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mds
  • Member for 7 years, 1 month
  • Last seen more than 2 years ago
4 votes

Dirichlet Convolution of the Mobius Function with Itself

4 votes

Exponential Generating Function Transformation

3 votes

An integral formula for the reciprocal gamma function

2 votes

Stirling numbers of the second kind: How to obtain a recurrence relation from a generating function?

2 votes
Accepted

What is known about the transformation of a power series in which $z^n$ is replaced with $z^{n^2}$?

2 votes

Generating function for gamma function (or factorial)

2 votes
Accepted

On the Lambert series for the Möbius function

2 votes

Text about connections between complex analysis and partition theory?

2 votes

Simplify $\Gamma\left(\frac27\right) \Gamma\left(\frac{11}{42}\right)/\;\Gamma\left(\frac1{21}\right)$ to elementary terms

1 vote

Questions related to Moebius Transform of Characteristic Function of the Primes

1 vote

Dirichlet convolution k times.

1 vote
Accepted

Möbius inversion formula for two functions f(x) and g(x)

1 vote

Generating function for the factorial sequence

1 vote

Multiple Dirichlet convolutions

1 vote

Is there a generating function for $\sqrt{n}$?

1 vote

Computing Residues of Contour Integral involving Gamma Function

1 vote

Is there a complex variant of Möbius' function?

1 vote

Ordinary generating function for Bernoulli polynomial

1 vote
Accepted

Transforming Exponential to Ordinary Generating Functions

1 vote

transforming ordinary generating function into exponential generating function

1 vote
Accepted

Positivity of an alternating series.

1 vote

Sum of almost-prime zeta functions

1 vote

Squared modulus of Dirichlet eta function.

1 vote

$(f'/f)(s)$ and $f(s)$

1 vote

Dirichlet series expansion?

0 votes

Möbius inversion formula for two functions f(x) and g(x)

0 votes

Dirichlet series with logs&zeta function: $f(n)= \sum_{d\mid n} \frac{\log(d)}d$

0 votes

Stirling number and cycles

0 votes

Divisor sum or möbius inversion of the Jacobi Symbol?

0 votes

$\sum_{n = 1}^\infty (-1)^n \frac{H_n}{n^s}$ in terms of $\sum_{n = 1}^\infty \frac{H_n}{n^s}$