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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'}.