%I #10 May 03 2024 12:43:10
%S 14,28,10,26,18,62,116,44,14,92,50,78,60,122,82,222,260,232,114,46,44,
%T 78,252,106,138,410,354,774,1064,218,2,1366,336,276,228,16,8,2,22,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N a(n) = A372358(A372443(n)).
%C These are the bitmasks (or symmetric differences) obtained when the n-th iterate of 27 with Reduced Collatz-function R [= A372443(n), where R(n) = A000265(3*n+1)] is xored with that term of A086893 that has the same binary length. The binary expansions of the terms of A086893 are always of the form 10101...0101 (i.e., alternating 1's and 0's starting and ending with 1) when the binary length is odd, and of the form 110101...0101 (i.e., 1 followed by alternating 1's and 0's, and ending with 1) when n is even. Note that for all n >= 1, R(A086893(2n-1)) = 1, and R(A086893(2n)) = 5 (with R(5) = 1), so the first zero here, a(39) = 0 indicates that the iteration will soon have reached the terminal 1, and indeed, A372443(41) = 1.
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%o (PARI)
%o A000265(n) = (n>>valuation(n,2));
%o A000523(n) = logint(n,2);
%o A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3);
%o A372358(n) = bitxor(A086893(1+A000523(n)),n);
%o A372443(n) = { my(x=27); while(n, x=A000265(3*x+1); n--); (x); };
%o A372446(n) = A372358(A372443(n));
%Y Column 14 of A372361.
%Y Cf. A000265, A075677, A086893, A372358, A372443.
%K nonn
%O 0,1
%A _Antti Karttunen_, May 03 2024