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<DIV>Also, if you multiply the rational terms in Geralds' resultant
sequence by 3, you get:</DIV>
<DIV>S={3,6,14,42,158,714,3758,22554,151934,1134762,9304142,83068986,802115870,...}</DIV>
<DIV>or</DIV>
<DIV>S(n) = A098830(n) + A098830(n+1)</DIV>
<DIV>where</DIV>
<DIV>A098830(n) = </DIV>
<DIV>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).</DIV>
<DIV> </DIV>
<DIV>This is Benoit Cloitre's entry:</DIV>
<DIV>
<DIV><A
href="http://www.research.att.com/projects/OEIS?Anum=A098830">http://www.research.att.com/projects/OEIS?Anum=A098830</A></DIV></DIV>
<DIV>and equals the antidiagonal sums of Ralf Stephan's array:</DIV>
<DIV>
<DIV><A
href="http://www.research.att.com/projects/OEIS?Anum">http://www.research.att.com/projects/OEIS?Anum=</A><A
href="http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A099594">A099594</A></DIV></DIV>
<DIV>from which I derived the above formula. </DIV>
<DIV> </DIV>
<DIV>This is so because Catalan(n) = C(2n,n)/(n+1)</DIV>
<DIV>so that Wouter's sum</DIV>
<DIV> Sum(k=1..Inf; k^n/CatalanNumber[k]) <BR>equals </DIV>
<DIV> Sum(k=1..Inf; (k+1)*k^n/C(2k,k)) = </DIV>
<DIV> Sum(k=1..Inf; k^n/C(2k,k)) + Sum(k=1..Inf;
k^(n+1)/C(2k,k)) </DIV>
<DIV>which is now in terms of the sum that Benoit addresses in
A098830.<BR> </DIV>
<DIV>Paul</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>On Sun, 05 Dec 2004 14:58:51 -0500 Gerald McGarvey <<A
href="mailto:Gerald.McGarvey@comcast.net">Gerald.McGarvey@comcast.net</A>>
writes:<BR>> Something to note, the terms can be rewritten as
follows:<BR>> <BR>> 2/1+(4*Pi)/(3^2*Sqrt[3])<BR>>
2/1+(16*Pi)/(3^3*Sqrt[3])<BR>> 14/3+(104*Pi)/(3^4*Sqrt[3])<BR>>
14/1+(936*Pi)/(3^5*Sqrt[3])<BR>> 158/3+(10584*Pi)/(3^6*Sqrt[3])<BR>>
238/1+(143496*Pi)/(3^7*Sqrt[3])<BR>>
3758/3+(2265624*Pi)/(3^8*Sqrt[3])<BR>>
7518/1+(40791816*Pi)/(3^9*Sqrt[3])<BR>>
151934/3+(824378904*Pi)/(3^10*Sqrt[3])<BR>>
378254/1+(18471328776*Pi)/(3^11*Sqrt[3])<BR>>
9304142/3+(454350385944*Pi)/(3^12*Sqrt[3])<BR>>
27689662/1+(12169555717896*Pi)/(3^13*Sqrt[3])<BR>>
802115870/3+(352528455936024*Pi)/(3^14*Sqrt[3])<BR>>
2776117230/1+(10980885962741256*Pi)/(3^15*Sqrt[3])<BR>>
92521462766/3+(365967121556087064*Pi)/(3^16*Sqrt[3])<BR>> <BR>> which
contains the following sequence:<BR>>
4,16,104,936,10584,143496,2265624,40791816,824378904,18471328776,454350385944,12169555717896,352528455936024,10980885962741256,365967121556087064
<BR>> <BR>> <BR>> <BR>> Gerald<BR>> <BR>> At 10:41 AM
12/5/2004, wouter meeussen wrote:<BR>> >the Sum(k=1..Inf;
k^n/CatalanNumber[k]) can be written in closed <BR>> form (W. <BR>>
>Gosper, 2000) as<BR>> ><BR>>
>Sum[Hypergeometric2F1[m+1,m+2,m+1/2,1/4]StirlingS2[n,m]/(2m-1)!!/2^m(m+1)!m!,{m,1,n}]<BR>>
><BR>> >and this simplifies to (for n= 0..14)<BR>> ><BR>>
>2+(4*Pi)/(9*Sqrt[3])<BR>> >2+(16*Pi)/(27*Sqrt[3])<BR>>
>(2*(567+52*Sqrt[3]*Pi))/243<BR>> >14+(104*Pi)/(27*Sqrt[3])<BR>>
>158/3+(392*Pi)/(27*Sqrt[3])<BR>> >238+(15944*Pi)/(243*Sqrt[3])<BR>>
>3758/3+(83912*Pi)/(243*Sqrt[3])<BR>>
>7518+(1510808*Pi)/(729*Sqrt[3])<BR>>
>151934/3+(30532552*Pi)/(2187*Sqrt[3])<BR>>
>378254+(228041096*Pi)/(2187*Sqrt[3])<BR>>
>9304142/3+(5609264024*Pi)/(6561*Sqrt[3])<BR>>
>27689662+(150241428616*Pi)/(19683*Sqrt[3])<BR>>
>802115870/3+(483578128856*Pi)/(6561*Sqrt[3])<BR>>
>2776117230+(15062943707464*Pi)/(19683*Sqrt[3])<BR>>
>92521462766/3+(167337504140872*Pi)/(19683*Sqrt[3])<BR>> ><BR>>
>Now, these expression *seem* to huddle uncomfortably close to <BR>>
integers:<BR>> >2,806133<BR>> >3,074844<BR>> >6,995495<BR>>
>20,986486<BR>> >79,000346<BR>> 009124<BR>>
>1879,002190<BR>> >11276,988463<BR>> >75966,991041<BR>>
>567381,021008<BR>> >4652071,037121<BR>> >41534492,955918<BR>>
>401057934,821915<BR>> >4164175845,053300<BR>>
>46260731383,985200<BR>> ><BR>> >but, the loss of accuracy
towards the end troubles me.<BR>> >Can anyone suggest a mathematical basis
for this small 'numirical' <BR>> ?<BR>> ><BR>> ><BR>>
>W.<BR>> <BR>> <BR>> </DIV>
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