This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A333970 #99 Aug 05 2023 21:19:50
%S A333970 2,2,3,2,4,2,5,2,3,6,2,3,7,2,3,4,8,2,3,9,2,5,10,2,11,2,3,4,6,12,2,4,
%T A333970 13,2,4,7,14,2,3,4,5,15,2,3,4,8,16,2,3,17,2,3,6,9,18,2,3,19,2,3,4,5,
%U A333970 10,20,2,3,5,7,21,2,3,5,11,22,2,3,5,23,2,3,4,5,6,8,12,24
%N A333970 Irregular triangle read by rows where the n-th row lists the bases 2<=b<=n+1 where n in base b contains the digit b-1.
%C A333970 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.*
%C A333970 This condition is necessary and sufficient.
%C A333970 Proof that Condition is Necessary:
%C A333970 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).
%C A333970 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)).
%C A333970 n = Sum_{i>=0} (A(i)*(b'^i)) +
%C A333970 Sum_{i>=1} (A(i)*(b^i - b'^i))
%C A333970 = 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)),
%C A333970 where A'(i) = A(i) (mod b').
%C A333970 Now n-Sum_{i>=0} ((A(i)-A'(i))*(b'^i))
%C A333970 - Sum_{i>=1} (A(i)*(b^i - b'^i))
%C A333970 = Sum_{i>=0} (A'(i)*(b'^i)).
%C A333970 Since A'(j) = A(j) (mod b') = (b-1) (mod b') = b'-1(due to equation (1) above and A(j) = b-1.
%C A333970 Hence there exists n' = Sum_{i>=0} (A'(i)*(b'^i)) > 0 containing digit b'-1 in base b'.
%C A333970 Table of n/b with cell containing T(n, b) = (n', b') for q = b/2. n' = Sum_{i>=0} (A'(i)*(b'^i))
%C A333970 n/b| 4 | 6 | 8 | 10 | 12
%C A333970 3 |(1,2)| | | |
%C A333970 4 | | | | |
%C A333970 5 | |(2,3)| | |
%C A333970 6 | | | | |
%C A333970 7 |(3,2)| |(3,4)| |
%C A333970 8 | | | | |
%C A333970 9 | | | |(4,5)|
%C A333970 10 | | | | |
%C A333970 11 |(1,2)|(5,3)| | |(5,6)
%C A333970 Example: For table n/b in comments containing (n',b') in its cells.
%C A333970 For n = 7:
%C A333970 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.
%C A333970 In base b = 8, n = 7 :- q = b' = 8/2 = 4, and n' = (7 mod (4))*(4)^0 = 3.
%C A333970 There are no other bases b >= 4 except 4, 8 for n = 7.
%C A333970 (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).
%C A333970 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'}.
%H A333970 Rémy Sigrist, <a href="/A333970/b333970.txt">Table of n, a(n) for n = 1..10027</a> (rows for n = 1..1014, flattened)
%e A333970 Triangle begins
%e A333970 Row Bases
%e A333970 n=1: 2
%e A333970 n=2: 2 3
%e A333970 n=3: 2 4
%e A333970 n=4: 2 5
%e A333970 n=5: 2 3 6
%e A333970 n=6: 2 3 7
%e A333970 n=7: 2 3 4 8
%e A333970 n=8: 2 3 9
%e A333970 n=9: 2 5 10
%e A333970 n=10: 2 11
%o A333970 (PARI) row(n) = {my(list = List()); for (b=2, n+1, if (vecmax(digits(n, b)) == b-1, listput(list, b));); Vec(list);} \\ _Michel Marcus_, Sep 11 2020
%Y A333970 Cf. A337535 (second column), A338295 (penultimate column), A337496 (row widths), A337536 (width 2), A337143 (width 3).
%Y A333970 Rows containing bases 3..11 respectively: A074940, A337250, A337572, A333656, A337141, A337239, A338090, A011539, A095778.
%K A333970 nonn,base,tabf,uned
%O A333970 1,1
%A A333970 _Devansh Singh_, Sep 03 2020
%E A333970 More terms from _Michel Marcus_, Sep 11 2020