The theorem prover Walnut (see link) can compute the following "linear representation" for a(n). Let v = [1,0,0,0,0,0,0,0,0,0,0,0], w = [0,1,1,3,2,1,5,3,8,5,2,9]^T, mu(0) =[[1,0,0,0,0,0,0,0,0,0,0,0], [0,0,1,0,0,0,0,0,0,0,0,0], [0,0,0,1,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,0,0,0,0,0,0], [0,0,0,0,0,0,0,1,0,0,0,0], [0,0,0,0,0,0,0,0,1,0,0,0], [0,0,0,0,0,0,0,0,0,0,1,0], [0,0,0,0,0,0,0,0,0,0,0,1], [0,0,0,0,0,0,0,-1,0,0,2,0], [0,0,-2,0,0,1,0,2,0,0,0,0], [0,0,-1,-1,0,1,0,1,1,0,-1,1], [0,0,-1,0,0,0,0,0,0,0,2,0]], mu(1) = [[0,1,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,1,0,0,0,0,0,0,0], [0,0,0,0,0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,1,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,-1,0,0,2,0,0], [0,-2,0,0,1,0,2,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0], [0,-1,0,0,0,0,0,0,0,2,0,0], [0,-4,0,0,2,0,4,0,0,-1,0,0]]. Then a(n) = v.mu(x).w, where x is the Zeckendorf (or Fibonacci) representation of n. This gives an algorithm for a(n) that runs in polynomial time in log n. -
Jeffrey Shallit, Aug 09 2025
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