On 2018-03-19 08:35, James Propp wrote:
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
In[26]:= Select[Fibonacci@Range@99999, squareq[4 # + 5] &] // tim During evaluation of In[26]:= 4767.121894,5 Out[26]= {1, 1, 5, 55, 89} In[24]:= squareq[n_Integer] := IntegerQ@√n Looks like a companion to 0,1,144. (Neither in OEIS) How about OEIS typographical finiteness conventions? E.g.: (Heegner) ..., 163. (Period means no more.) (Wieferich) 1093,3511,? means probably more (Pemutables) 991,1111111111111111111,11111111111111111111111.? maybe no more (Fibs) 4181,6765, fer sure more. --rwg
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?
In[2]:= RealDigits[1/9899]
Out[2]= {{{1, 0, 1, 0, 2, 0, 3, 0, 5, 0, 8, 1, 3, 2, 1, 3, 4, 5, 5, 9, 0, 4, 6, 3, 6, 8, 3, 2, 0, 0, 3, 2, 3, 2, 6, 4, 9, 7, 6, 2, 6, 0, 2, 2, 8, 3, 0, 5, 8, 8, 9, 4, 8, 3, 7, 8, 6, 2, 4, 1, 0, 3, 4, 4, 4, 7, 9, 2, 4, 0, 3, 2, 7, 3, 0, 5, 7, 8, 8, 4, 6, 3, 4, 8, 1, 1, 5, 9, 7, 1, 3, 1, 0, 2, 3, 3, 3, 5, 6, 9, 0, 4, 7, 3, 7, 8, 5, 2, 3, 0, 8, 3, 1, 3, 9, 7, 1, 1, 0, 8, 1, 9, 2, 7, 4, 6, 7, 4, 2, 0, 9, 5, 1, 6, 1, 1, 2, 7, 3, 8, 6, 6, 0, 4, 7, 0, 7, 5, 4, 6, 2, 1, 6, 7, 8, 9, 5, 7, 4, 7, 0, 4, 5, 1, 5, 6, 0, 7, 6, 3, 7, 1, 3, 5, 0, 6, 4, 1, 4, 7, 8, 9, 3, 7, 2, 6, 6, 3, 9, 0, 5, 4, 4, 4, 9, 9, 4, 4, 4, 3, 8, 8, 3, 2, 2, 0, 5, 2, 7, 3, 2, 5, 9, 9, 2, 5, 2, 4, 4, 9, 7, 4, 2, 3, 9, 8, 2, 2, 2, 0, 4, 2, 6, 3, 0, 5, 6, 8, 7, 4, 4, 3, 1, 7, 6, 0, 7, 8, 3, 9, 1, 7, 5, 6, 7, 4, 3, 1, 0, 5, 3, 6, 4, 1, 7, 8, 1, 9, 9, 8, 1, 8, 1, 6, 3, 4, 5, 0, 8, 5, 3, 6, 2, 1, 5, 7, 7, 9, 3, 7, 1, 6, 5, 3, 7, 0, 2, 3, 9, 4, 1, 8, 1, 2, 3, 0, 4, 2, 7, 3, 1, 5, 8, 9, 0, 4, 9, 3, 9, 8, 9, 2, 9, 1, 8, 4, 7, 6, 6, 1, 3, 7, 9, 9, 3, 7, 3, 6, 7, 4, 1, 0, 8, 4, 9, 5, 8, 0, 7, 6, 5, 7, 3, 3, 9, 1, 2, 5, 1, 6, 4, 1, 5, 7, 9, 9, 5, 7, 5, 7, 1, 4, 7, 1, 8, 6, 5, 8, 4, 5, 0, 3, 4, 8, 5, 2, 0, 0, 5, 2, 5, 3, 0, 5, 5, 8, 6, 4, 2, 2, 8, 7, 0, 9, 9, 7, 0, 7, 0, 4, 1, 1, 1, 5, 2, 6, 4, 1, 6, 8, 0, 9, 7, 7, 8, 7, 6, 5, 5, 3, 1, 8, 7, 1, 9, 0, 6, 2, 5, 3, 1, 5, 6, 8, 8, 4, 5, 3, 3, 7, 9, 1, 2, 9, 2, 0, 4, 9, 7, 0, 1, 9, 9, 0, 1, 0, 0, 0}}, -3}
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