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%I #57 Aug 14 2024 11:06:06
%S 0,1,1,3,1,3,3,7,1,3,5,7,3,7,7,15,1,3,6,7,5,11,13,15,3,7,11,15,7,15,
%T 15,31,1,3,6,7,9,13,14,15,5,11,21,23,13,27,29,31,3,7,14,15,11,23,27,
%U 31,7,15,23,31,15,31,31,63,1,3,6,7,12,13,14,15,9,19
%N a(n) is the least number whose binary expansion can be obtained by reversing one or more consecutive bits in the binary expansion of n.
%C This sequence has similarities with A087734; here we reverse some consecutive bits, there we negate some consecutive bits.
%H Rémy Sigrist, <a href="/A373886/b373886.txt">Table of n, a(n) for n = 0..8190</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F a(n) <= n with equality iff n belongs to A000225.
%F a^k(n) = A038573(n) for k sufficiently large (where a^k denotes the k-th iterate of the sequence).
%F a(2^k) = 1 for any k >= 0.
%F A000120(a(n)) = A000120(n).
%e The first terms, alongside their binary expansion, are:
%e n a(n) bin(n) bin(a(n))
%e -- ---- ------ ---------
%e 0 0 0 0
%e 1 1 1 1
%e 2 1 10 1
%e 3 3 11 11
%e 4 1 100 1
%e 5 3 101 11
%e 6 3 110 11
%e 7 7 111 111
%e 8 1 1000 1
%e 9 3 1001 11
%e 10 5 1010 101
%e 11 7 1011 111
%e 12 3 1100 11
%e 13 7 1101 111
%e 14 7 1110 111
%e 15 15 1111 1111
%e 16 1 10000 1
%p f:= proc(n) local L,nL,i,j,k,r,x;
%p L:= convert(n,base,2);
%p nL:= nops(L);
%p r:= n;
%p for i from 1 to nL-1 do
%p for j from i+1 to nL do
%p r:= min(r, n + add((L[j-k]-L[i+k])*2^(i+k-1),k=0..j-i));
%p od od;
%p r
%p end proc:
%p map(f, [$0..100]); # _Robert Israel_, Aug 13 2024
%o (PARI) a(n, base = 2) = { my (d = if (n, digits(n, base), [0])); setbinop((i, j) -> fromdigits(concat([d[1..i-1], Vecrev(d[i..j]), d[j+1..#d]]), base), [1..#d])[1]; }
%o (Python)
%o def a(n):
%o b = bin(n)[2:]
%o return min(int(b[:i]+b[i:j][::-1]+b[j:], 2) for i in range(len(b)) for j in range(i, len(b)+1))
%o print([a(n) for n in range(74)]) # _Michael S. Branicky_, Aug 13 2024
%Y Cf. A000120, A000225, A038573, A087734, A331856.
%K nonn,base,look,new
%O 0,4
%A _Rémy Sigrist_, Aug 10 2024
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