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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[%]


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