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%I #22 Oct 02 2020 09:03:39
%S 0,0,3,1,5,7,4,3,9,11,8,15,14,9,10,7,17,19,16,23,22,17,18,31,26,29,30,
%T 19,24,21,22,15,33,35,32,39,38,33,34,47,42,45,46,35,40,37,38,63,50,53,
%U 54,59,48,61,62,39,60,49,50,43,52,45,46,31,65,67,64,71,70,65,66,79,74,77,78,67,72,69,70,95,82,85,86,91,80,93,94,71,92,81,82,75,84,77,78,127,98,101,102,107,96
%N a(0) = 0; for n >= 1, a(n) is the result of inverting s-th bit (from right) in n, where s is the number of ones in the binary representation of n.
%C Iterations of a(n) always reach 0 (cf. A326730), see Problem 5 of IMO 2019.
%H Robert Israel, <a href="/A326729/b326729.txt">Table of n, a(n) for n = 0..10000</a>
%H International Mathematical Olympiad, <a href="https://www.imo-official.org/problems.aspx">Problem 5 of IMO 2019</a>.
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F For n>=1, a(n) = n XOR 2^(A000120(n)-1).
%F From _Robert Israel_, Oct 01 2020: (Start)
%F a(2*n+1) = 2*a(n).
%F a(2*n + 2^k) = 2*a(n)+2^k if 2^k > 2*n. (End)
%p f:= proc(n) local s;
%p s:= convert(convert(n,base,2),`+`);
%p Bits:-Xor(n,2^(s-1))
%p end proc:
%p f(0):= 0:
%p map(f, [$0..100]); # _Robert Israel_, Oct 01 2020
%o (PARI) A326729(n) = if(n==0,return(0)); bitxor(n,2^(hammingweight(n)-1));
%Y Cf. A000120, A326730, A326731, A326732.
%K base,nonn,look
%O 0,3
%A _Max Alekseyev_, Jul 22 2019
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