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

F(135) = 7,308,805,952,221,443,105,020,355,490

The difference in the is in the previous error.

F(136)=11,825,896,447,871,834,976,429,068,427 (I hope)

F(137) = 19,134,702,400,093,278,081,449,423,917

Don't Panic, it is.

F(138)=30,960,598,847,965,113,057,878,492,344

There'd be fewer mistakes if we just let you do them all, Marjin.

F(139) = 50,095,301,248,058,391,139,327,916,261

F(140) = 81,055,900,096,023,504,197,206,408,605

F(141) = 131,151,201,344,081,895,336,534,324,866

F(142) = 212,207,101,440,105,399,533,740,733,471

Keep adding some comments, or we will be called spammers

F(143) = 343,358,302,784,187,294,870,275,058,337

I'm actually working on a Fibonacci Series related art project, if anyone's interested in hearing about it... /

F(144)=555,565,404,224,292,694,404,015,791,808

Do tell, GT.

F(145) = 898,923,707,008,479,989,274,290,850,145



Well, it has to do with something I mentioned on the last page, about how the final digits of numbers in the sequence repeat after a while. F(60) ends in a '0', and F(61) ends in a '1', so that's like starting back at the beginning, if you only look at the final digits. The sequence of final digits looks like:

0, 1, 1, 2, 3, 5, 8, 3, 1, . . ., 2, 9, 1, 0, . . .

...where that last '0' is just like the first one. There are a total of 60 terms. The thing is, some pairs of terms never occur. For example, you never have a '0' followed by another '0', because if you did, every term after those would end with a '0'. If you start with (0, 10) as your seeds, that's what you'd have for your series.

Less trivially, you never have the pair (0, 5) either. The seeds (0, 5) would lead to a series whose final digits repeated:

0, 5, 5, 0, 5, 5, 0, 5, 5, . . .

... over and over again.



to be continued...

My calculator says 4181

D.

Sorry, ignore that - I missed 151 entries in the thread!

D.

898,923,707,008,479,989,274,290,850,145

Hmm... wonder how long this can continue...

Oops, didn't see someone had already posted that one.

F(146) = 1,454,489,111,232,772,683,678,306,641,953

And another digit added

F(146)=1,454,489,111,232,772,683,678,306,641,953

GT - There's 60 different pairings, counting (0,1) as different from (1,0). Do you get a nice pattern if you plot these or something?

F(147) = 2,353,412,818,241,252,672,952,597,492,098

"Do you get a nice pattern if you plot these or something?"


Well, you can make a grid, and plot those 60 pairings, maybe in some colour. Then you take a different seed, a pairing that doesn't show up among the 60, and see the cycle that it generates. (0,0) generates a trivial cycle, with 1 pairing. (0,5) generates a cycle with 3 pairings. (0,2) generates a 20-pairing cycle, containing only even numbers! Each of these cycles (there are 6 of them) can be plotted on the same grid in a different colour. How you choose the colours is another question.

Anyway, at some point, you realize that this is all based on the fact that we write our numerals in base 10. In different bases, you get different cycles of different lengths, and you draw differently sized grids with different patterns in them.

There are some neat correspondences across bases. Just looking at the cycles in base 10: the (0,5,5) sequence is just a multiple of the (0,1,1) sequence that appears in binary, and in every even base. Likewise, the 20-number cycle beginning with (0,2) is a multiple of a 20-number cycle beginning with (0,1) that shows up in base 5. Our original 60-number cycle doesn't show up in any smaller base, but it will be repeated in base 20, beginning with (0,2).

I think base 7 is my favourite. The 49 possible pairings are divided among 3 16-number cycles, plus the trivial case (0,0). It looks really good using black or white for the trivial one, and three metallic colours for the others.

F(148) = 3,807,901,929,474,025,356,630,904,134,051

Have you seen this one in tha Mandelbrot-set?
http://math.bu.edu/DYSYS/FRACGEOM2/node7.html

F(149)=6,161,314,747,715,278,029,583,501,626,149

That picture looks remarkably like the edge of a cast-zinc surface (and I've seen a few...) - it's amazing where these numbers will find an outlet.

B