h3. Explicaţie

In primul exemplu, numarul 6 se descompune astfel:

Numarul $6$ se descompune in $2$ termeni astfel:

Intr-un termen

* [1,6], cost=gcd([1]) + gcd([1,6])=1+1=2
* [6,1], cost=gcd([6]) + gcd([6,1])=6+1=7
* [2,3], cost=gcd([2]) + gcd([2,3])=2+1=3
* [3,2], cost=gcd([3]) + gcd([3,2])=3+1=4

* %{color:black}$[6$%$]$
 
In $2$ termeni
 
* $[1, 6]$
* $[6, 1]$
* $[2, 3]$
* $[3, 2]$
 
In $3$ termeni
 
* $[1, 1, 6]$
* $[1, 2, 3]$
* $[1, 3, 2]$
* $[1, 6, 1]$
* $[2, 1, 3]$
* $[2, 3, 1]$
* $[3, 1, 2]$
* $[3, 2, 1]$
* $[6, 1, 1]$

Deci $f(6,2)=2+7+3+4=16$

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