|
|
A288161
|
|
Denominator of half moments of Rvachëv function.
|
|
2
|
|
|
2, 18, 6, 1350, 270, 23814, 17010, 65063250, 7229250, 9762090030, 4437313650, 8267713725521250, 635977978886250, 81188783595533250, 297692206516955250, 22510683177794610356250, 1564913803803903393750, 40011216302189267004656036250, 10529267447944543948593693750
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
a(n) is equal to the denominator of the integral over (0,1) of n*t^(n-1)*up(t).
These numbers are the half moments of the Rvachëv function. The Rvachëv function is related to the Fabius function, up(x)=F(x+1) for |x|<1 and up(x)=0 for |x|>=1.
The sequence of numerators is not in the OEIS because it appears t coincide with A272755: Numerators of Fabius function F(1/2^n). In fact d(n) = n! 2^binomial(n,2)F(1/2^n). The coincidence depends on the fact that n! 2^binomial(n,2) divides the denominator of F(1/2^n). It is true that 2^binomial(n,2) divides this denominator, but I do not see any reason for n! to divide this denominator.
|
|
LINKS
|
|
|
FORMULA
|
Recurrence d(0)=1; d(n)=Sum_{k=0..n-1}(binomial(n+1,k)d(k))/((n+1)*(2^n-1)) with a(n) are the denominators of d(n).
It may also be defined to be the only sequence d(n) with d(0)=1 and such that the function f(x)=Sum_{n>=0} d(n) x^n/n! satisfies x*f(2x)=(e^x-1)*f(x).
|
|
EXAMPLE
|
The rationals d(n) are 1/2, 5/18, 1/6, 143/1350, 19/270, ...
|
|
MATHEMATICA
|
d[0] = 1;
d[n_] := d[n] =
Sum[Binomial[n + 1, k] d[k], {k, 0, n - 1}]/((n + 1)*(2^n - 1));
Table[Denominator[d[n]], {n, 1, 20}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,frac
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|