A160322 - OEIS

%I #10 Sep 25 2018 20:50:28

%S 2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,

%T 1,1,1,1,2,1,2,1,1,1,1,1,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,

%U 3,1,2,1,2,1,1,1,1,1,1,1,2,1,9,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2

%N a(n) = min(A160198(n), A160267(n)).

%C Let f be defined as in A159885. Then a(n) is the least k such that either f^k(2n+1))<2n+1 or A000120(f^k(2n+1)) < A000120(2n+1) or A006694((f^k(2n+1)-1)/2) < A006694(n).

%C In connection with A160198, A160267, A160322 we pose a new (3x+1)-problem: does there exist a finite number of sequences A_i(n), i=1,...,T, such that: 1) A_i(0)=0 and A_i(n)>0 for n>=1; 2) if B_i(n) denotes the least k for which A_i(n)>A_i((f^k(2n+1)-1)/2), then B(n)=min_{i=1,...,T}B_i(n)=1 for every n>=1? Note that this problem is weaker than (3x+1)-Collatz problem. Indeed, if the Collatz conjecture is true, then there exist nonnegative sequences A(n) for which A(0)=0 and A(n)>A((f(2n+1)-1)/2) for every n>=1 (see A160348). - _Vladimir Shevelev_, May 15 2009

%H Antti Karttunen, <a href="/A160322/b160322.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%F a(n) = min(A122458(n), A159885(n), A160266(n)).  - _Antti Karttunen_, Sep 25 2018

%o (PARI)

%o f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined for odd n only. Cf. A075677.

%o A006519(n) = (1<<valuation(n, 2));

%o A006694(n) = (sumdiv(2*n+1, d, eulerphi(d)/znorder(Mod(2, d))) - 1); \\ From A006694

%o A160322(n) = { my(v=A006694(n), u = (n+n+1), w = hammingweight(u), k=0); while((u >= (n+n+1))&&(hammingweight(u) >= w)&&(A006694((u-1)/2) >= v), k++; u = f(u)); (k); }; \\ _Antti Karttunen_, Sep 25 2018

%Y Cf. A000120, A006694, A122458, A159885, A159945, A160198, A160266, A160267.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, May 08 2009, May 11 2009

%E a(1) corrected and sequence extended by _Antti Karttunen_, Sep 25 2018