A348783 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.
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
Offset: 0
Examples
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) ...
Links
- Michael S. Branicky, Table of n, a(n) for n = 0..9999
Programs
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Mathematica
Array[FromDigits@*RotateLeft@*Reverse@*DigitCount,35,0] (* Giorgos Kalogeropoulos, Nov 15 2021 *)
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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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Python
def a(n): s = str(n) return int("".join(str(s.count(d)) for d in "9876543210").lstrip("0")) print([a(n) for n in range(35)]) # Michael S. Branicky, Nov 14 2021
Extensions
More terms from Michael S. Branicky, Nov 14 2021
Comments