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Kevin Ryde: If you're saying "a(i)" then may as well say the ranges instead of "S".  If you're saying S then may as well be S(k) = {S(k-1)+1, S(k)}, with +1 meaning +1 on each term of S(k-1) (if that's right).

Terms a(n); n >= 2 of this sequence can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, S(k) = {a(i)+1; a(i) in S(k-1), i = 2^(k-1)+1....2^k}, then adjoin the 2^(k-1) terms of S(k-1) as suffix. Thus S(1) = {2,1}, S(2) = {3,2,2,1}, and so on (see Example). Each step of the recursion gives the next 2^k terms, from a(2^k+1) to a(2^(k+1)) inclusive. - David James Sycamore, Jul 15 2024

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Michael De Vlieger: Minor edit.

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