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by pbsd
207 days ago
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This can be translated to the discrete domain pretty easily, just like the NTT. Pick a sufficiently large prime with order 15k, say, p = 2^61-1. 37 generates the whole multiplicative group, and 37^((2^61-2)/3) and 37^((2^61-2)/5) are appropriate roots of unity. Putting it all together yields f(n) = 5226577487551039623 + 1537228672809129301*(1669582390241348315^n + 636260618972345635^n) + 3689348814741910322*(725554454131936870^n + 194643636704778390^n + 1781303817082419751^n + 1910184110508252890^n) mod (2^61-1).
This involves 6 exponentiations by n with constant bases. Because in fizzbuzz the inputs are sequential, one can further precompute c^(2^i) and c^(-2^i) and, having c^n, one can go to c^(n+1) in average 2 modular multiplications by multiplying the appropriate powers c^(+-2^i) corresponding to the flipped bits. |
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