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A004720
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Delete all digits '1' from the sequence of nonnegative integers.
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11
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0, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 3, 4, 5, 6, 7, 8, 9, 20, 2, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3, 32, 33, 34, 35, 36, 37, 38, 39, 40, 4, 42, 43, 44, 45, 46, 47, 48, 49, 50, 5, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6, 62, 63, 64, 65, 66, 67, 68, 69, 70, 7, 72, 73, 74, 75
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OFFSET
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1,2
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COMMENTS
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More precisely, in A004176 the term becomes 0 if no digit remains, e.g., for 1 or 11, whereas here in such a case the integer is completely skipped (as in A004719, A004721, ... which are the analogs for deleting 0, 2, ...). - M. F. Hasler, Feb 01 2016
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LINKS
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EXAMPLE
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The first nonnegative integer, 0, remains as a(1).
The second nonnegative integer, 1, completely disappears upon removal of the digit 1.
The third nonnegative integer, 2, remains as a(2).
The number 10 becomes a(10)=0.
The number 11 completely disappears upon removal of both its digits '1'.
The number 12 becomes a(11)=2.
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MAPLE
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f:= proc(n) local L, i;
L:= subs(1=NULL, convert(n, base, 10));
if L = [] then NULL
else add(L[i]*10^(i-1), i=1..nops(L))
fi
end proc:
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MATHEMATICA
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f[n_] := Block[{a = DeleteCases[ IntegerDigits[n], 1]}, If[a != {}, FromDigits@ a, b]]; DeleteCases[ Array[f, 75, 0], b] (* Robert G. Wilson v, Feb 05 2016 *)
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PROG
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(PARI) for(n=0, 99, if(t=select(d->d!="1", Vec(Str(n))), print1(concat(t)", "))) \\ M. F. Hasler, Feb 01 2016
(Python)
l = len(str(n-1))
m = (10**l-1)//9
k = n + l - 2 + int(n+l-1 >= m)
return 0 if k == m else int(str(k).replace('1', '')) # Chai Wah Wu, Apr 20 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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