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%I #27 Sep 21 2023 11:10:13
%S 0,1,2,6,75
%N Smallest m such that A261923(m) = n.
%C A261923(m) != 5 for m <= 10^5.
%C The same for m <= 5*10^8. - _Michel Marcus_, Sep 20 2023
%C From _Michael S. Branicky_, Sep 21 2023: (Start)
%C a(5) <= 10718873460460617403023221866359404479.
%C a(n) exists for all n. Proof. Let b(i) be the binary representation of i. Let L be its length, and w = 0^L be a string of L 0's. Then a(n+1) <= u = b(1)wb(2)w...wb(a(n)-1)_2 since u's binary representation contains that of each number less than a(n) but not that of a(n). So, A261923(u) = 1 + A261923(a(n)). (End)
%o (Haskell)
%o import Data.List (elemIndex); import Data.Maybe (fromJust)
%o a262279 = fromJust . (`elemIndex` a261923_list)
%o (PARI) a(n) = my(k=0); while (A261923(k) != n, k++); k; \\ _Michel Marcus_, Sep 20 2023
%Y Cf. A056744, A261923.
%K nonn,more
%O 0,3
%A _Reinhard Zumkeller_, Sep 17 2015
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