

A224165


Number of cycles the reduced complex Collatz function using 3 + (2n1)i takes to hit 1 + i, or 0 if this never happens.


1



5, 5, 5, 13, 9, 5, 13, 5, 9, 5, 17, 9, 13, 13, 5, 9, 21, 9, 9, 17, 0, 5, 9, 5, 13, 13, 13, 13, 17, 5, 9, 9, 0, 9, 9, 13, 13, 5, 17, 17, 17, 17, 17, 17, 21, 9, 5, 5, 25, 13, 5, 0, 25, 13, 13, 17, 9, 5, 13, 9, 33, 9, 9, 0, 0, 21, 9, 33, 21, 9, 13, 5, 13, 9, 17
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OFFSET

1,1


COMMENTS

The complex Collatz function takes a complex number z to 3iz + (1+i). The resulting real part is divided by 2 until it's odd, and the same for the imaginary part.
The effect here is that say 4 + 14i is reduced to 1 + 7i.
The reduced complex Collatz function does all this in the same cycle.
Equals 0 for n = 21, 33, 52, 64, 65, 81, 82, 101, 103, 127, 129, 130, 163, 201, 204, 206, 253, 254...
This is conjectured to be infinite and the same as in A224067.
For example, 3 + 41i yields 61 + 5i, 7 + 91i, 137 + 5i, 1 + 103i, 77 + 1i, 1 + 115i, 173 + 1i, 1 + 65i, 97 + 1i, 1 + 145i, 109 + 1i, 1 + 41i, 61 + 1i, 1 + 91i, 137 + 1i, 1 + 103i, 77 + 1i, 1 + 115i, 173 + 1i, 1 + 65i, 97 + 1i, 1 + 145i, 109 + 1i, 1 + 41i, which becomes cyclic about 1 + 41i.


LINKS

Table of n, a(n) for n=1..75.
Eric Weisstein, Mathworld: Collatz Problem
Wikipedia, Collatz Problem


EXAMPLE

a(5) = 13 because rcC(3 + 7i) yields 5 + 5i, 7 + 7i, 11 + 5i, 1 + 17i, 25 + 1i, 1 + 37i, 7 + 1i, 1 + 11i, 1 + 1i, 1 + 1i, 1 + 1i, 1 + 1i, which is 13 terms.


PROG

(JavaScript)
for (b=1; b<150; b+=2) {
c=1;
r[0]=3; i[0]=b;
while (r[c1]!=1  i[c1]!=1 && c<1000) {
i[c]=r[c1]*3+1;
r[c]=i[c1]*3+1;
while (r[c]%2==0) r[c]/=2;
while (i[c]%2==0) i[c]/=2;
c++;
}
document.write(c+", ");
}


CROSSREFS

Cf. A224067.
Sequence in context: A262947 A265821 A266953 * A131286 A183390 A013607
Adjacent sequences: A224162 A224163 A224164 * A224166 A224167 A224168


KEYWORD

nonn,less


AUTHOR

Jon Perry, Apr 02 2013


STATUS

approved



