

A339736


Numbers which, in every base B, either have a digit 0, do not have a digit B1, or only have one digit.


0



1, 2, 4, 6, 10, 12, 18, 28, 36, 40, 42, 66, 82, 88, 100, 102, 150, 180, 210, 226, 228, 256, 262, 268, 270, 408, 420, 456, 540, 732, 738, 808, 810, 906, 910, 1030, 1032, 1090, 1092, 1380, 1458, 1480, 1620, 1876, 1888, 2212, 3270, 3300, 3528, 4000, 4132, 4138, 4152, 4156, 4158, 4512, 4950, 5208, 5652, 5688, 6300
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OFFSET

1,2


COMMENTS

If the 0 digit requirement didn't exist, every number greater than one could be represented in base 2 with more than one digit and will contain the digit 1. If the singledigit requirement didn't exist, every number N greater than zero could be represented as "N" in base N+1. The sequence appears to be infinite. Every number other than the first one seems to be even.


LINKS

Table of n, a(n) for n=1..61.


EXAMPLE

3 is "11" in base 2 so it's not in the sequence.
4 is either "100" in base 2 (contains a 0), "11" in base 3 (doesn't contain a 2), "10" in base 4 (contains a 0) or "4" in base 5 and higher (only has one digit), so it's in the sequence.
5 is "12" in base 3 so it's not in the sequence.
45 is "231" in base 4 so it's not in the sequence.


PROG

(Haskell) convertToBase b 0 = []; convertToBase b n = convertToBase b (n `div` b) ++ [n `mod` b]; convertToDynamic n = head $ filter (not.(0`elem`)) $ map snd $ filter (\(b, n) > (b1)`elem` n) $ map (\b > (b, convertToBase b n)) [2..n+1]; main = print $ concat $ filter ((==1).length) $ map convertToDynamic [1..]
(PARI) isoknb(n, b) = my(d=digits(n, b)); (#d == 1)  (vecmin(d) == 0)  (vecmax(d) < b1);
isok(n) = for (b=2, n1, if (!isoknb(n, b), return (0))); return (1); \\ Michel Marcus, Dec 15 2020


CROSSREFS

Sequence in context: A324059 A055235 A083887 * A064374 A000885 A068336
Adjacent sequences: A339733 A339734 A339735 * A339737 A339738 A339739


KEYWORD

nonn,base


AUTHOR

Alex Stefanov, Dec 14 2020


STATUS

approved



