A123726 Denominators of fractional partial quotients appearing in a continued fraction for the power series Sum_{n>=0} x^(2^n - 1)/(n+1)^s.
1, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 36, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 49, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 36, 1, 4
Offset: 0
Examples
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.
Links
- Antti Karttunen, Table of n, a(n) for n = 0..65537
Programs
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Magma
[1] cat [(Valuation(n, 2) + 1)^2: n in [1..50]]; // G. C. Greubel, Nov 01 2018
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Mathematica
Join[{1}, Table[(1 + IntegerExponent[n, 2])^2, {n, 1, 50}]] (* G. C. Greubel, Nov 01 2018 *) -
PARI
{a(n)=denominator(subst(contfrac(sum(m=0,#binary(n),1/x^(2^m-1)/(m+1)),n+3)[n+1],x,1))} -
PARI
/* a(n) = (A007814(n)+1)^2: */ {a(n)=if(n==0,1,(valuation(n,2)+1)^2)}
Formula
a(n) = (A007814(n) + 1)^2 = A001511(n)^2 for n>=1, with a(0)=1, where A007814(n) is the exponent of the highest power of 2 dividing n.
Multiplicative with a(2^e) = (e + 1)^2, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Jul 31 2018
From Amiram Eldar, Dec 29 2022: (Start)
Dirichlet g.f.: zeta(s)*(4^s+2^s)/(2^s-1)^2.
Sum_{k=1..n} a(k) ~ 6*n. (End)
Extensions
Ref to A001511 added by Franklin T. Adams-Watters, Dec 22 2013
Comments