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A075680
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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.
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20
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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
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(4) = 5 because 7 is the fourth odd number and 5 iterations are needed: R(R(R(R(R(7)))))=1.
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MATHEMATICA
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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}]
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PROG
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(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+=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
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CROSSREFS
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Cf. A075684 for the largest number attained during the iteration.
Cf. A060445 which also counts intermediate even steps.
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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