[seqfan] Re: Fw: Closed form?

[David Wilson]

Prof. Dr. Alois Heinz wrote:
> I have no other "closed form" (yet) but your observation is
> very interesting.  It seems that there is a constant c=0.9233..
> with a(n)/f(n) -> c for n -> oo.
>
> Digits:=70:
> f:= proc(x) option remember;
>             evalf (sum (x^k/k!/2^binomial(k,2), k=0..infinity)) end:
> g:= proc(b, n) option remember; local t;
>    if b<0 then 0 elif b=0 or n=0 then 1
>  elif b>=n then add (g(b-t, n) *binomial (n+1, t) *(-1)^(t+1), t=1..n+1)
>  else g(b-1, n) +g(2*b, n-1) fi end:
> a:= proc(n) local t; t:= nops (convert (n, base, 2)); g(n /2^(t-1), t) end:
> for p from 25 to 30 do m:= 10^p: print (a(m)/f(m)) od:
>
> .9233 049132867869282854328449063186689783879094200987655866432714032601
> .9233 055492034789360128094855433414334325830219927073909182819254619767
> .9233 101656306179062025549798530255741041873743718112821787767740848994
> .9233 086902130197332283162644294024178639791138718439995819446220587922
> .9233 045005693638679516587374502832667457864006480990249814586610979267
> .9233 069322864489830441904416767875050659436261404406133336182581267821
>
>   
I added the spaces in the numbers above to exphasize my observations.

At n = 10^30, n is around 1359 digits (according to your calculations). 
For such large n, I would expect the jaggies in a(n) to have smoothed 
out, and at least the initial, say, 100 digits of a(n) to be growing in 
a steady fashion, similarly for f(n). The jaggies in a(n)/f(n) in the 
5th digit (it maxes at n = 10^27) therefore seem suspicious to me. I 
would expect a(n)/f(n) to drift in a constant direction, either up or 
down, in [10^25,10^30] range. I suspect roundoff errors, etc, but still, 
the fact that we get .9233 over this huge range almost vindicates my 
suspicions.

Thank you for your interest. If you make any discoveries of interest, 
please feel free to publish, I won't ask for attribution.