It’s nice to 89, in addition to being 10^2-10-1, is also a Fibonacci number. How many Fibonacci numbers are of the form n^2-n-1? Jim Propp On Monday, March 19, 2018, James Buddenhagen <jbuddenh@gmail.com> wrote:
Looks nicely fiby up till 90. (re: Allan Wechsler's 1/9899).
On Sun, Mar 18, 2018 at 12:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
What do you think of 1/9899?
On Sun, Mar 18, 2018 at 2:26 PM, Thane Plambeck <tplambeck@gmail.com> wrote:
i like the decimal expansion of 1/9801
On Sat, Mar 17, 2018 at 2:59 PM, Cris Moore <moore@santafe.edu> wrote:
Indeed, 14+28+57 = 99… because 7 is a divisor of 10101... and 07+69+23 = 99, because 13 is too.
- Cris
On Mar 17, 2018, at 3:13 PM, Simon Plouffe <
simon.plouffe@gmail.com>
wrote:
Hello,
we can remark that 142857 is symmetrical, 142 + 857 = 999
so one just has to remember half of the period in order to know all of the digits, 1/13 = 076923 , 076+923 = 999 too.
this phenomena is true whenever p <> 2 or p <> 11 and the period is even.
best regards, Simon Plouffe
Le 2018-03-17 à 21:59, Cris Moore a écrit :
well,
1/13 = 0769/9997 = 0.0769 (1 + 0.0003 + 0.00000009 + …) = 0.0769
2307 …
although this doesn’t quite break up the repetend 076923. You
could use the clumsier
1/13 = 076/988 = 0.076 (1 + 0.012 + 0.000144) = 0.076 + 0.000912 +
0.0000109… = 0.076923…
but this involves a bunch of carrying.
Cris
> On Mar 17, 2018, at 2:36 PM, James Propp <jamespropp@gmail.com>
wrote:
> > I just figured out for myself a probably well-known trick for > deriving/remembering the decimal expansion of 1/7: 1/7 = 14/98 = 14/(100-2) > = .14/(1-.02) = .14 + .0028 + .000056 + .00000112 + ... = .142857... > > > Are there other examples where the repetend of the decimal expansion of 1/n > in splits into blocks that are related to this sort of fashion? > > > Jim Propp > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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