I searched the forum but found nothing related to this issue. Any help would be greatly appreciated.
I am concerned with Generalized Fibonacci sequences which use two different initial values. However, these two values are not arbitrarily chosen as much as derived. By that I mean, I start with a very large number, calculate what *would* be the preceding number in a *true* Fibonacci sequence utilizing a formula based on Binet's. (However, I find that a simple rounded product of phi suffices for a successor just as well in my case). From these two large numbers, I then subtract until I generate a list of generalized Fibonacci numbers that eventually end at some point where the number would begin to become negative if continued.
Here is an example:
15975329436
9873296573
6102032863
3771263710
2330769153
1440494557
890274596
550219961
340054635
210165326
129889309
80276017
49613292
30662725
18950567
11712158
7238409
4473749
2764660
1709089
1055571
653518
402053
251465
150588
100877
49711
Were this a true Fibonacci sequence, I would be able to chose any number in this sequence, multiply by phi and round for a surprisingly accurate value matching the next number in the sequence. However, since this is a generalized sequence, that method yields unreliable results...but oddly unreliable.
When comparing the values derived from a simple rounded phi product to the actual next value in the generalized sequence, the difference between the expected value and the actual value creates an alternating positive and negative pseudo-Fibonacci sequence. (Actually, I suspect there are two sine functions overlapping; one for the positive, the other for the negative.)
Here is the example showing the differences:
15975329436 15975329436 0
9873296573 9873296572 1
6102032863 6102032863 0
3771263710 3771263709 1
2330769153 2330769153 0
1440494557 1440494555 2
890274596 890274598 -2
550219961 550219957 4
340054635 340054640 -5
210165326 210165316 10
129889309 129889323 -14
80276017 80275992 25
49613292 49613331 -39
30662725 30662661 64
18950567 18950669 -102
11712158 11711991 167
7238409 7238677 -268
4473749 4473313 436
2764660 2765364 -704
1709089 1707949 1140
1055571 1057414 -1843
653518 650535 2983
402053 406878 -4825
251465 243656 7809
150588 163222 -12634
100877 80434 20443
49711
As you can see, the differences are quite large as the generalized values decrease, but approach unity at the highest value.
Currently the nearest formula I have been able to discern utilizing Cornell's Eureqa application on just the positive values is the following. Clearly this is incorrect.
2643978.8 + 1.3097997*x + 2495350.5*sin(1569598.3 + 0.43991414*x) + 0.31525132*x*sin(0.32958981*x) - 615003.63*sin(0.026585015*x) - 118125.93*sin(0.32958981*x)
Attempting to use the Eureqa application on both the positive and negative data yielded something completely unruly.
It is my hope that a simple successor formula may be discerned that will allow me to provide only one value in the Generalized sequence (e.g. 653518 above) and this successor formula would be able to provide the next highest value (i.e. 1055571 in this case) in order to recreate the G-sequence to reach my original large value.
Certainly, I could merely record the lowest two values and recreate the sequence in this manner, but a similar "rounded product of phi" solution would serve my purposes better.
In my naivete I suspect the solution may involve a combination of phi and a trigonometric function.
Thank you,
Scott A. Rossell
San Diego, California, USA
I am concerned with Generalized Fibonacci sequences which use two different initial values. However, these two values are not arbitrarily chosen as much as derived. By that I mean, I start with a very large number, calculate what *would* be the preceding number in a *true* Fibonacci sequence utilizing a formula based on Binet's. (However, I find that a simple rounded product of phi suffices for a successor just as well in my case). From these two large numbers, I then subtract until I generate a list of generalized Fibonacci numbers that eventually end at some point where the number would begin to become negative if continued.
Here is an example:
15975329436
9873296573
6102032863
3771263710
2330769153
1440494557
890274596
550219961
340054635
210165326
129889309
80276017
49613292
30662725
18950567
11712158
7238409
4473749
2764660
1709089
1055571
653518
402053
251465
150588
100877
49711
Were this a true Fibonacci sequence, I would be able to chose any number in this sequence, multiply by phi and round for a surprisingly accurate value matching the next number in the sequence. However, since this is a generalized sequence, that method yields unreliable results...but oddly unreliable.
When comparing the values derived from a simple rounded phi product to the actual next value in the generalized sequence, the difference between the expected value and the actual value creates an alternating positive and negative pseudo-Fibonacci sequence. (Actually, I suspect there are two sine functions overlapping; one for the positive, the other for the negative.)
Here is the example showing the differences:
15975329436 15975329436 0
9873296573 9873296572 1
6102032863 6102032863 0
3771263710 3771263709 1
2330769153 2330769153 0
1440494557 1440494555 2
890274596 890274598 -2
550219961 550219957 4
340054635 340054640 -5
210165326 210165316 10
129889309 129889323 -14
80276017 80275992 25
49613292 49613331 -39
30662725 30662661 64
18950567 18950669 -102
11712158 11711991 167
7238409 7238677 -268
4473749 4473313 436
2764660 2765364 -704
1709089 1707949 1140
1055571 1057414 -1843
653518 650535 2983
402053 406878 -4825
251465 243656 7809
150588 163222 -12634
100877 80434 20443
49711
As you can see, the differences are quite large as the generalized values decrease, but approach unity at the highest value.
Currently the nearest formula I have been able to discern utilizing Cornell's Eureqa application on just the positive values is the following. Clearly this is incorrect.
2643978.8 + 1.3097997*x + 2495350.5*sin(1569598.3 + 0.43991414*x) + 0.31525132*x*sin(0.32958981*x) - 615003.63*sin(0.026585015*x) - 118125.93*sin(0.32958981*x)
Attempting to use the Eureqa application on both the positive and negative data yielded something completely unruly.
It is my hope that a simple successor formula may be discerned that will allow me to provide only one value in the Generalized sequence (e.g. 653518 above) and this successor formula would be able to provide the next highest value (i.e. 1055571 in this case) in order to recreate the G-sequence to reach my original large value.
Certainly, I could merely record the lowest two values and recreate the sequence in this manner, but a similar "rounded product of phi" solution would serve my purposes better.
In my naivete I suspect the solution may involve a combination of phi and a trigonometric function.
Thank you,
Scott A. Rossell
San Diego, California, USA