I'm writing a program to compute a value of periodic function for any arbitrary large argument:
$f(k) = (\sum_{n=1}^{2^k} n)\mod\ (10^9 + 7)$, where $n,k \in \mathbb{N} $
I know that $ f(k + P) = f(k) $, which limits the value of $k$ I would have to compute.
What confuses me is that Wolfram Alpha says that it have a period of $ \frac {2{i}\pi}{\log(2)} $. Is it even possible to program that period?
Since your function is actually $f(k) = (2^{2k-1} + 2^{k-1})\pmod{10^9+7}$, it suffices to tabulate values of $2^k$.
$10^9+7$ is prime, so by Fermat's theorem, we know that $2^{10^9+6}\equiv 1\pmod{10^9+7}$.
So suppose you want to calculate $f(k)$. Calculating powers is expensive, so we will calculate $z=2^{k-1}\pmod{10^9+7}$ and then $f(k) = z\cdot(2z+1) \pmod{10^9+7}$.
First calculate $k' = k-1\pmod{10^9+6}$. Then calculate $2^{k-1} \equiv 2^{k'}\pmod{10^9+7}$ using the usual fast recursive exponentiation algorithm:
pow2(k):
result = 1
power = 1
while k > 0:
power = (power * 2) mod 1000000007
if k is odd, result = (result * power) mod 1000000007
k = int(k/2)
return result
(In C, use k >>= 1 instead of k=int(k/2), power <<= 1, etc.)
We have $k < 10^9 + 7$, so this will iterate at most about 30 times. If this is too slow, you can cache the results for $0 ≤ k ≤ 10^6$ and then it will iterate at most ten times per call.
Let $z = \mathrm{pow2}(k-1\pmod{10^9+6})$, and then your answer is $(2z+1)\cdot z\pmod{(10^9+6}$.
Wolfram|Alpha appears to have gone temporarily insane.
