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Given $a,b\in\mathbb Z^+$, and let $F_{a,b}:\mathbb N\to\mathbb N$ be a function such that $F_{a,b}(0)=0$ and $F_{a,b}(n+1)=a\cdot F_{a,b}(n)+b\cdot F_{a,b}(n-1)$. $F_{1,1}$ correspond to the Fibonacci function if $F_{1,1}(1)=1$.

Conjecture which I would like to see proved:

If $a,b$ are co-prime and $F_{a,b}(1)\in \mathbb Z^+$, then $F_{a,b}(\gcd(m,n))=\gcd(F_{a,b}(n),F_{a,b}(m))$

I guess that some formulation of the conjecture is true for all homogeneous linear recurrence relations with constant coefficients of degree > 1, but I have only (randomly) tested it for degree 2.

See also A generalization of a divisibility relation for Fibonacci numbers

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    I think you need two initial values, not just one, for this type of recurrence. But maybe in your question you mean to just allow $F_{a,b}(1)$ to be some specific natural, is that it? – 2017-02-15
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    @coffeemath. Yes, $F_{a,b}(1)\in\mathbb Z^+$ is required. – 2017-02-15
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    Have you taken a look at the literature on strong divisibility sequences? I suspect that this is contained in there... somewhere. For example, Theorem 0 (i) from Wayne L. McDaniel, *The g.c.d. in Lucas sequences and Lehmer number sequences* ( https://pdfs.semanticscholar.org/44cc/9285b47e5bfe9f9c91e24e6699dd7a552beb.pdf ) seems to cover the cases where the congruence has nonzero discriminant. – 2017-02-15
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    @darijgrinberg. Yes, I think you are right. Thanks! – 2017-02-15

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