0,6

A proof/ derivation: The Fibonacci Markov Matrices can be defined as: M[0]={{F[0],F[1]},{F[1],F[2}} M[1]={{F[1],F[2]},{F[2],F[3}} ... M[n]={{F[n],F[n+1]},{F[n+1],F[n+2}} And Det[M[n]]=1 So that you can form the Poincare bilinear/ Moebius function: f(z)=(F[n]*x+F[n+1])/(F[n+1]*x+F(n+2])

Now the 2*m modular group function is known to be: M[(F[n]*x+F[n+1])/(F[n+1]*x+F(n+2])]=(F[n+1]*x+F(n+2])^(2*m)*M[x] It is also a fact that if M is an elliptical invariant that: M[(F[n]*x+F[n+1])/(F[n+1]*x+F(n+2])]=M[x] Or equivalently: (F[n+1]*x+F(n+2])^(2*m)-1=0 Expanding F[n+2]=F[n]+F[n+1] And dividing through by F[n+1]^(2*m): (F[n+1]*(x+1)+F[n])^(2*m)-1=0 ((x+1-F[n]/F[n+1])^(2*m)-1/F[n+1]^(2*m]=0 That gives the limit: Limit[((x+1-F[n]/F[n+1])^(2*m)-1/F[n+1]^(2*m],n->Infinity]=(x+1+1/Phi)=0 xLimitRoot=-1-1/Phi

So not only can we form a Modular group in SPL(2,Z) using the Fibonacci numbers, but we can find solutions for all the points up to a limiting point at infinity. And the 2*m polynomial start with the Pascal's triangle like : (x+1)^(2*m)-1=0 If we set: y=F[n+1]*(x+1)+F[n]= F[n+1]*x+F(n+2]

The roots of y^(2*m)-1=0 are the even cyclotomic group roots: yroot[k,m]=Exp[i*2*Pi*k/(2*m)]=F[n+1]*(x+1)+F[n] Solving for x gives general solution: xroot[k,m,n]=(Exp[i*2*Pi*k/(2*m)]-F[n])/F[n+1]-1

The modular form of weight 2*m defined: McKean and Moll, Ellipic Curves, Function Theory,Geometry, Arithmetic, Cambridge University Press, New York, 199, page 172

The elliptical invariant definition is in: Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986, page 124

F[n_] := F[n] = F[n - 1] + F[n - 2] xroot[k_, m_, n_] := (Exp[I*2*Pi*k/(2*m)] - F[n])/F[n + 1] - 1 a(n,m) = Abs[F[n+1]*xroot[2*m, m, n]]

{0},

{0, 1},

{0, 1, 2},

{0, 1, 2, 4},

{0, 1, 2, 4, 7},

{0, 1, 2, 4, 7, 12}

F[0] = 0; F[1] = 1; F[n_] := F[n] = F[n - 1] + F[n - 2] a = Table[F[n], {n, 0, 50}]; xroot[k_, m_, n_] := (Exp[I*2*Pi*k/(2*m)] - a[[n]])/a[[n + 1]] - 1 aout = Table[Table[Abs[a[[n + 1]]*xroot[2*m, m, n]], {n, 1, m}], {m, 1, 12}] b = Flatten[aout]

Cf. A000071, A000045.

Sequence in context: A023858 A011118 A131644 this_sequence A140531 A117316 A109189

Adjacent sequences: A115343 A115344 A115345 this_sequence A115347 A115348 A115349

nonn,uned

Roger Bagula (rlbagulatftn(AT)yahoo.com), Mar 07 2006