cp's OEIS Frontend

A372453 a(n) = A372443(n) - A086893(1+A372447(n)).

%I A372453 #10 May 05 2024 10:42:10
%S A372453 6,-12,10,-6,-14,22,-52,36,6,-76,18,-58,20,-38,-78,54,-260,104,-46,38,
%T A372453 36,-58,84,-22,138,-134,-286,254,-984,58,2,-1362,-336,-276,92,-16,8,2,
%U A372453 -18,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N A372453 a(n) = A372443(n) - A086893(1+A372447(n)).
%C A372453 These are the differences obtained when the term of A086893 that has the same binary length as A372443(n) is subtracted from the latter. Here A372443(n) gives the n-th iterate of 27 with Reduced Collatz-function R, where R(n) = A000265(3*n+1).
%C A372453 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 A372453 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F A372453 a(n) = A372443(n) - A086893(1+A000523(A372443(n))).
%e A372453 The term of A086893 that has same binary length as A372443(0) = 27 is 21 [as 21 = 10101_2 in binary, and 27 = 11011_2 in binary], therefore a(0) = 27-21 = 6.
%e A372453 The term of A086893 that has same binary length as A372443(1) = 41 is 53, therefore a(1) = 41-53 = -12.
%o A372453 (PARI)
%o A372453 A000265(n) = (n>>valuation(n,2));
%o A372453 A000523(n) = logint(n,2);
%o A372453 A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3);
%o A372453 A372443(n) = { my(x=27); while(n, x=A000265(3*x+1); n--); (x); };
%o A372453 A372453(n) = { my(x=A372443(n)); (x - A086893(1+A000523(x))); };
%Y A372453 Cf. A000265, A000523, A086893, A372443.
%Y A372453 Cf. also A372446.
%K A372453 sign
%O A372453 0,1
%A A372453 _Antti Karttunen_, May 05 2024