Solving Project Euler problem #258 using PARI/GP

I first encountered PARI/GP when lumyk posted a solution for Project Euler Problem 258:

“A sequence is defined as:

  • gk = 1, for 0 ≤ k ≤ 1999
  • gkgk-2000gk-1999, for k ≥ 2000.

Find gk mod 20092010 for k = 1018.”

One solution was using matrix multiplication in 19 hours (!!!) and I was curious if lumyk’s claim that PARI/GP could solve it in 2s (!!!) was true.

so I tried it.

Solving Project Euler Problem #258 using PARI/GP

yup. 2 seconds. Now, that’s very, very fast. Once you’ve solved the problem, you can check out the forum for other algorithms for this lagged Fibonacci problem.


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