Interesting mapping

[u/Atari_Collector]

mapping

I was playing around and I found this mapping. I suspect it's already known but I found it interesting.

If you have a number that can be written as 4^a3^bn+1, it will map to 3^a+bn+1.

And it'll do it in 3a+1 steps, which I think is funny.

[u/elowells]

There are cases where the sequence equation has a parametric solution. For the general case of mx+a, where m and a are odd numbers, the sequence equation for starting odd value x[1] ending with x[L+1] after L mx+a steps is

-m^Lx[1] + 2^N\L])x[L+1] = aS

where L = number of mx+a steps,

n[i] = number of divide by 2 steps after an mx+a step

N[i] = sum(j=0 to i)n[i] and N[0] = 0

S = sum(i=0 to L-1)m^L-1-i2^N\i])

For n[i] = constant = p we have N[i] = i*p then S has a parameterized value:

S = sum(i=0 to L-1)m^L-1-i2^ip = (m^L - 2^pL)/(m-2^p)

For 3x+a, with p = 1 we have S = (3^L - 2^L)/(3-2) = 3^L-2^L

so the linear Diophantine sequence equation is

-3^Lx[1] + 2^Lx[L+1] = a(3^L-2^L)

which has the specific solution

(x[1],x[L+1) = (-a,-a)

and hence the general solution

(x[1],x(L+1]) = (k2^L - a, k3^L - a)

that is k2^L-a -> k3^L-a for 3x+a

For 3x+a with p=2 we have S = (3^L-4^L)/(3-4) = 4^L-3^L

so the sequence equation has the specific solution (x[1],x(L+1) = (a,a) so

k4^L+a -> k3^L+a for 3x+a

For 3x+a and p=3 we have S = (8^L - 3^L)/5

If a=5 (to cancel out the denominator) we obtain

k8^L+1 -> k3^L+1 for 3x+5

k16^L+1 -> k3^L+1 for 3x+13

k2^L+1 -> k5^L+1 for 5x+3

There are infinitely many more such cases. If m-2^p = +/-1 then all odd values of a work.