A224165 Number of cycles the reduced complex Collatz function using 3 + (2n-1)i takes to hit 1 + i, or 0 if this never happens.

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

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[c-1]!=1 || i[c-1]!=1 && c<1000) {
i[c]=r[c-1]*3+1;
r[c]=-i[c-1]*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 * A330318 A334934 A131286

Adjacent sequences: A224162 A224163 A224164 * A224166 A224167 A224168

Keyword

nonn,less

Author

Jon Perry, Apr 02 2013

Status

approved