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A326732 Number of iterations of A326731(x) starting at x = n to reach 0.

%I A326732 #15 Jul 23 2019 18:26:43
%S A326732 0,1,1,2,1,4,2,3,1,6,4,7,2,5,3,4,1,8,6,11,4,9,7,10,2,7,5,8,3,6,4,5,1,
%T A326732 10,8,15,6,13,11,16,4,11,9,14,7,12,10,13,2,9,7,12,5,10,8,11,3,8,6,9,4,
%U A326732 7,5,6,1,12,10,19,8,17,15,22,6,15,13,20,11,18,16,21,4,13,11,18,9,16,14,19,7,14,12,17,10,15,13,16,2,11,9,16,7
%N A326732 Number of iterations of A326731(x) starting at x = n to reach 0.
%C A326732 The sequence is well-defined, see Problem 5 of IMO 2019.
%H A326732 International Mathematical Olympiad, <a href="https://www.imo-official.org/problems.aspx">Problem 5 of IMO 2019</a>.
%H A326732 Rémy Sigrist, <a href="/A326732/a326732.png">Scatterplot of the ordinal transform of the first 2^16 terms</a>
%F A326732 a(n) = 2*A230877(n) - A000120(n)^2.
%F A326732 a(n) = 1 iff n is a power of 2.
%o A326732 (PARI) A326732(n) = my(b=binary(n)); 2*sum(i=1, #b, i*b[i]) - vecsum(b)^2;
%Y A326732 Cf. A000120, A326729, A326730, A326731.
%K A326732 base,nonn
%O A326732 0,4
%A A326732 _Max Alekseyev_, Jul 22 2019