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

Author

Vladimir Shevelev, May 07 2009

Keywords

• nonn
• look

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

Crossrefs

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

Programs

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

Mathematica

A006519[n_] := Do[If[fi[[1]] == 2, Return[2^fi[[2]]], Return[1]], {fi, FactorInteger[n]}];
f[n_] := With[{n2 = 3 n/2 + 1/2}, n2/A006519[n2]];
A006694[n_] := Sum[EulerPhi[d]/MultiplicativeOrder[2, d], {d, Divisors[2n + 1]}] - 1;
a[n_] := Module[{ref, k, fk}, ref = A006694[n]; k = 1; fk = f[2n + 1]; While[True, If[A006694[(fk - 1)/2] < ref, Return[k]]; fk = f[fk]; k++]];
Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Aug 28 2024, after R. J. Mathar *)

PARI

f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2);
A006519(n) = (1<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

Extensions

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