Re: Lucas 1874 Fibonacci as binomial sum generalization problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg114815] Re: Lucas 1874 Fibonacci as binomial sum generalization problem*From*: Roger Bagula <roger.bagula at gmail.com>*Date*: Fri, 17 Dec 2010 23:48:41 -0500 (EST)*References*: <iecqtn$bjn$1@smc.vnet.net>

I solved the polynomial the hard way: -1 - x^5 + x^6 I matched a vector Matrix Markov ( operational definition naming) to the sequence by successive solutions ( the end result has a different starting set of vector but the right end limiting ration): m = {{0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 1}} v[0] = Table[a[n], {n, 0, 5}] v[n_] := v[n] = m.v[n - 1] Table[v[n][[1]], {n, 0, 50}] Table[N[v[n + 1][[1]]/v[n][[1]]], {n, 11, 100}] CharacteristicPolynomial[m, x] It took me 5 approximate solutions before it settled down to this. There has to be a better way in Mathematica? It was the long set of zeros at the beginning that gave me the idea that they were like starting vectors in (k+1)x(k+1) matrices. So what is the k=7 with near 1.255422883905122 ratio? {0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 7, 8, 9, 10, 11, 12, 28, 35, 43, 52, 62, 73, 105, 133, 168, 211, 263, 325, 413} Clear[t, n, m, k, a] t[n_, m_, k_] = Binomial[n - k*(m - 1), m - 1] a[n_, k_] = Sum[Binomial[n - k*(m - 1), m - 1], {m, 1, Floor[n/k]}] Table[Table[a[n, k], {n, 0, 30}], {k, 1, 31}] TableForm[%] Table[Table[N[a[n + 1, k]/a[n, k]], {n, 0, 100}], {k, 1, 21}] 1-x^6+x^7 and 1-x+x^7 Both have too low a ratio. Roger Bagula m71 = {{0, 1, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 1}} CharacteristicPolynomial[m71, x] Table[x /. NSolve[CharacteristicPolynomial[m71, x] == 0, x][[i]], {i, 1, 6}] Abs[%] m72 = {{0, 1, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 0}} CharacteristicPolynomial[m72, x] Table[x /. NSolve[CharacteristicPolynomial[m72, x] == 0, x][[i]], {i, 1, 6}] Abs[%]