For given $A, B \in Z$, consider the sequences
$P(n) = P(n-2) + P(n-1), P(0) = 1, P(1) = 0,$
$Q(n) = Q(n-2) + Q(n-1), Q(0) = 0, Q(1) = 1,$
and the cumulative sequences $\Sigma P(n), \Sigma Q(n), A.\Sigma P(n) + B.\Sigma Q(n)$
and the $GCD$ sequence defined by $GCD(\Sigma P(n) + \Sigma Q(n))$.
$$\array{ A.P(n)+B.Q(n) & n & P(n) & Q(n) & ΣP(n) & ΣQ(n) & GCD(ΣP(n), ΣQ(n)) \\ A & 0 & 1 & 0 & 1 & 0 & 1 \\ B & 1 & 0 & 1 & 1 & 1 & 1 \\ A + B & 2 & 1 & 1 & 2 & 2 & 2 \\ A + 2B & 3 & 1 & 2 & 3 & 4 & 1 \\ 2A + 3B & 4 & 2 & 3 & 5 & 7 & 1 \\ 3A + 5B & 5 & 3 & 5 & 8 & 12 & 4 \\ 5A + 8B & 6 & 5 & 8 & 13 & 20 & 1 \\ 8A +13B & 7 & 8 & 13 & 21 & 33 & 3 \\ 13A +21B & 8 & 13 & 21 & 34 & 54 & 2 \\ 21A +34B & 9 & 21 & 34 & 55 & 88 & 11 \\ \dots & \dots & \dots& \dots& \dots& \dots& \dots \\ }$$
For $n \in Z, z > 1, z \in Z_{+}$ consider the equation,
$$A.\Sigma P(n) + B.\Sigma Q(n) = z$$
When $GCD(\Sigma P(n), \Sigma Q(n)) = 1$, we can determine the minimal $A, B$ using the Euclidean Algorithm.
Questions:
- When $GCD(\Sigma P(n), \Sigma Q(n)) = 1$, is $\log(z) \gt \log(n) + \log(A) + \log(B)$ for any $n>1, z \in Z$?
- When $GCD(\Sigma P(n), \Sigma Q(n)) = g \ne 1$ and $g | z$, is $\log(z) \gt \log(g) + \log(n) + \log(A) + \log(B)$ for any $n>1, z \in Z$. (Note: $g | z$ is required for the equation $A.\Sigma P(n) + B.\Sigma Q(n) = z$ to have a solution in integers)
Given a natural number $n>1$ and integer $z$, and integers $A$ and $B$ such that $z=A\Sigma P(n)+B\Sigma Q(n)$, the inequality $$\log z>\log n+\log A+\log B,$$ is equivalent to $z>nAB$, or equivalently $$(\Sigma P(n)- nB)(\Sigma Q(n)-nA)<\Sigma P(n)\Sigma Q(n).$$ Clearly $\Sigma Q(n)>n$ for $n\geq3$ and so for $A=1$ and any $B>\tfrac{\Sigma P(n)}{n}$ we have $$(\Sigma P(n)- nB)(\Sigma Q(n)-nA)=(\Sigma P(n)-nB)(\Sigma Q(n)-n)<0<\Sigma P(n)\Sigma Q(n).$$ So for any integer $n\geq3$ you can take any integer $B>\tfrac{\Sigma P(n)}{n}$ and take $$z=\Sigma P(n)+B\Sigma Q(n).$$ Note that the pair of coefficients $(A,B)=(1,B)$ has $|A|+|B|$ minimal because $\Sigma Q(n)\geq\Sigma P(n)$.