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Re: Fibonacci based sum that is b-normal on binary numbers

Roger Bagula wrote:
> This sum and it's b-normal sequence is due
>  to work of a friend who doesn't like me to use his name here or elsewhere
> ..He came up with two very nice sums using Fibonacci numbers.
> I used the Binet  function in them and got very good agreement.
> So I tried them in a b-normal.
> I had to modify the result some to get this result.
> I get a new sum that appears irrational
> and an iteration that is b-normal .
> I think that using the Binet function in this makes it
> a new sequence sum.
> I thought that this was a very remarkable result.
> Clear[x,a,digits,f,fib]
> (* convergent sum based on Fibonacci sequence to make a binary b-normal 
> iteration *)
> digits=200
> fib[n_Integer?Positive] :=fib[n] =    fib[n-1]+fib[n-2]
> fib[0]=0;fib[1] = fib[2] = 1;
> sfib=Sum[fib[n]/((n+1)*2^(n+1)),{n,0,digits}]
> N[sfib,digits]
> x[n_]:=x[n]=Mod[2*x[n-1]+fib[n-1]/(2*n),1]
>     x[0]=0
> a=Table[N[x[n],digits],{n,0,digits}]
> ListPlot[a,PlotJoined->True,PlotRange->All]
> b=Sort[Table[N[x[n],digits],{n,0,digits}]];
> ListPlot[b,PlotJoined->True,PlotRange->All]   
> Respectfully, Roger L. Bagula
> tftn at, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
> alternative email: rlbtftn at
> URL :
...just another way to compute

sfib = 1/10*(5*Log[4] + Sqrt[5]*Log[1/2*(7 - 3*Sqrt[5])])

Peter Pein

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