Mally, it appears that you are using the inverse of the ratio. For Fibonacci numbers, the ratio is normally F[sub]m+1[/sub]/F[sub]m[/sub]. In these generalized sequences R[sub]n,m[/sub] which is the sum of the previous n numbers, the ratio I considered is R[sub]n,m+1[/sub]/R[sub]n,m[/sub].

So by definition the ratio is always greater than 1. From the formulas I've shown above, it cannot be greater or equal to 2.

Some examples that show the correct ratio and the value found using the formula I developed:

[code]
n=2 ratio: 1.618033988749895 formula: 1.66015625
n=3 ratio: 1.839286755214161 formula: 1.84423828125
n=4 ratio: 1.927561975482925 formula: 1.9281005859375
n=5 ratio: 1.965948236645485 formula: 1.96600341796875
n=10 ratio: 1.999018632710101 formula: 1.999018633039668
n=15 ratio: 1.999969475434503 formula: 1.999969475434504
n=20 ratio: 1.999999046316588 formula: 1.999999046316588[/code]

I used the program Derive in order to extend the asymptotic formula.

I found:

[tex]\large 2-\left(\frac{1}{2}\right)^n-\frac{n}{2}\left(\frac{1}{4}\right)^n-\left(\frac{3}{8}n^2+\frac{1}{8}n\right)\left(\frac{1}{8}\right)^n -\left(\frac{1}{3}n^3+\frac{1}{4}n^2+\frac{1}{24}n\right)\left(\frac{1}{16}\right)^n - \\ - \left(\frac{125}{384}n^4+\frac{75}{192}n^3+\frac{55}{384}n^2+\frac{1}{64}n\right)\left(\frac{1}{32}\right)^n + O\left(n^5\left(\frac{1}{64}\right)^n\right)[/tex]

I tried to find the next term, but the program took too much time and I had to stop it.

The table is now:

[code]
n=2 ratio: 1.618033988749895 formula: 1.63677978515625
n=3 ratio: 1.839286755214161 formula: 1.840293884277344
n=4 ratio: 1.927561975482925 formula: 1.927605926990509
n=5 ratio: 1.965948236645485 formula: 1.965949894860387
n=10 ratio: 1.999018632710101 formula: 1.999018632710136
n=15 ratio: 1.999969475434503 formula: 1.999969475434503
n=20 ratio: 1.999999046316588 formula: 1.999999046316588[/code]

Reciprocals
 

[QUOTE=alpertron]Mally, it appears that you are using the inverse of the ratio. For Fibonacci numbers, the ratio is normally F[sub]m+1[/sub]/F[sub]m[/sub]. In these generalized sequences R[sub]n,m[/sub] which is the sum of the previous n numbers, the ratio I considered is R[sub]n,m+1[/sub]/R[sub]n,m[/sub].

So by definition the ratio is always greater than 1. From the formulas I've shown above, it cannot be greater or equal to 2.

Some examples that show the correct ratio and the value found using the formula I developed:

[code]
n=2 ratio: 1.618033988749895 formula: 1.66015625
n=3 ratio: 1.839286755214161 formula: 1.84423828125
n=4 ratio: 1.927561975482925 formula: 1.9281005859375
n=5 ratio: 1.965948236645485 formula: 1.96600341796875
n=10 ratio: 1.999018632710101 formula: 1.999018633039668
n=15 ratio: 1.999969475434503 formula: 1.999969475434504
n=20 ratio: 1.999999046316588 formula: 1.999999046316588[/code][/QUOTE]
:rolleyes: No problem, Alpertron! You are absolutely right and your ratio tends to 2 as you can see if you work further.

However the ratio Fn/Fn+1 (less than 1) seems to be favoured by the 14 yr. old Mark Feinberg, Mark Barr, and Martin Gardner, the Scientific American
celebrated columnist, who writes about them in his book 'Mathematical Circus'.

Barring Martin Gardner, the former two tackled these problems and arrived at
their results pre - computer era and so possibly it was easier to compute, I dont know as Im not skilled in programming.

Mark Barr who tackled the general series gave the ratio Fn,m/ Fn,m+1 = 0.5
So the reciprocal tends to 2.0 as you have so clearly worked out.

By the way there are interesting formulae derived from the Fibonacci series when phi, phi-1, and - phi are linked together. I think I gave these in one of the Threads but off hand I cant recall them.

Perhaps you can find a non trivial link between these constants. I mean the Fib Series, The Trib., The tetra, and 5, 6 terms... to n,n+1.......etc. I have not come across it in any publication anywhere.
Look forward to more interesting derivations from you like the values of a and b in a^b=b^a in another thread that is current in puzzles. :bow:
Mally :coffee:

[QUOTE=mfgoode]:rolleyes: No problem, Alpertron! You are absolutely right and your ratio tends to 2 as you can see if you work further.[/QUOTE] No need to work further, since it is implied by the formula I stated above. When n tends to infinite, the formula tends to 2.

[QUOTE=mfgoode]However the ratio Fn/Fn+1 (less than 1) seems to be favoured by the 14 yr. old Mark Feinberg, Mark Barr, and Martin Gardner, the Scientific American
celebrated columnist, who writes about them in his book 'Mathematical Circus'.[/QUOTE]
At least this shows that I haven't read those sources and that my work is original.

[QUOTE=mfgoode]Barring Martin Gardner, the former two tackled these problems and arrived at
their results pre - computer era and so possibly it was easier to compute, I dont know as Im not skilled in programming.[/QUOTE]
There is no need for computers. I found the formulas before computing the tables. The problem is the number of calculations to be performed to find the last two terms, so I decided to use a symbolic math program.

For example, from the original formula

x[sup]n+1[/sup]-2x[sup]n[/sup]+1 = 0

the root with largest real part is the requested ratio. Dividing by x[sup]n[/sup]:

x-2+1/x[sup]n[/sup] = 0

Since x > 1 by definition, the third term vanishes as n tends to infinite, so:

x-2 = 0 which implies that x = 2, as expected.

[QUOTE=mfgoode]Mark Barr who tackled the general series gave the ratio Fn,m/ Fn,m+1 = 0.5
So the reciprocal tends to 2.0 as you have so clearly worked out.[/QUOTE]
Well, notice that the formula I stated above is much more precise than just this limit.

[QUOTE=mfgoode]Perhaps you can find a non trivial link between these constants. I mean the Fib Series, The Trib., The tetra, and 5, 6 terms... to n,n+1.......etc. I have not come across it in any publication anywhere.
[/QUOTE]
Apart from being roots of x[sup]n+1[/sup]-2x[sup]n[/sup]+1 = 0 I do not see any other (trivial) connection.

Remembering Mark Feinberg
 

[QUOTE=mfgoode;82414]:rolleyes: No problem, Alpertron! You are absolutely right and your ratio tends to 2 as you can see if you work further.

However the ratio Fn/Fn+1 (less than 1) seems to be favoured by the 14 yr. old Mark Feinberg, Mark Barr, and Martin Gardner, the Scientific American
celebrated columnist, who writes about them in his book 'Mathematical Circus'.

Barring Martin Gardner, the former two tackled these problems and arrived at
their results pre - computer era and so possibly it was easier to compute, I dont know as Im not skilled in programming.

Mark Barr who tackled the general series gave the ratio Fn,m/ Fn,m+1 = 0.5
So the reciprocal tends to 2.0 as you have so clearly worked out.

By the way there are interesting formulae derived from the Fibonacci series when phi, phi-1, and - phi are linked together. I think I gave these in one of the Threads but off hand I cant recall them.

Perhaps you can find a non trivial link between these constants. I mean the Fib Series, The Trib., The tetra, and 5, 6 terms... to n,n+1.......etc. I have not come across it in any publication anywhere.
Look forward to more interesting derivations from you like the values of a and b in a^b=b^a in another thread that is current in puzzles. :bow:
Mally :coffee:[/QUOTE]

Thank you for pointing out my late brother's work. He wrote 3 papers in Fibonacci Quarterly before he was killed getting a ride from a friend on a motorcycle as a sophomore at Penn. I was 11 when he was doing his first project, coining the term Tribonacci but more substantively working out the math, including the idea of nested pyramids from a trinomial expansion, projected on the surface (summing), then taking the diagonal, similar to Vern Hoggatt's trick with the binomial and Fibonacci. I wrote a paper in the same journal later myself but I was an old man of 17 by then. :smile:

Mark Feinberg
 

:sad:

Thank you afeinberg for giving us (me) this feed back.

I’m very sorry to read of the tragic event in which you lost your brother Mark.

Your very expression of gratitude is in itself a reward for me to acknowledge the works of others Mark, included. This was the spirit of Fr. Marin Mersenne who meticulously passed on the works of other Mathematicians.

My mind goes back to the days of Jimmy Rogers and Gene Autry the singing cowboys.
I couldn’t say it better than in their song ‘That little kid sister (substitute brother) of mine’

“They needed a new star up yonder
And they couldn’t find a brighter light to shine,
God decided he was meant for
A star and so He sent for,
That little kid brother of mine”

I look forward to some more posts from you,
Regards,
Mally. :coffee: