

A326729


a(0) = 0; for n >= 1, a(n) is the result of inverting sth bit (from right) in n, where s is the number of ones in the binary representation of n.


4



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, 19, 24, 21, 22, 15, 33, 35, 32, 39, 38, 33, 34, 47, 42, 45, 46, 35, 40, 37, 38, 63, 50, 53, 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
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OFFSET

0,3


COMMENTS

Iterations of a(n) always reach 0 (cf. A326730), see Problem 5 of IMO 2019.


LINKS

Robert Israel, Table of n, a(n) for n = 0..10000
International Mathematical Olympiad, Problem 5 of IMO 2019.
Index to sequences related to Olympiads.


FORMULA

For n>=1, a(n) = n XOR 2^(A000120(n)1).
From Robert Israel, Oct 01 2020: (Start)
a(2*n+1) = 2*a(n).
a(2*n + 2^k) = 2*a(n)+2^k if 2^k > 2*n. (End)


MAPLE

f:= proc(n) local s;
s:= convert(convert(n, base, 2), `+`);
Bits:Xor(n, 2^(s1))
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Oct 01 2020


PROG

(PARI) A326729(n) = if(n==0, return(0)); bitxor(n, 2^(hammingweight(n)1));


CROSSREFS

Cf. A000120, A326730, A326731, A326732.
Sequence in context: A077020 A107920 A169998 * A171998 A159285 A021080
Adjacent sequences: A326726 A326727 A326728 * A326730 A326731 A326732


KEYWORD

base,nonn,look


AUTHOR

Max Alekseyev, Jul 22 2019


STATUS

approved



