How can we calculate $\displaystyle\sum\limits_{k=1}^{2^{16}} \binom{2k}{k}(3\times 2^{14} +1)^k (k-1)^{2^{16}-1}$ mod $(2^{16} +1)$? I am aware that $2^{16} +1$ is a prime.
Summation mod$ (2^{16} +1)$
4
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prime-numbers
modular-arithmetic