In another question, posted here by jordan, we are asked whether it is possible to cover the numbers $\{1,2,\ldots,100\}$ with $20$ geometric sequences of real numbers. Naturally, we would like to extend the question:
Problem: What is the minimum number $n$ of geometric progressions $A_1, A_2,\ldots,A_{n}$ of rational* numbers such that $$ \{1,2,\ldots,100\} \subseteq A_1 \cup A_2\cup \ldots\cup A_{n}? $$
In the other question, 6005 obtained a lower bound of $31 \leq n$ with an argument about square free integers. We can also obtain an upper bound of $43$ as follows. Consider these $5$ sequences: $[1, 2, 4, 8, 16, 32, 64]$ and $[6, 12, 24, 48, 96]$ and $[5, 10, 20, 40, 80]$ and $[3, 9, 27, 81]$ and $[7, 21, 63]$. Together, these cover $7 + 5 + 5 + 4 + 3 = 24$ terms. The remaining $76$ terms can be covered in at most $38$ sequences, by an argument made here. So we have the bound:
$$31 \leq n \leq 43$$
Can anyone do better?
*We need only consider rational ratios by arguments made in answers to the original question.
(Update) We have a winner!! Thanks to the cumulative efforts of the answerers below, we have arrived at $n = 36$. The upper bound is thanks to jpvee, and the lower bound is due to san. Hooray!
The exact value of n is 36:
First, consider all progressions of length 4 or more:
Length 7: 1,2,4,8,16,32,64
Length 6: 3,6,12,24,48,96
Length 5: 5,10,20,40,80$\qquad \ \ $ 1,3,9,27,81$\qquad \ \ $ 16,24,36,54,81
Length 4: 2,6,18,54$\qquad \ \ $ 27,36,48,64$\qquad \ \ $ 7,14,28,56$\qquad \ \ $ 9,18,36,72$\qquad \ \ $ 11,22,44,88$\qquad \ \ $ 8,12,18,27
These progressions cover the 33 numbers
1,2,3,4,5,6,7,8,9,10,11,12,14,16,18,20,22,24,27,28,32,36,40,44,48,54,56,64,72,80,81,88,96
Moreover, two of the length 5 geometric progressions have only 3 numbers disjoint from the progressions of length 6 and 7. Hence, the best way to cover these 33 numbers is to cover 30 of them with 6 geometric progressions:
The progressions of length 7, 6 one of 5 and 3 progressions of length 4:
Length 7: 1,2,4,8,16,32,64
Length 6: 3,6,12,24,48,96
Length 5: 5,10,20,40,80
Length 4: 7,14,28,56$\qquad \ \ $ 9,18,36,72$\qquad \ \ $ 11,22,44,88
There are 35 numbers which are not in a geometric progression of length three or more:
29,31,37,39,41,43,47,51,53,55,57,58,59,61,62,65,67,69, 71,73,74,77,78,79,82,83,85,86,87,89,91,93,94,95,97
So you need 6 progressions to cover 30 numbers, 18 progressions to cover 35 numbers (in fact you cover 36), and the remaining 35 numbers must be covered with at least 12 progressions of length 3 (even if you covered 36 numbers with the 18 progressions in the item before you need 12 to cover 34 numbers).
So you need at least 36 progressions. This bound is achieved in the answer of jpvee, which is therefore optimal. Since the method of counting which numbers are covered by certain progressions is also of jpvee, and my only contribution was to count how many numbers are covered by progressions of length 4 or more, the bounty should be awarded to jpvee.