I created https://oeis.org/A267193 for the "complement obverse" function. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sun, Jan 24, 2016 at 11:51 AM, Mike Stay <metaweta@gmail.com> wrote:
The complement obverse of a number is the complement of the number with the digits read in the reverse direction.
On Sat, Jan 23, 2016 at 9:11 PM, rwg <rwg@sdf.org> wrote:
http://www.sciencedirect.com/science/article/pii/0022314X9290042N A bit strange: Converting to -1 numerators, In[380]:= Clear[negcfL]; negcfL[(n_Integer | n_Rational)] :=Prepend[negcfL[1/(# - n)], #] &@Ceiling@n
In[383]:= negcfL[1895759871/2^32]
Out[383]= negcfL[1, 2, 5, 5, 2, 2, 3, 2, 2, 5, 2, 2, 3, 2, 2, 3, 2, 2, 5, 5, 2, 2, 5, 5, 2, 2, 3, 2, 2, 3, ComplexInfinity]
eschews 4, seems to have bigger terms, but converges slower.
Does anybody remember what "complement obverse" means? I once jokingly requested converting to complement obverse, excess 9 biquinary. What the heck was I talking about? --rwg Remember excess 3?
On 2016-01-23 14:42, Eugene Salamin via math-fun wrote:
Uncountably many real numbers share that property of having small partial quotients, even partial quotients that are only 1 or 2.
-- Gene
From: Warren D Smith <warren.wds@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, January 23, 2016 10:19 AM Subject: [math-fun] continued fractions for a/2^w with only small partial quotients
Here is a special case of theorem 1 of Harald Niederreiter: Dyadic fractions with small partial quotients, Monatshefte f"ur Mathematik 101,4 (December 1986) 309-315
Let p[0]=1, p[1]=3, p[2]=7, p[3]=113 and if n>=3 let p[n+1]=2^(2^n)*p[n]-1. (This sequence is not in, but should be in, the oeis.) Then the rational number p[n]/2^(2^n) has continued fraction with all partial quotients either 1, 2, or 3. For n large this rational number approaches 0.441390990978106856291920262469474778296688624079447195658789...
1/2 = [0; 2]
3/4 = [0; 1, 3]
7/16 = [0; 2, 3, 2]
113/256 = (7*16+1)/2^8 = [0; 2, 3, 1, 3, 3, 2]
28927/65536 = (113*256-1)/2^16 = [0; 2, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 2]
1895759871/2^32 = (28927*2^16-1)/2^32 = [0; 2, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 2]
8142226647014178815/2^64 = (1895759871*2^32-1)/2^64 = [0; 2, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 2]
150197571147608796277790585648096215039/2^128 = (8142226647014178815*2^64-1)/2^128 = [0; 2, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3,
1,
3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 2]
51109585015884376828428305273936708056145924221155280299477400051792199286783/2^256
= (150197571147608796277790585648096215039*2^128-1)/2^256 = [0; 2, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 1, 3, 3, 3, 1, 3, 3, 1, 3, 2]
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