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A339736
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Numbers which, in every base B, either have a digit 0, do not have a digit B-1, or only have one digit.
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1
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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
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1,2
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COMMENTS
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Without the digit 0 requirement, every number greater than one could be represented in base 2 with more than one digit and would contain the digit 1. Without the single-digit requirement, every number m greater than zero could be represented as "m" in base m+1.
The sequence appears to be infinite. Every term other than a(1)=1 is even because every odd number 2k+1 can be represented as "1k" in base k+1 (for k>0).
Every term is 1 less than a prime number because, for all numbers m, if m+1 is not prime then, in base b = largest divisor of m+1, m is a two digit number ending in b-1 (since m+1 is a two digit number ending in 0), thus proving m is not in the sequence. (End)
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LINKS
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EXAMPLE
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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 (has only 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.
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MATHEMATICA
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Select[Range@6300, (k=2; While[FreeQ[l=IntegerDigits[#, k], k-1]||MemberQ[l, 0]||Length@l==1&&k<#, k++]; k-1==#)&] (* Giorgos Kalogeropoulos, Sep 13 2021 *)
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PROG
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(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) -> (b-1)`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) < b-1);
isok(n) = for (b=2, n-1, if (!isoknb(n, b), return (0))); return (1); \\ Michel Marcus, Dec 15 2020
(PARI) bad(n, b)=my(d=Set(digits(n, b))); b-d[#d]==1 && d[1];
ok(n, L)=for(b=2, L, if(bad(n, b), return(0))); 1;
list(lim)=my(v=List([1])); forprime(p=3, lim+1, if(ok(p-1, sqrtint(p)+1), listput(v, p-1))); Vec(v) \\ Charles R Greathouse IV, Sep 17 2021
(Python)
from sympy.ntheory.digits import digits
def okb(n, b):
d = digits(n, b)[1:]
return (len(d) == 1) or (0 in d) or (b-1 not in d)
def ok(n):
return all(okb(n, b) for b in range(2, n+1))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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