|
Search: id:A160198
|
|
| |
|
| 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 4, 1, 3, 1, 2, 1, 2, 1, 3, 1, 3, 1, 3, 1, 2, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
Let f(2n+1)=A000265(3n+2) be defined as in A159885. Then a(n) is the least number k of iterations such that either f^k(2n+1))<2n+1 or A000120(f^k(2n+1))<A000120(2n+1).
Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n. [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 05 2009]
|
|
MAPLE
|
A000265 := proc(n) option remember ; local a; a := n ; while a mod 2 = 0 do a := a/2 ; end do; a; end proc:
f := proc(n) local m ; m := (n-1)/2 ; A000265(3*m+2) ; end:
A000120 := proc(n) local d; add(d, d=convert(n, base, 2)) ; end proc:
A159885 := proc(n) local k, twon1; k := 0 ; twon1 := 2*n+1 ; while ( A000120(twon1) > A000120(n) ) do twon1 := f(twon1) ; k := k+1 ; end do; k ; end proc:
A122458 := proc(n) local tx1, a; a := 0 ; tx1 := 2*n+1 ; while tx1 >= 2*n+1 do if tx1 mod 2 = 0 then tx1 := tx1/2 ; else tx1 := 3*tx1+1 ; a := a+1 ; fi; end do; a ; end proc:
A160198 := proc(n) min(A159885(n), A122458(n)) ; end: seq(A160198(n), n=1..130) ; # R. J. Mathar, May 15 2009
|
|
CROSSREFS
|
Cf. A122458, A159885, A159945
Sequence in context: A081729 A080215 A060500 this_sequence A131718 A131017 A049046
Adjacent sequences: A160195 A160196 A160197 this_sequence A160199 A160200 A160201
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 04 2009
|
|
EXTENSIONS
|
a(1) corrected and sequence extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 15 2009
|
|
|
Search completed in 0.002 seconds
|
