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Comment from Devansh Singh, Dec 07 2020: (Start)
0. Take some maximum n = N for finding out a(n).
1. Iterate b in ascending order. 3<=b<=(N+1)/2.
2. Start with n = b-1 and iterate n as follows.
3. n in base b, contains r (b-1) digits and s (b-2) digits in continuation from the right i.e. n_b = ..(b-k)(b-2)(b-2)..(b-2)(b-1)..(b-1), so a(n+i) = b for i = 0 to (b^(r+s) - b^r)/(b-1) inclusive.
4. Take next n equal to n+i+b with n <= N.
5. Repeat step 3 and 4 till n<=N.
6. Repeat all steps(1 to 6) and skip setting a(n) in step 3 for those values of n for which a(n) is already set.
This procedure gives a(n) for those values of n which are not in A337536. For those values of n, a(n) = n+1.
For example: Start with b=3 so n>=2. a(2) = 3 where (r,s) for n=2 is (1,0) so (b^(r+s) - b^r)/(b-1) = 0, then next n = 2+0+3 = 5. a(5) = 3 where (r,s) for n=5 is (1,1) so (b^(r+s) - b^r)/(b-1) = 3 and a(5)=a(6)=a(7)=a(8)=3. Now next n = 5+3+3 = 8+3 =11. If 11 is maximum n for searching a(n) then take next b = 4. Repeat the same procedure with b = 4 skipping those values of n for which a(n) is already set.(End)
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Comment from Devansh Singh, Dec 07 2020: (Start)
0. Take some maximum n = N for finding out a(n).
1. Iterate b in ascending order. 3<=b<=(N+1)/2.
2. Start with n = b-1 and iterate n as follows.
For b = 3, 4, 5, ... taken in ascending order and n taken in ascending order from {n>=b-1: b|(n+1)} in base b, contains r (b-1) digits and s (b-2) digits in continuation from the right (i.e. n_b = ..(b-k)(b-2)(b-2)..(b-2)(b-1)..(b-1)), then , so a(n+i) = b for i = 0 to (b^(r+s) - b^r)/(b-1) inclusive. For base b+m, m>0 skip those values of n for which a(n) is already set. - _Devansh Singh_, Dec 07 2020
4. Take next n equal to n+i+b with n <= N.
5. Repeat step 3 and 4 till n<=N.
6. Repeat all steps(1 to 6) and skip setting a(n) in step 3 for those values of n for which a(n) is already set.
This procedure gives a(n) for those values of n which are not in A337536. For those values of n, a(n) = n+1.
For example: Start with b=3 so n>=2. a(2) = 3 where (r,s) for n=2 is (1,0) so (b^(r+s) - b^r)/(b-1) = 0, then next n = 2+0+3 = 5. a(5) = 3 where (r,s) for n=5 is (1,1) so (b^(r+s) - b^r)/(b-1) = 3 and a(5)=a(6)=a(7)=a(8)=3. Now next n = 5+3+3 = 8+3 =11. If 11 is maximum n for searching a(n) then take next b = 4. Repeat the same procedure with b = 4 skipping those values of n for which a(n) is already set.(End)
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For n>1 , a(n) is the least base b>2 such that the digits of n in base b contain the digit b-1; a(1)=1.
For b in {b= 3, 4, 5, ...} taken in ascending order: and n taken in ascending order from {n>=b-1: b|(n+1)} in base b , contains r (b-1) digits and s (b-2) digits in continuation from the right (n_b = ..(b-k)(b-2)(b-2)..(b-2)(b-1)..(b-1)) , then a(n+i) = b for i = 0 to (b^(r+s) - b^r)/(b-1) inclusive. For base b+m, m>0 skip those values of n for which a(n) is already set. - Devansh Singh, Dec 07 2020
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For fixed b in {b=3,4,5..} taken in ascending order: n taken in ascending order from {n>=b-1: b|(n+1)} in base b contains r (b-1) digits and s (b-2) digits in continuation from the right(n_b = ..(b-k)(b-2)(b-2)..(b-2)(b-1)..(b-1)) then a(n+i) = b for i = 0 to (b^(r+s) - b^r)/(b-1) inclusive. For base b+m, m>0 skip those values of n for which a(n) is already set. - Devansh Singh, Dec 07 2020