OFFSET
1,1
COMMENTS
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).
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
LINKS
FORMULA
PROG
(PARI)
A006519(n) = (1<<valuation(n, 2));
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, May 08 2009, May 11 2009
EXTENSIONS
a(1) corrected and sequence extended by Antti Karttunen, Sep 25 2018
STATUS
approved