4

What is the greatest common denominator of $t(c^a)$ and $t(c^b)$, if $t(n) := k_1 f_1^n + k_2 f_2^n $?

I already found out that the gcd is always a member of $t(n), n \in N $.

$t(n)$ was originally defined of as $t(n) = 10 t(n-1) + t(n-2)$ and $t(0) = 1, t(1) = 10$. WolframAlpha gave me the equation from my first sentence. (The link gives the values for $k_1, k_2, f_1, f_2$.)

This resource gives many useful identities for both gcds and Fibonacci numbers. I tried to use Example 251 but for that I would need to adapt Theorem 102 and subsequently Theorem 50, which I failed to do. Example 251 proves that $gcd(f_m, f_n) = f_{gcd(n, m)}$, where $f_n$ is the nth Fibonacci number.

Here is a table of some results. There clearly is a pattern, but I can't find it. I fixed c=2 and iterated over a=[0,14], b=[0,14]. The value shown is $i$ s.t. $t(i) = gcd(t(c^a), t(c^b))$

1 0 0 0  0  0  0   0   0   0    0    0    0    0     0
0 2 0 2  0  2  0   2   0   2    0    2    0    2     0
0 0 4 0  0  0  4   0   0   0    4    0    0    0     4
0 2 0 8  0  2  0   2   0   8    0    2    0    2     0
0 0 0 0 16  0  0   0   0   0    0    0   16    0     0
0 2 0 2  0 32  0   2   0   2    0    2    0    2     0
0 0 4 0  0  0 64   0   0   0    4    0    0    0     4
0 2 0 2  0  2  0 128   0   2    0    2    0    2     0
0 0 0 0  0  0  0   0 256   0    0    0    0    0     0 
0 2 0 8  0  2  0   2   0 512    0    2    0    2     0
0 0 4 0  0  0  4   0   0   0 1024    0    0    0     4
0 2 0 2  0  2  0   2   0   2    0 2048    0    2     0
0 0 0 0 16  0  0   0   0   0    0    0 4096    0     0
0 2 0 2  0  2  0   2   0   2    0    2    0 8192     0
0 0 4 0  0  0  4   0   0   0    4    0    0    0 16384

Possibly related: https://math.stackexchange.com/a/468493/13597

Comment:

So as not to have too many comments below Aleks Vlasev's answer, I will write this here.

I found that $gcd(c^a + 1, c^b + 1) = (c \mod 2)$, if either a or b is odd (and the other even). The only difficult case to remain now, is if both a and b are even.

The remaining question has been moved to its own question here: $\gcd(c^a + 1, c^b + 1)$ for even $a$ and $b$?

Community
  • 1
Unapiedra
  • 220
  • There is something wrong with your table, because a gcd of two integers is never zero (except if both are 0). – gammatester Oct 15 '13 at 11:52
  • Read the paragraph above the table carefully: The table shows i such that i is the solution of t(i) = gcd(..., ...) and t(0) = 1. So wherever the value is 0 the gcd is 1. – Unapiedra Oct 15 '13 at 11:58

1 Answers1

2

I think you are having difficulty with this because of issues of indexing. The right way to look at the recurrence while keeping gcds in mind is to assign $T(1) = 1$ and $T(2) = 10$ in the recursive definition. Then you have the simple formula $$\gcd(T(n),T(m)) = T(\gcd(n,m))$$

in analogy to the formula for the Fibonacci numbers. Example 251 should be a good guide on how to prove this. This formula tells us that

$$\gcd(T(c^n),T(c^m)) = T(\gcd(c^n,c^m)) =T(c^{\min(n,m)})$$

If you'd still like a solution to your formulation we have to translate indexing to this one via $t(n) = T(n+1)$. Then we have

$$\gcd(t(c^n),t(c^m)) = \gcd(T(c^n+1),T(c^m+1)) = T(\gcd(c^n+1,c^m+1))$$

Using the relationship between $T$ and $t$, the latter becomes

$$ T(\gcd(c^n+1,c^m+1)) = t(\gcd(c^n+1,c^m+1)-1)$$

and finally

$$\gcd(t(c^n),t(c^m)) = t(\gcd(c^n+1,c^m+1)-1)$$

Now, any remaining difficulty will stem from trying to calculate the latter gcd. It's difficult even for $c = 2$.

EDIT: For odd $n$ and $m$ you have

$$\gcd(c^n+1, c^m+1) = c^{\gcd(n,m}+1$$

The matters get complicated once you have one or both of $n,m$ even.

Alexander Vlasev
  • 3,120