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]