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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)

The choice b <= n in the name of this sequence comes from the fact that base n+1 has the desired property for all n > 1.

For n > 2, a(n) <= (n+1)/2.

If a number n has base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)) contains digit b-1, where b = q*(k+1)/k, k>=1 , and Sum_{i>=0} ((A(i)(mod b-q))*((b-q)^i)) > 0 then there exists n' < n such that that n' in base b-q = b' contains digit b'-1 at the same place as n in base b and 0 <= (A(i)-A'(i))/b' <= (k+1)-((A'(i)+1)/b') (A'(i) is digit of n' in base b')for all i>=0.*

This condition is necessary and sufficient.

Proof that Condition is Necessary:

Since b-1 = b-q+q-1 and b' = q/k (as b = q*(k+1)/k). Therefore (b-1) (mod b') = (b'+q-1) (mod b') = (q-1) (mod b') = b'-1 :-(1).

n in base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)).Then n = Sum_{i>=0} (A(i)*(b^i)) = Sum_{i>=0} (A(i)*((b-q+q)^i)).

n = Sum_{i>=0} (A(i)*(b'^i)) +

Sum_{i>=1} (A(i)*(b^i - b'^i))

= Sum_{i>=0} (A'(i)*(b'^i)) + Sum_{i>=0} ((A(i)-A'(i))* (b'^i)) + Sum_{i>=1} (A(i)*(b^i - b'^i)),

where A'(i) = A(i) (mod b').

Now n-Sum_{i>=0} ((A(i)-A'(i))*(b'^i))

- Sum_{i>=1} (A(i)*(b^i - b'^i))

= Sum_{i>=0} (A'(i)*(b'^i)).

Since A'(j) = A(j) (mod b') = (b-1) (mod b') = b'-1(due to equation (1) above and A(j) = b-1.

Hence there exists n' = Sum_{i>=0} (A'(i)*(b'^i)) > 0 containing digit b'-1 in base b'.

Table of n/b with cell containing T(n, b) = (n', b') for q = b/2. n' = Sum_{i>=0} (A'(i)*(b'^i))

n/b| 4 | 6 | 8 | 10 | 12

3 |(1,2)| | | |

4 | | | | |

5 | |(2,3)| | |

6 | | | | |

7 |(3,2)| |(3,4)| |

8 | | | | |

9 | | | |(4,5)|

10 | | | | |

11 |(1,2)|(5,3)| | |(5,6)

Example: For table n/b in comments containing (n',b') in its cells.

For n = 7:

In base b = 4, n = 13 :- q = b' = 4/2 = 2, and n' = (3 mod (2))*(2)^0 + (1 mod(2))*(2)^1 = 1+2 = 3.

In base b = 8, n = 7 :- q = b' = 8/2 = 4, and n' = (7 mod (4))*(4)^0 = 3.

There are no other bases b >= 4 except 4, 8 for n = 7.

(n, b) maps to (0, 1) if b is prime. Following this and comment in A337536 we can say that all of the terms of A337536 will map to (0, 1) only, except A337536(2).

For above (n, b) -> (n', b') one possible (n, b) pair for (n', b') is { Sum_{i>=0} ((A'(i)+b') *((2*b')^i)), 2*b'}.

This sequence is the list of indices k such that A337496(k)=3.

Conjecture: this sequence is finite and full. a(26) > 3.8*10^12 if it exists.

All terms of this sequence increased by 1 are either prime numbers, or prime numbers squared, or 2 times a prime number because if b is a strict divisor of k+1, the digit for the units in the expansion of k in base b is b-1 so it must be 2 or the third base. In fact k+1 could have been equal to 8=2*4 but 7 is not a term of the sequence (7 = 111_2 = 21_3 = 13_4 = 7_8).