1,3

a(n) appears to equal n-1 for n not a multiple of 3.

The matrix entries of M^n, with n >= 0, are M^n(1, 1) = 2^(n-1)*F(n+3) = A063782(n), M^n(2, 2) = 2^(n-1)*F(n-3) = A319196(n), M^n(1, 2) = M^n(2, 1) = 2^(n-1)*F(n) = A085449(n), where i = sqrt(-1), F = A000045, and F(-1) = 1, F(-2) = -1, F(-3) = 2. Proof by Cayley-Hamilton, with S(n, -i) = (-i)^n*F(n+1), where S(n, x) is given in A049310. - Wolfdieter Lang, Oct 08 2018

The above conjecture is true. From the preceding formulas for the elements of M^n this claims that the Fibonacci numbers F(n-3), F(n) and F(n+3) are always odd for n == 1 or 2 (mod 3). This is true because F(n) is even iff n == 0 (mod 3) (see e.g. Vajda, p.73), and each of the three indices is == 1 or 2 (mod 3) for n == 1 or 2 (mod 3), respectively. - Wolfdieter Lang, Oct 09 2018

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 73.

For n = 3, the matrix {{3,1},{1,-1}}^3 = {{32,8},{8,0}} and the largest power of 2 appearing in the factorization of any entry is 2^5 = 32. Hence, a(3) = 5.

Join[{0, 1, 5}, Table[Max[ IntegerExponent[Flatten[MatrixPower[{{3, 1}, {1, -1}}, n]], 2]], {n, 4, 40}]]

(PARI) a(n) = vecmax(apply(x->if (x, valuation(x, 2), 0), [3, 1; 1, -1]^n)); \\ Michel Marcus, Sep 09 2018

Cf. A000045, A049310, A063782, A085449, A130481, A319196.

Sequence in context: A004494 A193090 A004162 * A109681 A196406 A070367

Adjacent sequences: A319050 A319051 A319052 * A319054 A319055 A319056

Greg Dresden, Sep 09 2018

approved