CF of Reciprocal Sum of A112373

Sun Dec 11 18:14:46 CET 2005

The terms of the continued fraction (starting with the second term) appear 
to be
an interleaving of A112373 and the ratios of successive terms of A112373.

By the way, the ratios of the ratios of A112373 also appear to be integers.

Gerald

At 10:40 PM 12/10/2005, Gerald McGarvey wrote:

>The ratios of successive terms in A112373 is 1, 2, 6, 78, 73086, 
>4999703411742, 1710009514450915230711940280907486, ...
>If these are all integers (I believe they are), then A112373 is a series 
>of the form s = - (x/a - x^2/(ab) + x^3(abc) - x^4/(abcd) ...) where in 
>this case x = -1, and so a formula by Euler can be applied, cf. page 31 of
>'On the Transformation of Infinite Series to Continued Fractions'  by 
>Leonhard Euler
>Translated by Daniel W. File
>http://www.math.ohio-state.edu/~sinnott/ReadingClassics/continuedfractions.pdf
>
>Starting with a(0)=.5 instead of a(0)=1 the CF is
>[3, 3, 1, 4, 4, 20, 80, 1620, 129600, 209953620, 27209989152000, 
>5712835722623340193620, ...]
>With a(0)=2 the CF is
>[3, 11, 1, 5, 1, 1, 2, 1, 1, 77, 1, 1, 233, 1, 1, 73085, 1, 1, 17102123, 
>1, 1, 4999703411741, ...]
>
>Notes about CFs for some other related series...
>
>For m > 2, the CFs for (y^m+y^2)/x look related.
>(used x=1.;y=1.;s=1/x+1/y; for(n=1,10, z=(y^m+y^2)/x; x=y; y=z; 
>s=s+1/z;);contfrac(s) )
>e.g.
>m=4: [2, 1, 1, 4, 2, 200, 20, 321602000, 80200, 
>53353214040377436196491224020000, 2068556928401602000,...]
>m=5: [2, 1, 1, 8, 2, 23328, 36, 767667136004966560896, 30233736, 
>11428202543809917040939553535324671293831057914382382411094812574379924574115058355844062625584214528,...]
>m=6: [2, 1, 1, 16, 2, 10690688, 68, 
>87818313168587795575060012891154023826432, 49433743624,...]
>m=7: [2, 1, 1, 32, 2, 20037321216, 132, 
>39297387807850612366968992508128169281051499010505152979708234780909568,...]
>
>x=1.;y=1.;s=1/x+1/y; for(n=1,10, z=(y^6+y^4)/x; x=y; y=z; 
>s=s+1/z;);contfrac(s) gives
>
>[2, 1, 1, 19, 2, 20483199, 1, 1, 19, 1, 1, 
>3691655198239376495977121567478906879999999, 2, 19, 1, 1, 20483199, 2, 19, 
>1, 1,
>123574533747542722214957698182141192745653850510053101282509714858557754904985512229823666795783898517889568869081119394312092597448934456687819075878231584557769579194924580526323512417294588125382157621211132327430507724799999999999999999999999999, 
>2, 19, 2, 20483199, 1, 1, 19, 2, 
>3691655198239376495977121567478906879999999, 1, 1, 19, 1, 1, 20483199, 2, 
>19, 2]
>
>x=1.;y=1.;s=1/x+1/y; for(n=1,10, z=(y^6+y^4)/x^2; x=y; y=z; 
>s=s+1/z;);contfrac(s) gives
>
>[2, 1, 1, 19, 2, 10241599, 1, 1, 19, 1, 1, 
>2884105623624512887482629698371583999999, 2, 19, 1, 1, 10241599, 2, 19, 1, 
>1, 
>5486954368333562346037641264953313150459684198888964050292391859772355695426782383112496022886203196889588144368973696806007710117004903186837938942910307608577691248476679717135469522458859153219230105599999999999999999999, 
>2, 19, 2, 10241599, 1, 1, 19, 2, 2884105623624512887482629698371583999999, 
>1, 1, 19, 1, 1, 10241599, 2, 19, 2]
>
>x=1.;y=1.;s=1/x+1/y; for(n=1,10, z=(y^6+y^4)/x^3; x=y; y=z; 
>s=s+1/z;);contfrac(s) gives
>
>[2, 1, 1, 19, 2, 5120799, 1, 1, 19, 1, 1, 
>2253207518456650693347377807411199999, 2, 19, 1, 1, 5120799, 2, 19, 1, 1, 
>974526622181924528515971755878699631752575949091079700857729706271643386255863455225741167482292092038787498907791283425110727869940932773342817410934199506590456858043707859521031372799999999999999, 
>2, 19, 2, 5120799, 1, 1, 19, 2, 2253207518456650693347377807411199999, 1, 
>1, 19, 1, 1, 5120799, 2, 19, 2]
>
>x=1.;y=1.;s=1/x+1/y;for(n=1,21,z=(y^3+y)/x;x=y;y=z;s=s+1/z;)      a few 
>huge terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,21,z=(y^3+y^2+y)/x;x=y;y=z;s=s+1/z;)  a few 
>huge terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,20,z=(y^3+y^2+ y)/x;x=y;y=z;s=s+1/z;) a few 
>huge terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,20,z=(y^3+2*y^2)/x;x=y;y=z;s=s+1/z;)  many 
>huge terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,20,z=(2*y^3+3*y^2)/x;x=y;y=z;s=s+1/z;) many 
>huge terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,20,z=(2*y^3)/x;x=y;y=z;s=s+1/z;)     many huge 
>terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,10,z=(y^4+y^3)/x;x=y;y=z;s=s+1/z;)   many huge 
>terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,10,z=(y^5+y^4)/x;x=y;y=z;s=s+1/z;)   many huge 
>terms
>x=1.;y=1.;s=1/x+1/y;for(n=1,10,z=(y^5+y^4)/x^2;x=y;y=z;s=s+1/z;) many huge 
>terms, repeats
>x=1.;y=1.;s=1/x+1/y;for(n=1,10,z=(y^5+y^2)/x;x=y;y=z;s=s+1/z;) many huge 
>terms, repeats
>x=1.;y=1.;s=1/x+1/y;for(n=1,10,z=(y^6+y^4)/x;x=y;y=z;s=s+1/z;) many huge 
>terms, repeats
>
>Regards,
>Gerald
>
>At 10:09 AM 12/9/2005, Mitch Harris wrote:
>>>From: Jeffrey Shallit > From: "Paul D Hanna" > > Seqfans,
>>> >        This may or may not be surprising, but I find it interesting neve=
>>> > rtheless.
>>> > The sum of the reciprocal terms of Andrew Hone's sequence A112373:
>>> >  x Sum_{n>0} 1/A112373(n) =
>>> > >      2.584401724019776724812076147...
>>> > where A112373 is defined by: =
>>> > > a(n+2) = (a(n+1)^3+a(n+1)^2)/a(n) with a(0)=1, a(1)=1 =
>>> > >  =
>>> > > is a constant with an interesting Continued Fraction:
>>> >  =
>>> > > x = [2; 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742,
>>> >  342022190843338960032, 1710009514450915230711940280907486, 584861200495=
>>> > 456320274313200204390612579749188443599552,...]
>>> >  =
>>> > > I wonder if the terms of the above CF has any recurrence pattern?
>>>Yes:  the even-indexed terms (such as 12 = 2*6) appear to be the product 
>>>of the previous two terms.
>>>The odd-indexed terms (such as 78=6*12 + 6) appear to be the product
>>>of the previous two terms, plus the term two behind.
>>>It would be nice to prove this.  Probably an easy induction, but
>>>I don't have time right now.  Very beautiful.
>>
>>Paul's conjecture, that the sequence b(n) defined by
>>
>>   b(2n) = a(n)
>>   b(2n+1) = a(n+1)/a(n),
>>
>>matches your observation, that
>>
>>   1) b(2n) = b(2n-1)b(2n-2)
>>   2) b(2n+1) = (b(2n)+1)b(2n-1),
>>
>>is not hard to prove (using the definition of a(n)):
>>
>>   1) b(2n-1)b(2n-2) = a(n)/a(n-1) * a(n-1) = a(n) = b(2n).
>>   2) (b(2n)+1)b(2n-1) = (a(n)+1)a(n)/a(n-1)
>>                       = (a(n)^2+a(n))a(n)/(a(n-1)a(n))
>>                       = a(n+1)/a(n) = b(2n+1)
>>
>>(and base cases by inspection)
>>
>>As to the CF conjecture, i.e. that b(n) is the CF, from computation
>>of the CF of the partial sums (that is \sum_{k=0..n} 1/a(n)), the last
>>element of the CF always seems to be a(n), but the next to last isn't
>>always a(n-1), so a straightforward induction probably won't work exactly 
>>right. Any ideas on how to attack this otherwise (or get around this 
>>"probably")?
>>
>>Mitch