functional equations/solving a functional equation

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[math]

very well explained, thanks for sharing

[b.]

I got 22. I did the following:
1. f(12)=f(2*6)
2. f(2*6)=6+f(10)
3. f(10)=f(2*5)=5+f(8)
4. f(8)=f(2*4)=4+f(6)
5. f(6)=f(2*3)=3+f(4)
6. f(4)=f(2*2)=2+f(2)
7. f(2)=2
8. f(4)=2+2=4
9. f(6)=3+4=7
10. f(8)=4+7=11
11. f(10)=5+11=16
12. f(12)=6+16=22

[wesleysuen]

The “n=1” part is weird unless you include it in the equation: f(2n)=1 for n=1.

[damiennortier]

For f(12), we need f(10), f(8), (6), f(4), f(2) so f(12) = 6 + 5 + 4 + 3 + 2 + 1 = 6*7/2 that the general formul
F(x) = x/2*(x+1)/2/2 = (x+1)*x/8