OFFSET
0,2
LINKS
Antti Karttunen, Table of n, a(n) for n = 0..65537
FORMULA
EXAMPLE
Surprisingly, the following analog of the Riemann zeta function:
Z(x,s) = Sum_{n>=0} x^(2^n-1)/(n+1)^s = 1 + x/2^s + x^3/3^s +x^7/4^s+..
may be expressed by the continued fraction:
Z(x,s) = [1; f(1)^s/x, -f(2)^s/x, -f(3)^s/x,...,f(n)^s*(-1)^e(n)/x,...]
such that the (2^n-1)-th convergent = Sum_{k=0..n} x^(2^k-1)/(k+1)^s,
where f(n) = (b(n)+2)/(b(n)+1)^2 and e(n) = A073089(n+1) for n>=1,
and b(n) = A007814(n) the exponent of highest power of 2 dividing n.
Case s=1.
Sum_{n>=0} x^(2^n-1)/(n+1) = [1; 2/x, -(3/4)/x, -2/x, -(4/9)/x, 2/x,
(3/4)/x, -2/x, -(5/16)/x, 2/x, -(3/4)/x, -2/x, (4/9)/x, 2/x, (3/4)/x,
-2/x, -(6/25)/x, 2/x, -(3/4)/x, -2/x, -(4/9)/x, 2/x, (3/4)/x, -2/x,...].
Note that (2^n-1)-th convergents exactly equal n-th partial sums:
[1;2/x] = 1 + x/2;
[1;2/x,-(3/4)/x,-2/x] = 1 + x/2 + x^3/3;
[1;2/x,-(3/4)/x,-2/x,-(4/9)/x,2/x,(3/4)/x,-2/x] = 1 +x/2 +x^3/3 +x^7/4.
Case s=2.
Sum_{n>=0} x^(2^n-1)/(n+1)^2 = [1; 4/x, -(9/16)/x, -4/x, -(16/81)/x,
4/x, (9/16)/x, -4/x, -(25/256)/x, 4/x, -(9/16)/x, -4/x, (16/81)/x, 4/x,
(9/16)/x, -4/x, -(36/625)/x, 4/x, -(9/16)/x, -4/x, (16/81)/x, 4/x ...].
Note that (2^n-1)-th convergents exactly equal n-th partial sums:
[1;4/x] = 1 + x/4;
[1;4/x,-(9/16)/x,-4/x] = 1 + x/4 + x^3/9;
[1;4/x,-(9/16)/x,-4/x,-(16/81)/x,4/x,(9/16)/x,-4/x]=1+x/4+x^3/9+x^7/16.
Case s=3.
Sum_{n>=0} x^(2^n-1)/(n+1)^3 = [1; 8/x,-(27/64)/x,-8/x,-(64/729)/x,8/x,
(27/64)/x,-8/x,-(125/4096)/x, 8/x,-(27/64)/x,-8/x, (64/729)/x, 8/x,
(27/64)/x,-8/x,-(216/15625)/x, 8/x, -(27/64)/x, -8/x, (64/729)/x ...].
Likewise, the (2^n-1)-th convergents exactly equal n-th partial sums.
It is conjectured that these patterns continue for all s.
PROG
CROSSREFS
KEYWORD
cofr,frac,sign
AUTHOR
Paul D. Hanna, Oct 12 2006
STATUS
approved