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Delete the decimal digits of n that appear exactly once; write 0 if all digits disappear.
4

%I #9 Nov 21 2018 03:20:36

%S 0,0,0,0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,22,0,0,0,0,0,0,0,0,0,0,

%T 33,0,0,0,0,0,0,0,0,0,0,44,0,0,0,0,0,0,0,0,0,0,55,0,0,0,0,0,0,0,0,0,0,

%U 66,0,0,0,0,0,0,0,0,0,0,77,0,0,0,0,0,0,0,0,0,0,88,0,0,0,0,0,0,0,0,0,0,99,0,11,0,0,0,0,0,0,0,0,11,111,11,11,11,11,11,11,11,11,0,11,22,0,0,0,0,0,0,0,0,11,0,33,0,0,0,0,0,0,0,11,0,0,44,0,0,0,0,0,0,11,0,0,0,55,0,0,0,0,0,11,0,0,0,0,66,0,0,0,0,11,0,0,0,0,0,77,0,0,0,11,0,0,0,0,0,0,88,0,0,11,0,0,0,0,0,0,0,99

%N Delete the decimal digits of n that appear exactly once; write 0 if all digits disappear.

%C A companion sequence to A320486.

%C Digits that appear exactly once in n are erased. Leading zeros are erased unless the result is 0. If all digits are erased, we write 0 for the result (A321800 is another version, which uses -1 for the empty string).

%C More than the usual number of terms are shown in order to reach some interesting examples.

%H Chai Wah Wu, <a href="/A321797/b321797.txt">Table of n, a(n) for n = 0..10000</a>

%e 12321 becomes 1221, 1123 becomes 11, 11231 becomes 111, and 100223 becomes 22 (as we don't accept leading zeros). Note that 12345 disappears immediately and we get 0.

%o (Python)

%o def A321797(n):

%o return int('0'+''.join(d if str(n).count(d) != 1 else '' for d in str(n)))

%Y Cf. A320486, A321800.

%K nonn,base

%O 0,12

%A _Chai Wah Wu_, Nov 19 2018