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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.
20
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
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
KEYWORD
easy,nonn
AUTHOR
T. D. Noe, Sep 25 2002
STATUS
approved