cp's OEIS Frontend

Consider the lexicographically earliest sequence of nonnegative integers that does not contain the arithmetic mean of any pair of terms (such sequences are called 3-free sequences as they do not contain 3-term arithmetic progressions): 0,1,3,4 and so on. This sequence is Stanley sequence S(0,1). Remove numbers in the Stanley sequence from nonnegative integers and repeat the process of finding the next earliest 3-free sequence, which is sequence A323398. The next iteration produces sequence A323418. The fourth iteration produces this sequence.

When represented in ternary this sequence consists of integers ending in 10, 11, 20 or 21, and there is exactly one digit 2 before that that might be followed by zeros.

All integers n >= 4 may trivially be expressed using only the digits 0 and 1 in three different bases: 2, n-1 (as '11') and n (as '10'). The numbers in this sequence cannot be expressed using only 0 and 1 in any other base.

The only positive integers that may be expressed using only the digits 0 and 1 in fewer than three different bases are 2 and 3, for which the values {2, n-1, n} are not all distinct or are not all valid bases.

An equivalent definition: For each term a(n) of this sequence, there are at most three integers k >= 2 for which a(n) is a sum of distinct nonnegative integer powers of k.

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).

Complement of the Moser-de Bruijn sequence (A000695).

Numbers whose base 4 digits contain either 2 or 3.