CF of Reciprocal Sum of A112373

Wed Dec 21 12:43:08 CET 2005

Dear Gerald,

The paper by Euler is marvellous - thanks for providing the reference.
However, the continued fraction that results from Euler's formula
(either the case x=-1 of Theorem VIII on p.30, as was earlier
quoted below, or Theorem VI on p.25 - beware typos) does not converge
to the correct value for the parameter values being considered.

For a series

s=1/a + 1/(ab)+1/(abc)+1/(abcd)+...       (*)

the formula at the top of p.26 (after setting alpha=a, beta=-b, gamma=-c,
etc.) gives a continued fraction

s=1/(a- a/(b+1 -b/(c+1 -c/(d+1-...)))).   (**)

Now for the terms of A112373, consider the sum

s=1/a(2)+1/a(3)+1/a(4)+1/a(5)+... =0.584401720...
                                  =[0;1,1,2,2,6,12,78,936,73086,...]

with a(2)=2,a(3)=12,a(4)=936,a(5)=68408496,... (I stripped away
the integer part 1/a(0)+1/a(1)=2 from the beginning of the sum of inverses).

Now, because the ratios a(n+1)/a(n) are always integers, the integer part
of the sum of inverses of the terms of A112373 clearly has
the form (*) with a=2, b=6, c=78, d=73086,...

But, when I calculate the convergents of the continued fraction (**)
using  these values, I get the values 0.5, 0.4285714286, 0.4277879342,
0.4277879235,... which are decreasing and not towards the correct
value of 0.5844...

Since many of Euler's manipulations are purely formal, it seems to me that
his formula may not apply for these values (essentially he was evaluating
an alternating series, whereas (*) has all positive terms after inserting
minus signs in the formula on p.26). If somebody else has time to check
these numbers, please let me know.

In any case, the standard continued fraction for the sum of inverses of
A112373(n), with all the beautiful properties found by Paul Hanna and
Jeffrey Shallit, has a completely different form, and so requires a
different proof. It seems like an ideal amusement for my Christmas holiday!

Best wishes to all of seqfan over the festive season.
Andy

On Sat, 10 Dec 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
>
>