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%I #41 Jan 02 2023 09:01:30
%S 1,1,2,6,3,5,7,12,4,14,6,11,8,8,13,13,5,10,15,15,7,7,12,12,9,17,9,71,
%T 14,14,14,68,6,19,11,11,16,16,16,24,8,70,8,21,13,13,13,67,10,18,18,18,
%U 10,10,72,72,15,23,15,23,15,15,69,69,7,20,20,20,12,12,12,66,17,74,17
%N Number of iterations of A064455 to reach 2 (or 1 in the case of 1).
%C Similar to 3x+1 series (A008908). Does this sequence converge to 2 for all values of n (true for all values of n up to 100000)? The inverse sequence using next n = n-int(n/2) for n even and n+int(n/2) for n odd leads to 3 (?) possible end sequences (1), (5, 7, 10) and (17, 25, 37, 55, 82, 41, 61, 91, 136, 68, 34)
%C Starting with a number n, the next value generated is n+int(n/2) if n is even, n-int(n/2) if n is odd; a(n) is the number of iteration for the initial value n to reach the limit of 1 to 2
%C Collatz's 3N+1 function as isometry over the dyadics is N->N/2 if even, but N->(3N+1)/2 if odd, including the (necessary) halving into each tripling step. Counting steps until reaching 1 in this way leads to this sequence instead of A008908. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
%C The value at each step of a trajectory starting with n (n>1) is equal to the value plus one at the same step of the row starting with (n-1) of the irregular triangle of the abbreviated (Terras-modified) Collatz sequence (A070168). - _K. Spage_, Aug 07 2014
%H Antti Karttunen, <a href="/A064433/b064433.txt">Table of n, a(n) for n = 1..19683</a>
%H M. del P. Canales Chacon and M. J. Vielhaber, <a href="http://eccc.hpi-web.de/eccc-reports/2004/TR04-057/index.html">Structural and Computational Complexity of Isometries and Their Shift Commutators</a>, Electr. Colloq. on Computational Cpx., ECCC TR04-057, 2004. [From Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009]
%F a(n) = A006666(n-1) + 1. - _K. Spage_, Aug 04 2014
%e a(4) = 6. Starting with 4, 4 is even so the next number is 4+int(4/2) = 6, 6 is even so next number is 6+int(6/2) = 9, 9 is odd so next number is 9-int(9/2) = 5, 5 is odd so next number is 5-int(5/2) = 3, 3 is odd so next number is 3-int(3/2)=2, so giving a sequence of 4,6,9,5,3,2: 6 numbers.
%e a(5) = 3. Starting with 5, A064455(5) = 3, A064455(3) = 2, so giving a trajectory of 5,3,2: 3 numbers. - _K. Spage_, Aug 07 2014
%t Table[Length@ NestWhileList[If[EvenQ@ #, 3 #/2, (# + 1)/2] &, n, # != 1 + Boole[n > 1] &], {n, 75}] (* _Michael De Vlieger_, Sep 24 2016 *)
%o (PARI) A064455(n) = {if(n%2, (n + 1)/2, 3*n/2)}
%o A064433(n) = {my(c=1); if(n==1, 1, while(n!=2, n=A064455(n); c++); c)} \\ _K. Spage_, Aug 07 2014
%Y Cf. A006666, A008908, A064455, A070168.
%K nonn,easy,look
%O 1,3
%A Jonathan Ayres (Jonathan.ayres(AT)btinternet.com), Oct 01 2001
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