[seqfan] series coefficients of EllipticK[x]

[Wouter Meeussen]

hi,

I'm half way there!
Thanks to  Jeffrey Shallit <shallit at graceland.uwaterloo.ca>,
who properly identified 

 Fold[Flatten[{  #1,  #2, #1   } ]& , 0 ,(   Range[5]) ]
as
  Table[ Nu2[k],{k, 63}]

with Nu2[k] defined as the number of trailing zero's in the binary expansion
of n :
Thanks, Jeffry!
Mathematica definition:
 Nu2[n_]:=  -1+ Length[Last[Split[#]]]& @IntegerDigits[2n,2]

>From there, we get easily to:
 
FoldList[Plus,1,2+Flatten[Fold[{#1,#2,#1}&,0,Range[5]]]]
 ==
Table[ Nu2[k!]+2k +1,{k,0,63}] 
 ==
{1,3,6,8,12,14,17,19,24,26,29,31,35,37,40,42,48,50,53,55,59,61,64,66,71,73,76,
  78,82,84,87,89,96,98,101,103,107,109,112,114,119,121,124,126,130,132,135,
  137,143,145,148,150,154,156,159,161,166,168,171,173,177,179,182,184}

et voila, 
the denominator of the n-th coefficient in the series decomposition of
EllipticK[x]/Pi is 2^(Nu2[n!]+2n +1).

************************************
bear with me for the numerator, it gets hairy!

the first few terms of the numerator were :

{1,1,9,75,3675,59535,2401245,57972915,13043905875,418854310875,30241281245175,
  1212400457192925}

now, if we apply the well known magic wand (details only under duress):

Table[(  (2n)!/n!/n! /2^w[[n+1]]  )^2 /2^  Nu2[n!] n! ,{n,0,11}]

{1,1,9,75,3675,59535,2401245,57972915,13043905875,418854310875,30241281245175,
  1212400457192925}

or, more compact :

      (2^(-Nu2[n!] - 2*wmn[1 + n])*(2*n)!^2)/n!^3

with the still-to-be-symbolised function wmn[n_]:=w[[n]]  that represents
the list:

w=Flatten[FoldList[{ #2 ,Rest[#1],#2,1+Rest[#1]} &,{0},Range[6]]]
==
{0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,1,2,2,3,2,3,
  3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,1,2,2,3,2,3,3,4,2,3,3,4,
  3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,
  4,5,4,5,5,6,4,5,5,6,5,6,6, ...}

that can also be (trivially) rewritten as :

NestList[Flatten[{First[#]+1,Rest[#],1+#}]&,{0},6]//ColumnForm
and tabulared nicely:
{0,
1,1,
2,1,2,2,
3,1,2,2,3,2,3,3,
4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,
5,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,
6,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,
4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,
7, ...}

Now,
is anyone familiar with this last series?
Then we could wrap it all up.
How well known is Nu2[ n ] ????

wouter.

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