

A075680


For odd numbers 2n1, 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
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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.


LINKS



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)
(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;
(PARI) a(n)=my(s); n+=n1; 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. A075684 for the largest number attained during the iteration.
Cf. A060445 which also counts intermediate even steps.


KEYWORD

easy,nonn


AUTHOR



STATUS

approved



