A354914 The least cost to reach n using additions and multiplications, where multiplication is free.

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

Author

Stan Wagon, Jun 11 2022

Keywords

• nonn

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.

Examples

			For n = 23, the least cost a(23) is 4, via the sequence 1, 2, 3, 4, 8, 16, 19, 23.
		

Links

• Stan Wagon, Table of n, a(n) for n = 1..5000
• H. M. Bahig, On a generalization of addition chains: Addition-multiplication chains, Discrete Mathematics 308 (2008), 611-616.
• Stan Wagon, Optimal paths for n up to 100

Crossrefs

Cf. A355015, A230697.