On Mon, Mar 19, 2018 at 2:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
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
Yikes! I completely missed that the crossref to A004023 (= 2, 19, 23, 317, 1031, 49081, 86453, 109297, 270343,) held the rest of A003459! Rich says they're probably infinite! --rwg (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