A159945 Let f be defined as in A159885. Then a(n) = max{A000120(f^i(2n+1)): 1 <= i <= A159885(n)}.

2, 1, 3, 3, 2, 2, 4, 3, 4, 1, 3, 4, 3, 3, 5, 4, 4, 3, 5, 8, 2, 2, 4, 3, 4, 2, 4, 4, 4, 4, 6, 3, 4, 3, 5, 8, 4, 4, 6, 5, 6, 1, 3, 3, 3, 3, 5, 8, 4, 3, 6, 6, 3, 3, 5, 4, 5, 3, 5, 5, 5, 5, 7, 8, 4, 4, 5, 8, 5, 4, 6, 8, 6, 3, 5, 5, 6, 5, 7, 8, 6, 5, 8, 7, 2, 2, 4, 3, 4, 2, 4, 4, 4, 4, 6, 8, 6, 3, 5, 5, 4, 4, 7, 5, 6

Offset: 1

Author

Vladimir Shevelev, Apr 27 2009

Keywords

• nonn

Comments

Problem: find an upper estimate for a(n).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A000120, A006519, A007814, A075677, A159885.

Programs

PARI

A006519(n) = (1<A006519((3*((n-1)/2))+2); \\ Defined for odd n only. Cf. A075677.
A159945(n) = { my(w=hammingweight(n), m = 0, n = (n+n+1)); for(k=1,oo,n = f(n); m = max(m,hammingweight(n)); if(hammingweight(n) <= w, return(m))); }; \\ Antti Karttunen, Sep 22 2018

Extensions

More terms from Antti Karttunen, Sep 22 2018