OFFSET

1,1

COMMENTS

Each term in a geometric progression must be an integer, but the ratio between two consecutive terms can be a rational number. This means that geometric progressions like (9,15,25) are allowed.

The difference between consecutive terms is at least 2, because we can always cover 2 extra numbers with a single geometric progression.

The original problem discussed in the links gives a(36) >= 100. In fact one cannot cover {1,2,...,101} with 36 geometric progressions, so a(36) = 100.

{1,2,...,1000} can be covered with 362 geometric progressions, so a(362) >= 1000.

{1,2,...,10000} can be covered with 3620 geometric progressions, so a(3620) >= 10000.

It seems that a(n) is approximately n*e.

Finding the smallest n for a given k is a set covering problem with a binary variable for each geometric progression and a constraint for each number from 1 to k. - Rob Pratt, Jul 23 2019

LINKS

Rob Pratt, Table of n, a(n) for n = 1..362

All-Russian Mathematical Olympiad 1995, Grade 11, problem 1

Dmitry Kamenetsky, Java helper program

Math Overflow, Covering a set with geometric progressions

Math StackExchange, Is it possible to cover {1,2,...,100} with 20 geometric progressions?

Math StackExchange, Cover {1,2,...,100} with minimum number of geometric progressions?

EXAMPLE

1 to 2 can be covered with 1 geometric progression: (1,2). So a(1) = 2. Note that we cannot cover 1 to 3 with 1 geometric progression.

1 to 5 can be covered with 2 geometric progressions: (1,2,4) and (3,5). So a(2) = 5. Note that we cannot cover 1 to 6 with 2 geometric progressions.

1 to 8 can be covered with 3 geometric progressions: (1,2,4,8), (3,5), (6,7). So a(3) = 8.

1 to 10 can be covered with 4 geometric progressions: (1,2,4,8), (1,3,9), (5,6), (7,10). So a(4) = 10.

1 to 13 can be covered with 5 geometric progressions: (1,2,4,8), (3,6,12), (5,7), (9,10), (11,13). So a(5) = 13.

1 to 16 can be covered with 6 geometric progressions: (1,2,4,8,16), (3,6,12), (5,7), (9,10), (11,13), (14,15). So a(6) = 16.

CROSSREFS

KEYWORD

nonn

AUTHOR

Dmitry Kamenetsky, Jul 12 2019

EXTENSIONS

a(37)-a(362) from Rob Pratt, Jul 23 2019

STATUS

approved