A075680 For odd numbers 2n-1, the minimum number of iterations of the reduced Collatz function R required to yield 1. The function R is defined as R(k) = (3k+1)/2^r, with r as large as possible.

0, 2, 1, 5, 6, 4, 2, 5, 3, 6, 1, 4, 7, 41, 5, 39, 8, 3, 6, 11, 40, 9, 4, 38, 7, 7, 2, 41, 10, 10, 5, 39, 8, 8, 3, 37, 42, 3, 6, 11, 6, 40, 1, 9, 9, 33, 4, 38, 43, 7, 7, 31, 12, 36, 41, 24, 2, 10, 5, 10, 34, 15, 39, 15, 44, 8, 8, 13, 32, 13, 3, 37, 42, 42, 6, 3, 11, 30, 11, 18, 35, 6, 40, 23

Offset

1,2

Comments

See A075677 for the function R applied to the odd numbers once. The 3x+1 conjecture asserts that a(n) is a finite number for all n. The function R applied to the odd numbers shows the essential behavior of the 3x+1 iterations.

Bisection of A006667. - T. D. Noe, Jun 01 2006

Links

T. D. Noe, Table of n, a(n) for n = 1..5000

Example

a(4) = 5 because 7 is the fourth odd number and 5 iterations are needed: R(R(R(R(R(7)))))=1.

Mathematica

nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[m=n; cnt=0; If[n>1, While[m=nextOddK[m]; cnt++; m!=1]]; cnt, {n, 1, 200, 2}]

Prog

(Haskell)
a075680 n = snd $ until ((== 1) . fst)
            (\(x, i) -> (a000265 (3 * x + 1), i + 1)) (2 * n - 1, 0)
-- Reinhard Zumkeller, Jan 08 2014
(Perl)
sub a {
  my $v = 2 * shift() - 1;
  my $c = 0;
  until (1 == $v) {
    $v = 3 * $v + 1;
    $v /= 2 until ($v & 1);
    $c += 1;
  }
  return $c;
} # Ruud H.G. van Tol, Nov 16 2021
(PARI) a(n)=my(s); n+=n-1; while(n>1, n+=n>>1+1; if(n%2==0, n>>=valuation(n, 2)); s++); s \\ Charles R Greathouse IV, Dec 22 2021

Crossrefs

Cf. A075677.

Cf. A075684 for the largest number attained during the iteration.

Cf. A000265.

Cf. A060445 which also counts intermediate even steps.

Sequence in context: A162750 A330984 A353648 * A192024 A249283 A176035

Adjacent sequences: A075677 A075678 A075679 * A075681 A075682 A075683

Keyword

easy,nonn

Author

T. D. Noe, Sep 25 2002

Status

approved