[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] -- Warren D. Smith
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] -- Warren D. Smith
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]
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]
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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]
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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It appears Niederreiter's theorem can be improved. Consider the following sequence of rationals (numerators in hexadecimal) 1/2^1 5/2^3 4F/2^7 4F01/2^15 4F00FFFF/2^31 4F00FFFF00000001/2^63 4F00FFFF00000000FFFFFFFFFFFFFFFF/2^127 ... in each case, the denominator is 2^(2^k-1) and the numerator is got by either (i) subtracting 1 from previous numerator then appending 1111...1111 in binary, or (i) appending 0000...0001 to the previous numerator's binary form (the append always exactly doubles bitlength). where we do (i) if k=odd and (ii) if k=even. In other words, the numerator is 2^(2^(k-1)) times the preceding numerator, plus (-1)^k. All these rationals have only 1s and 2s in their continued fraction expansion. No 3s. There are many other example sequences of the same ilk, where no 3's occur. This one just happens to be the simplest. Their continued fraction expansions are [0; 2] = [0; 1, 1] (Note, any final "2" can be replaced by "1, 1") [0; 1, 1, 1, 2] [0; 1, 1, 1, 1, 1, 1, 2, 1, 2] [0; 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2] [0; 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2] [0; 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2] [0; 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2] the rule for going from one CF expansion to the next is either you (i) copy its after-semicolon part (in form ending 2) twice, or (ii) copy it once (in form ending 1,1) then subtract 1 from the final entry; then append "1", then append a REVERSE-ORDER copy (of the form ending 2). In other words, Niederreiter's construction actually is stronger than he realized, since it creates infinite chains of rationals with denominator=power of 2, having only 1s and 2s in their continued fraction expansions. This is amazing. But then I found this paper: https://web.williams.edu/Mathematics/sjmiller/public_html/book/papers/vdp/Po... FOLDED CONTINUED FRACTIONS by A. J. van der Poorten and J. Shallit which seems to know about this sort of thing already. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
participants (5)
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Eugene Salamin -
Mike Stay -
Neil Sloane -
rwg -
Warren D Smith