CF of Reciprocal Sum of A112373

Sun Dec 11 04:40:28 CET 2005

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