100000000
9 -> 1000000000
10 -> 11
11 -> 20
12 -> 110
...
2222222222 -> 1000 (ten 2's, zero 1's, zero 0's, so 10 0 0)
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MATHEMATICA
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PROG
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(Python)
def a(n):
s = str(n)
return int("".join(str(s.count(d)) for d in "9876543210").lstrip("0"))
(PARI) apply( {A348783(n)=if(n, eval(concat([Str(#[0|d<-n, d==i])|i<-+-[-vecmax(n=digits(n))..0]])), 1)}, [0..66]) \\ M. F. Hasler, Nov 15 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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