Luca's numbers....

The easy way to solve this is $\color{#3D99F6}{COMPUTE}$ the numbers manually, we see that they will be

$2,1,3,4,7,11,18,29,47,76,123$ . (Observe that $L_{10} = 123$ , that's looking a good number :P 123 )

The sum of all these is $\displaystyle \sum_{n=0} ^{10} L_n = 2+1+3+4+7+11+18+29+47+76+123 = \boxed{321}$

The reason why I chose the number $10$ in the problem is that $L_{10}$ is $\color{#20A900}{123}$ and sum of terms till $L_{10}$ is $\color{#20A900}{321 }$ , mirror images ! That's a coincidence ....

There is an easy way of solve this which is just noting the pattern of the summation.

$S = L_0 + L_1 + L_2 + L_3 + L_4 + L_5 + L_6 + L_7 + L_8 + L_9 + L_{10}$ sum every pair using the recurrence $S = L_2 + L_4 + L_6 + L_8 + L_{10} + L_{10}$ add $L_1$ to both sides

$S + L_1 = L_1 + L_2 + L_4 + L_6 + L_8 + L_{10} + L_{10}$

use the recurrence again repeatedly until completely simplified

$S + L_1 = L_3+ L_4 + L_6 + L_8 + L_{10} + L_{10}$ $S + L_1 = L_5 + L_6 + L_8 + L_{10} + L_{10}$ $S + L_1 =L_7 + L_8 + L_{10} + L_{10}$ $S + L_1 = L_9+ L_{10} + L_{10}$ $S + L_1 = L_{11} + L_{10}$ $S + L_1 = L_{12}$ $S = L_{12} - L_1$

Giving us $S=322-1 = 321$

Using dynamic programming:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

memo = {0:2,1:1}
def luca(n):
    if n in memo:
        return memo[n]
    memo[n] = luca(n - 1) + luca(n - 2)
    return memo[n]
total = 0
for n in xrange(11):
    total += luca(n)
print total