Nov 27 revised expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zerosadded 164 characters in body Nov 26 comment expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros@blabler can you elaborate please ! Nov 26 comment expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zerosbecause the sum of the individual terms is divergent. for instance, the sum of the $\frac{x}{n}$ term is the divergent harmonic series. Nov 25 asked expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros Nov 18 asked expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros Nov 2 comment Integer factoring, prime factors, and a Lambert series.lets begin with $i=1$. $c_{1,m}=1$ for all $m$ is a prime. $c_{1,m}=2$ for all $m$ is a semi-prime. for $8=2.4=2.2.2$ we have 6 factors appearing in the three representations. for $12=2.6=3.4=2.2.3$ we have 8 factors appearing in the four representations. for $i=2$, $c_{2,m}=1$ for all $m=2r$, and $r$ is odd. for $m=4$, we have $4=2.2$, so we have two 2's appearing in the two different representation. for $m=8$, we have $8=2.4=2.2.2$, so we have four 2's appearing the three different representations. for $m=12$, we have $12=2.6=3.4=2.2.3$, so we have three 2's appearing the trepresentations. Nov 2 comment Integer factoring, prime factors, and a Lambert series.exactly !! that's why i said it is misleading in this case ! but in order for the definition of $c_{i,m}$ to be consistent we have to pick a convention ! i am still struggling with a better way to define $c_{i,m}$. Do you have any suggestions !? Nov 2 comment Integer factoring, prime factors, and a Lambert series.maybe the case where $i=1$ is a bit misleading. but, as you stated : $8=2.4=2.2.2$, or $8.1=1.2.1.4=1.2.1.2.1.2$. so maybe i should've said -in this case- $c_{1,m}$ counts the factors of $m$ in each product representation. in the case where $i=2$ , and $m=12$, we have: $12=2.6=3.4=3.2.2$, so 2 appears as a factor 3 times in the different representations of $12$. Nov 2 revised Integer factoring, prime factors, and a Lambert series.edited title Nov 2 revised Integer factoring, prime factors, and a Lambert series.added 2 characters in body Nov 2 asked Integer factoring, prime factors, and a Lambert series. Sep 15 revised Solution of an integral equationadded 30 characters in body Sep 15 asked Solution of an integral equation Sep 6 comment Simplifying a product over roots of unityah.. ok ... yes, your $z$ doesn't lie on the negative real line . Sep 6 comment Simplifying a product over roots of unityI'm afraid I don't take your meaning ! Sep 6 asked Simplifying a product over roots of unity Sep 1 comment A Fourier-type integral of a piecewise functionthe first summation above can be written as: $$\sum_{k=1}^{\infty}\frac{\zeta(k+1)}{k!}\log^{k}(x)$$, and the second summation truncates at the integer nearest to $\log_{2}(x)$. i don't know if this could help ! Sep 1 comment A Fourier-type integral of a piecewise functionUnfortunately, i wasn't able to prove that the integral exists. the summation of the fractional parts is very erratic, but after experimenting with it, it seems to be bounded for all $x$ ! i tried $$\sum_{n=1}^{\infty}\frac{\left \{ x^{1/n} \right \}}{n}=\sum_{n=1}^{\infty}\frac{x^{1/n}-1}{n}-\sum_{n=1}^{\infty}\frac{\left \lfloor x^{1/n} \right \rfloor-1}{n}$$ in order to obtain a bound on the summation, but in vain ! Sep 1 revised A Fourier-type integral of a piecewise functionadded 215 characters in body Sep 1 comment A Fourier-type integral of a piecewise functionFor purely heuristic reasons, i believe it's evaluable.