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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; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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]
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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
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CROSSREFS
| Cf. A122458, A159885, A159945
Sequence in context: A080215 A060500 A187284 * A131718 A131017 A049046
Adjacent sequences: A160195 A160196 A160197 * A160199 A160200 A160201
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KEYWORD
| nonn
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AUTHOR
| Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 04 2009
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EXTENSIONS
| a(1) corrected and sequence extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 15 2009
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