# A sequence that avoids both arithmetic and geometric progressions

Sequences that avoid arithmetic progressions have been studied, e.g., “Sequences Containing No 3-Term Arithmetic Progressions,” Janusz Dybizbański, 2012, journal link.

I started to explore sequences that avoid both arithmetic and geometric progressions,
i.e., avoid $$x,x+c,x+2cx, x+c, x+2c$$ and avoid $$y,cy,c2yy, c y, c^2 y$$ anywhere in the sequence
(not necessarily consecutively).
Starting with $$(1,2)(1,2)$$, one cannot extend with $$33$$ because $$(1,2,3)(1,2,3)$$ forms an
arithemtical progression, and one cannot extend with $$44$$ because $$(1,2,4)(1,2,4)$$ is a geometric
progression. But $$(1,2,5)(1,2,5)$$ is fine.

Continuing in the same manner leads to the following “greedy” sequence:
$$1,2,5,6,12,13,15,16,32,33,35,39,40,42,56,81,84,85,88,1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88,$$
$$90,93,94,108,109,113,115,116,159,189,207,208,222,…90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, \ldots$$

This sequence is not in the OEIS.
Here are a few questions:

Q1. What is its growth rate?

Q2. Does $$∑∞i=11/si\sum_{i=1}^\infty 1/s_i$$ converge? (Where $$sis_i$$ is the $$ii$$-th term of the above
sequence.)

Q3. If it does, does it converge to e?
Update: No. The sum appears to be approximately $$2.73>e2.73 > e$$, as per
@MichaelStocker and @Turambar.

That is wild numerical speculation. The first 457 terms (the extent
of the graph above) sum to 2.70261.

Addendum. 11Jul2014. Starting with $$(0,1)(0,1)$$ rather than $$(1,2)(1,2)$$ renders
a direct hit on OEIS A225571.

$$HINT\color{brown}{\textbf{HINT}}$$

Denote the target sequence $${F3(n)}\{F_3(n)\}$$ and let us try to estimate the probability $$P(N)}P(N)\}$$ that natural number belongs to $${F3}.\{F_3\}.$$

Suppose
$$F3(1)=1,F3(2)=2,P(1)=P(2)=P(5)=P(6)=1,P(3)=P(4)=P(7)=P(8)=0,F_3(1)=1,\quad F_3(2)=2,\quad P(1)=P(2)=P(5) = P(6) = 1,\\ P(3)=P(4)=P(7)=P(8)=0,\tag1$$
$$V(N)=N∑i=1P(i),F3(N)=[V−1(N)].V(N)=\sum\limits_{i=1}^{N}P(i),\quad F_3(N) = \left[V^{-1}(N)\right].\tag2$$

Let $$Pa(N)P_a(N)$$ is the probability that N does not belong to arithmetic progression and $$Pg(N)P_g(N)$$ is the similar probability for geometric progressions.

Suppose
$$P(N)=Pa(N)Pg(N)).P(N) = P_a(N)P_g(N)).\tag3$$

$$Arithmetic Probability estimation.\color{brown}{\textbf{Arithmetic Probability estimation.}}$$

Suppose
$$Pa(N)=[N/2]∏k=1Pa(N,k),P_a(N)=\prod\limits_{k=1}^{[N/2]}P_a(N,k),\tag4$$
where $$Pa(N,k)P_a(N,k)$$ is the probability that arithmetic progression $${N−2k,N−k,N}\{N-2k,N-k, N\}$$ does not exist for any $$j.j.$$
Suppose
$$Pa(N,k)=(1−P(N−2k))((1−P(N−k)).P_a(N,k) = \big(1-P(N-2k)\big)\big((1-P(N-k)\big).\tag5$$

$$Geometric Probability estimation.\color{brown}{\textbf{Geometric Probability estimation.}}$$

Suppose
$$Pg(N)=[√Nn]∏k=1Pg(N,k),P_g(N)=\prod\limits_{k=1}^{\left[\,\sqrt Nn\,\right]}P_g(N,k),\tag6$$
where $$Pg(N,k)P_g(N,k)$$ is the probability that geometric progression $$(Nk2,Nk,N}\left(\dfrac{N}{k^2}, \dfrac Nk, N\right\}$$ with the denominator $$kk$$ does not exist for all $$i,j.i,j.$$

Taking in account that the geometric progression can exist only if $$k2| N,k^2\,| \ N,$$ suppose
$$Pg(N,k)=(1−1k2P(Nk2))(1−P(Nk)).P_g(N,k) = \left(1-\dfrac1{k^2}P\left(\dfrac{N}{k^2}\right)\right)\left(1-P\left(\dfrac{N}{k}\right)\right).\tag7$$

$$Common model.\color{brown}{\textbf{Common model.}}$$

Common model can be simplified to next one,
$${P(N)=1−[N/2−1]∏k=1(1−(1−P(N−2k))(1−P(N−k)))×[√N]∏k=1(1−(1−1k2P(Nk2))(1−P(Nk)))P(1)=P(2)=P(5)=P(6)=1,P(3)=P(4)=P(7)=P(8)=0V(N)=N∑i=1P(i),F3(n)=[V−1(n)].\begin{cases} P(N) = 1-\prod\limits_{k=1}^{[N/2-1]}\Big(1-\big(1-P(N-2k)\big)\big(1-P(N-k)\big)\Big)\\ \times\prod\limits_{k=1}^{\left[\sqrt N\right]}\left(1-\left(1-\dfrac1{k^2}P\left(\dfrac{N}{k^2}\right)\right)\left(1-P\left(\dfrac{N}{k}\right)\right)\right)\\[4pt] P(1)=P(2)=P(5) = P(6) = 1,\quad P(3)=P(4)=P(7)=P(8)=0\\ V(N)=\sum\limits_{i=1}^{N}P(i),\quad F_3(n) = \left[V^{-1}(n)\right].\tag8 \end{cases}$$

Looking the solution in the form of
{P(N)=Pv(N),wherev=[N−1mod42],P0(N)={1,ifN<9cN−s,otherwiseP1(N)={0,ifN<9dN−t,otherwise\left\{\begin{align} &P(N)=P_v(N),\quad\text{where}\quad v=\left[\dfrac{N-1 \mod 4}2\right],\\[4pt] &P_0(N)= \begin{cases} 1,\quad\text{if}\quad N<9\\ cN^{-s},\quad\text{otherwise} \end{cases}\\[4pt] &P_1(N)= \begin{cases} 0,\quad\text{if}\quad N<9\\ dN^{-t},\quad\text{otherwise} \end{cases} \end{align}\right.\tag9

then
$$\begin{cases} &P_0(N) =1-&\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_0(N-4k)\big)\big(1-P_1(N-2k)\big)\Big)\\ &&\times\Big(1-\big(1-P_1(N-4k-2)\big)\big(1-P_0(N-2k-1)\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}P_0\left(\dfrac{N}{(2k-1)^2}\right)\right) \left(1-P_1\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}P_1\left(\dfrac{N}{4k^2}\right)\right) \left(1-P_0\left(\dfrac{N}{2k}\right)\right)\right)\\[4pt] &P_1(N) = 1- &\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_1(N-4k)\big)\big(1-P_0(N-2k)\big)\Big)\\ &&\Big(1-\big(1-P_0(N-4k-2)\big)\big(1-P_1(N-2k-1)\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}P_1\left(\dfrac{N}{(2k-1)^2}\right)\right) \left(1-P_0\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}P_0\left(\dfrac{N}{4k^2}\right)\right) \left(1-P_1\left(\dfrac{N}{2k}\right)\right)\right). \end{cases}\tag{10}\begin{cases} &P_0(N) =1-&\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_0(N-4k)\big)\big(1-P_1(N-2k)\big)\Big)\\ &&\times\Big(1-\big(1-P_1(N-4k-2)\big)\big(1-P_0(N-2k-1)\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}P_0\left(\dfrac{N}{(2k-1)^2}\right)\right) \left(1-P_1\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}P_1\left(\dfrac{N}{4k^2}\right)\right) \left(1-P_0\left(\dfrac{N}{2k}\right)\right)\right)\\[4pt] &P_1(N) = 1- &\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_1(N-4k)\big)\big(1-P_0(N-2k)\big)\Big)\\ &&\Big(1-\big(1-P_0(N-4k-2)\big)\big(1-P_1(N-2k-1)\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}P_1\left(\dfrac{N}{(2k-1)^2}\right)\right) \left(1-P_0\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}P_0\left(\dfrac{N}{4k^2}\right)\right) \left(1-P_1\left(\dfrac{N}{2k}\right)\right)\right). \end{cases}\tag{10}$$

Taking in account $$(9),(9),$$ can be written
$$\begin{cases} &P_0(N) =1-&\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k-1)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-c(N-4k)^{-s}\big)\big(1-d(N-2k)^{-t}\big)\Big)\\ &&\times\Big(1-\big(1-d(N-4k-2)^{-t}\big)\big(1-c(N-2k-1)^{-s}\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}c\left(\dfrac{N}{(2k-1)^2}\right)^{-s}\right) \left(1-d\left(\dfrac{N}{2k-1}\right)^{-t}\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}d\left(\dfrac{N}{4k^2}\right)^{-t}\right) \left(1-c\left(\dfrac{N}{2k}\right)^{-s}\right)\right)\\[4pt] &P_1(N) = 1- &\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-d(N-4k)^{-t}\big)\big(1-(N-2k)^{-s}\big)\Big)\\ &&\Big(1-\big(1-(N-4k-2)^{-s}\big)\big(1-d(N-2k-1)^{-t}\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}d\left(\dfrac{N}{(2k-1)^2}\right)^{-t}\right) \left(1-c\left(\dfrac{N}{2k-1}\right)^{-s}\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}c\left(\dfrac{N}{4k^2}\right)^{-s}\right) \left(1-d\left(\dfrac{N}{2k}\right)^{-t}\right)\right). \end{cases}\tag{11}\begin{cases} &P_0(N) =1-&\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k-1)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-c(N-4k)^{-s}\big)\big(1-d(N-2k)^{-t}\big)\Big)\\ &&\times\Big(1-\big(1-d(N-4k-2)^{-t}\big)\big(1-c(N-2k-1)^{-s}\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}c\left(\dfrac{N}{(2k-1)^2}\right)^{-s}\right) \left(1-d\left(\dfrac{N}{2k-1}\right)^{-t}\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}d\left(\dfrac{N}{4k^2}\right)^{-t}\right) \left(1-c\left(\dfrac{N}{2k}\right)^{-s}\right)\right)\\[4pt] &P_1(N) = 1- &\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-d(N-4k)^{-t}\big)\big(1-(N-2k)^{-s}\big)\Big)\\ &&\Big(1-\big(1-(N-4k-2)^{-s}\big)\big(1-d(N-2k-1)^{-t}\big)\Big)\\ &&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1} \left(1-\left(1-\dfrac1{(2k-1)^2}d\left(\dfrac{N}{(2k-1)^2}\right)^{-t}\right) \left(1-c\left(\dfrac{N}{2k-1}\right)^{-s}\right)\right)\\[4pt] &&\times\left(1-\left(1-\dfrac1{4k^2}c\left(\dfrac{N}{4k^2}\right)^{-s}\right) \left(1-d\left(\dfrac{N}{2k}\right)^{-t}\right)\right). \end{cases}\tag{11}$$

Model $$(11)(11)$$ should be checked theoretically and practically, but it gives the approach to the required estimations.

The next steps are estimation of parameters $$c,d,s,tc,d,s,t$$ and using of obtained model.