h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

F(118)=2,046,711,111,473,984,623,491,759

Error checking? Wozzat?

F(119) = 3,311,648,143,516,982,017,180,081

Trying to get the numbers following the original rule, instead of exchanging a 6 for a 4 when a previous number is incorrect.

F(120)= 5,358,359,254,990,966,640,671,840

(to be continued...)

8,670,007,398,507,948,657,850,921

What really appals me is that 13 posts have been made in this thread with no approach to the correct Fibonacci number.

I blame the Powers.

B

F(122)=14,028,366,653,498,915,298,522,761

Fourteen septillion... Are we getting close to the end yet? Better yet, when do they turn around and start getting smaller again?

F(123)=22,698,374,052,006,863,956,375,682

F(124)=31,368,381,450,514,812,614,227,603

Here, that wasn't right. Should be:


F(124)=36,726,740,705,505,779,254,898,443

...or something to that effect.

...or shouldn't it be:
F(124)=36,726,740,705,505,779,254,899,443

cause I was not sure about F(121).
LOL, I'm getting really crazy after this thread...

F(125)=59,425,114,757,512,643,211,274,125

Awww... the last 3 digits of F(125) are '125'. Isn't that cute?



replying to Hoovooloo's post 37...

You can reach 471151 with seeds F(0)=541 and F(1)=143, which are smaller than his F(0)=753, F(1)=307. I'll bet his are the second smallest.

As far as algorithms, I'd say you take the number you want to have crop up in your series, and call it F(n). Then:

F(n-1) = round[F(n) * (phi - 1)]

or: = round[F(n) / phi]

...which amounts to the same thing

Once you have F(n) and F(n-1), you can work backwards easily until it blows up on you and you get a negative. (I guess you could keep working backwards then, but it doesn't feel very 'Fibonacci'.... whoever heard of someone trying to get -398 rabbits to breed?)

F(126)=96,151,855,463,018,422,466,173,568

Was it something I said?

155,576,970,220,531,065,677,447,693

I need bigger Post-It notes ...

Gif

F(128) = 251.728.825.683.549.488.150.424.261

The numbers in the second and third triplets from the right have been in error for some time, so I'll give two numbers to set it right again.
F(127) should have ended with ....681.649.693

F(129) = 407.305.795.904.080.553.832.073.954

F(130) = 659,034,621,587,630,041,982,498,215

Marjin... how d'you do that?



BTW, the final digits follow a cycle that repeats every 60 steps, so:

F(N) = F(N+60) (mod 10)

...where that equal sign should really be a congruence sign.




Next poster breaks the octillion barrier!

OK, then...

F(131)=1,066,340,417,491,710,595,814,572,169

B

F(132) = 1,725,375,039,079,340,637,797,070,384

Warning: American attempting to hijack the thread again.

*evil laughter*

F(133)=2,791,715,456,571,051,234,611,642,553

F(134) = 4,517,090,495,650,391,871,408,712,937

Warning, Bagpuss' last number should have ended in: 233,611,642,553

F(134)=4,517,090,495,650,391,872,408,712,937


Number crunchin' in the USA!

well then, use Marjin's F(134), not mine.


Marjin, are you double checking our calculations, or do you have a list? Either way, I'm glad someone's keeping us honest.