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A354914
The least cost to reach n using additions and multiplications, where multiplication is free.
3
0, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 3, 1, 2, 2, 3, 2, 3, 3, 4, 2, 2, 3, 2, 3, 3, 3, 3, 1, 2, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 4, 4, 2, 3, 2, 3, 3, 4, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 3, 3, 4, 3, 4, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 4, 2, 3, 3, 3, 2
OFFSET
1,3
COMMENTS
Start with 1. Apply multiplication or addition to any values (not necessarily distinct) already attained to get a finite sequence of integers ending in n. The cost of addition is one unit, but multiplication is free. Then a(n) is the cost of the least expensive path to n.
The problem is folklore. It is not hard to prove that the cost function is unbounded. The values given were produced by Joseph DeVincentis, Stan Wagon, and Al Zimmermann.
LINKS
EXAMPLE
For n = 23, the least cost a(23) is 4, via the sequence 1, 2, 3, 4, 8, 16, 19, 23.
CROSSREFS
Sequence in context: A072086 A180094 A333870 * A366927 A103748 A104231
KEYWORD
nonn,hard
AUTHOR
Stan Wagon, Jun 11 2022
STATUS
approved