令p=q=n,得f2(n)=f(2n).
原式=
2f2(1) |
f(1) |
2f(4) |
f(3) |
2f(6) |
f(5) |
2f(8) |
f(7) |
=2f(1)+
2f(1)f(3) |
f(3) |
2f(1)f(5) |
f(5) |
2f(1)f(7) |
f(7) |
=8f(1)=24.
故答案为:24.
f2(1)+f(2) |
f(1) |
f2(2)+f(4) |
f(3) |
f2(3)+f(6) |
f(5) |
f2(4)+f(8) |
f(7) |
2f2(1) |
f(1) |
2f(4) |
f(3) |
2f(6) |
f(5) |
2f(8) |
f(7) |
2f(1)f(3) |
f(3) |
2f(1)f(5) |
f(5) |
2f(1)f(7) |
f(7) |