A160266 Let f and its k-fold iteration f^k be defined as in A159885. a(n) is the least k for which A006694( (f^k(2n+1)-1)/2 ) < A006694(n).

2, 1, 1, 2, 4, 2, 1, 1, 6, 1, 2, 1, 1, 5, 1, 1, 1, 6, 1, 4, 3, 1, 2, 1, 1, 2, 1, 1, 10, 5, 1, 1, 8, 1, 1, 1, 1, 1, 2, 1, 40, 1, 1, 1, 1, 1, 6, 3, 1, 7, 17, 1, 36, 1, 1, 2, 1, 1, 1, 20, 1, 1, 1, 1, 8, 1, 1, 18, 13, 1, 5, 1, 2, 6, 1, 1, 1, 1, 1, 1, 6, 1, 9, 11, 2, 9, 1, 2, 9, 4, 6, 1, 1, 1, 9, 7, 1, 7, 29, 2, 2, 1

Offset

1,1

Comments

Conjecture. For every n>=1, there exists a finite value of a(n). It is easy to see that this conjecture is equivalent to the well known Collatz 3n+1 conjecture.

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to 3x+1 (or Collatz) problem

Maple

A006519 := proc(n) local i ; for i in ifactors(n)[2] do if op(1, i) = 2 then return op(1, i)^op(2, i) ; fi ; od: return 1 ; end proc:
f := proc(twon1) local threen2 ; threen2 := 3*twon1/2+1/2 ; threen2/A006519(threen2) ; end proc:
A160266 := proc(n) local ref, k, fk ; ref := A006694(n) ; k := 1 ; fk := f(2*n+1) ; while true do if A006694( (fk-1)/2 ) < ref then return k; end if; fk := f(fk) ; k := k+1 ; end do ; end proc:
seq(A160266(n), n=1..120) ; # R. J. Mathar, Feb 02 2010

Prog

(PARI)
f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2);
A006519(n) = (1<<valuation(n, 2));
A006694(n) = (sumdiv(2*n+1, d, eulerphi(d)/znorder(Mod(2, d))) - 1); \\ From A006694
A160266(n) = { my(w=A006694(n), n = (n+n+1), k=0); while(A006694((n-1)/2) >= w, k++; n = f(n)); (k); }; \\ Antti Karttunen, Sep 22 2018

Crossrefs

Cf. A006694, A159885, A159945, A160198, A122458, A160267.

Sequence in context: A359627 A127309 A097853 * A322134 A023504 A157905

Adjacent sequences: A160263 A160264 A160265 * A160267 A160268 A160269

Keyword

nonn,look

Author

Vladimir Shevelev, May 07 2009

Extensions

More terms from R. J. Mathar, Feb 02 2010

Status

approved