unsure of convergence to Integer sequence

[Paul D. Hanna]

Also, if you multiply the rational terms in Geralds' resultant sequence
by 3, you get:
S={3,6,14,42,158,714,3758,22554,151934,1134762,9304142,83068986,802115870
,...}
or
S(n) = A098830(n) + A098830(n+1)
where
A098830(n) = 
Sum_{k=0..n}Sum_{j=0..n-k}(j+1)^k*Sum_{i=0..j}(-1)^(n-k+j-i)*C(j,i)*(j-i)
^(n-k).
 
This is Benoit Cloitre's entry:
http://www.research.att.com/projects/OEIS?Anum=A098830
and equals the antidiagonal sums of Ralf Stephan's array:
http://www.research.att.com/projects/OEIS?Anum=A099594
from which I derived the above formula. 

This is so because Catalan(n) = C(2n,n)/(n+1)
so that Wouter's sum
   Sum(k=1..Inf; k^n/CatalanNumber[k]) 
equals 
   Sum(k=1..Inf; (k+1)*k^n/C(2k,k)) = 
   Sum(k=1..Inf; k^n/C(2k,k))  + Sum(k=1..Inf; k^(n+1)/C(2k,k))  
which is now in terms of the sum that Benoit addresses in A098830.
  
Paul


On Sun, 05 Dec 2004 14:58:51 -0500 Gerald McGarvey
<Gerald.McGarvey at comcast.net> writes:
> Something to note, the terms can be rewritten as follows:
> 
> 2/1+(4*Pi)/(3^2*Sqrt[3])
> 2/1+(16*Pi)/(3^3*Sqrt[3])
> 14/3+(104*Pi)/(3^4*Sqrt[3])
> 14/1+(936*Pi)/(3^5*Sqrt[3])
> 158/3+(10584*Pi)/(3^6*Sqrt[3])
> 238/1+(143496*Pi)/(3^7*Sqrt[3])
> 3758/3+(2265624*Pi)/(3^8*Sqrt[3])
> 7518/1+(40791816*Pi)/(3^9*Sqrt[3])
> 151934/3+(824378904*Pi)/(3^10*Sqrt[3])
> 378254/1+(18471328776*Pi)/(3^11*Sqrt[3])
> 9304142/3+(454350385944*Pi)/(3^12*Sqrt[3])
> 27689662/1+(12169555717896*Pi)/(3^13*Sqrt[3])
> 802115870/3+(352528455936024*Pi)/(3^14*Sqrt[3])
> 2776117230/1+(10980885962741256*Pi)/(3^15*Sqrt[3])
> 92521462766/3+(365967121556087064*Pi)/(3^16*Sqrt[3])
> 
> which contains the following sequence:
>
4,16,104,936,10584,143496,2265624,40791816,824378904,18471328776,45435038
5944,12169555717896,352528455936024,10980885962741256,365967121556087064 
> 
> 
> 
> Gerald
> 
> At 10:41 AM 12/5/2004, wouter meeussen wrote:
> >the Sum(k=1..Inf; k^n/CatalanNumber[k]) can be written in closed 
> form (W. 
> >Gosper, 2000) as
> >
>
>Sum[Hypergeometric2F1[m+1,m+2,m+1/2,1/4]StirlingS2[n,m]/(2m-1)!!/2^m(m+1
)!m!,{m,1,n}]
> >
> >and this simplifies to (for n= 0..14)
> >
> >2+(4*Pi)/(9*Sqrt[3])
> >2+(16*Pi)/(27*Sqrt[3])
> >(2*(567+52*Sqrt[3]*Pi))/243
> >14+(104*Pi)/(27*Sqrt[3])
> >158/3+(392*Pi)/(27*Sqrt[3])
> >238+(15944*Pi)/(243*Sqrt[3])
> >3758/3+(83912*Pi)/(243*Sqrt[3])
> >7518+(1510808*Pi)/(729*Sqrt[3])
> >151934/3+(30532552*Pi)/(2187*Sqrt[3])
> >378254+(228041096*Pi)/(2187*Sqrt[3])
> >9304142/3+(5609264024*Pi)/(6561*Sqrt[3])
> >27689662+(150241428616*Pi)/(19683*Sqrt[3])
> >802115870/3+(483578128856*Pi)/(6561*Sqrt[3])
> >2776117230+(15062943707464*Pi)/(19683*Sqrt[3])
> >92521462766/3+(167337504140872*Pi)/(19683*Sqrt[3])
> >
> >Now, these expression *seem* to huddle uncomfortably close to 
> integers:
> >2,806133
> >3,074844
> >6,995495
> >20,986486
> >79,000346
> 009124
> >1879,002190
> >11276,988463
> >75966,991041
> >567381,021008
> >4652071,037121
> >41534492,955918
> >401057934,821915
> >4164175845,053300
> >46260731383,985200
> >
> >but, the loss of accuracy towards the end troubles me.
> >Can anyone suggest a mathematical basis for this small 'numirical' 
> ?
> >
> >
> >W.
> 
> 
> 
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