OFFSET
1,1
COMMENTS
These could be called "nine-free numbers".
From David A. Corneth, Aug 31 2020: (Start)
This sequence has density 0. Conjecture: this sequence is finite and full. a(9) > 10^100 if it exists.
Suppose we want to see if 22792 = 1011021011_3 is a term. Since it has a digit of 2 in base 3, we can see that it is not. The next number that does not have the digit 2 in base 3 is 1011100000_3 = 22842, so we can proceed from there. In a similar way we can skip numbers based on bases b > 3. (End)
All terms of this sequence increased by 1 (except a(2)=3) are prime. - François Marques, Aug 31 2020
From Devansh Singh, Sep 19 2020: (Start)
If n is one less than an odd prime and we are interested in bases 3 <= b <= n-1 such that n in base b contains the digit b-1, then divisor of b (except 1) -1 cannot be the last digit since divisor of b divides n+1, which is not possible as n+1 is an odd prime.
If the last digit is 1, then b is odd as 1 = 2-1 and 2 cannot divide b as n+1 is an odd prime.
If the last digit is 0, then b-1 is the last digit of n-1 in base b.
b <= n/2 for even n,b <= (n+1)/2 for odd n.
This sequence is equivalent to the existence of only one prime generating polynomial = F(x) (having positive integer coefficients >=0 and <=b-1 for F(b)) such that F(2) = p.
There is no other prime generating polynomial = G(x) (having positive integer coefficients >=0 and <= b-1 for G(b)) that generates p for 2 < x = b <= (p-1)/2.
(End)
LINKS
David A. Corneth, PARI program
EXAMPLE
2 is a term because 2 = 10_2 = 2_3, so both have the digit b-1, and there are no other bases where this happens.
4 is a term because 4 = 100_2 = 4_5, so both have the digit b-1, and there are no other bases where this happens.
PROG
(PARI) isok(n, b) = vecmax(digits(n, b)) == b-1;
b(n) = if (n==1, return (1)); my(b=3); while(!isok(n, b), b++); b; \\ A337535
is(n) = b(n) == n+1;
(PARI) \\ See Corneth link \\ David A. Corneth, Aug 31 2020
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Michel Marcus, Aug 31 2020
STATUS
approved