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The following series seems convergent for all $s\in \mathbb{C}$:

$$\displaystyle f(s):=\sum_{n=1}^\infty \frac{(-1)^n}{(n+s)^{n+s}}$$

The function itself does not appear to have any real or complex zeros, however numerical evidence suggests that all zeros of:

$$f(s) \pm f(1-s)$$

reside on the line $\Re(s)=\frac12$. Their density apparently increases in a quite regular manner.

enter image description here

1) Could this be proven? When using the finite series $\sum_{n=1}^N$, the claim seems to hold for each $N$.

2) Could there exist a functional equation between $f(s)$ and $f(1-s)$? The only relation I found so far is that $f(-1)=f(1)$.

EDIT: A partial result on the second question. Based on Joro's findings below and realising that for integers and half-integers an infinite number of terms could be cancelled out when adding or substracting $f(s)$ and $f(1-s)$, the following reflective formulae hold:

For all integers $s \in \mathbb{Z}$:

$$f(s)+f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

For all half-integers $s+\dfrac{1}{2}$ with $s \in \mathbb{Z}$:

$$f(s)-f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

The non-alternating series $\displaystyle g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$ has one formula for integers and half-integers:

$$g(s)-g(1-s) +\sum_{n=0}^{2(s-1)} \frac{1}{(s-n)^{s-n}} =0$$

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    $\begingroup$ If you want to define this for all $s\in\mathbb C$ directly by the series, you have to be more precise about the powers (a choice of logarithm is involved). $\endgroup$ – Christian Remling Oct 4 '15 at 15:42
  • $\begingroup$ Christian, you are referring to the branch cut of the natural log, right? For the Maple package I used to check convergence for complex values, this is the principle branch between $\pi$ and $-\pi$. $\endgroup$ – Agno Oct 4 '15 at 16:14
  • $\begingroup$ Yes, $(n+s)^{n+s}=\exp ((n+s)\log (n+s))$, and you have to say which value of $\log$. $\endgroup$ – Christian Remling Oct 4 '15 at 16:44
  • $\begingroup$ I doubt my computation is correct, but one should expect some relation like $f(1-s)=-f(-s)+\dfrac{1}{(1+s)^{(1+s)}}$ to hold. $\endgroup$ – Sylvain JULIEN Oct 4 '15 at 16:46
  • $\begingroup$ If you always take values from one fixed interval of length $2\pi$ for the imaginary part of the $\log$, then you have no chance of producing a holomorphic $f$. $\endgroup$ – Christian Remling Oct 4 '15 at 16:48
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EDIT Due to confusing XRay and programming mistake, I erroneously claimed real zeros. Root finding didn't found any zeros off the line.


Very partial answer for functional equation.

For integer $s$, $\{1,f(s),f(1-s)\}$ appear linearly dependent over the rationals with high precision.

Let $K=f(s),K_2=f(1-s),K_3=1/s^s,K_4=(K-K_2+K_3)^2$.

$$s=2, 1+4K+4K_2=0$$ $$s=3, 131-108K-108K_2=0$$ $$s=4, 36059+ 6912K+ 6912K_2=0$$ $$s=5, -695877463+ 21600000K+ 21600000K_2=0$$ $$s=6, 168087904001+ 583200000K+ 583200000K_2=0$$ $$s=7, -1639334733641495543+ 480290277600000K+ 480290277600000K_2=0$$

And in addition for $s=3/2, K^{2} - 2 \, K K_{2} + K_{2}^{2} + 2 \, K K_{3} - 2 \, K_{2} K_{3} + K_{3}^{2} - 2=0$.

...And with the help of Agno's observation, $s=5/2,8503056K_4^4 - 61096032K_4^3 + 202411224K_4^2 - 171663192K_4 + 252460321$

I suspect for $s$ half integer there is non-trivial algebraic dependency.

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  • $\begingroup$ Really nice, Joro! The coefficients 4, 108, 6912,... seem to follow oeis.org/A107048 , so the next one for $s=6$ would be $\pm$2332800000. The constants at the beginning don't seem to appear in a Sloane integer series. $\endgroup$ – Agno Oct 5 '15 at 7:22
  • $\begingroup$ @Agno for $s=6$ I get: $$168087904001+ 583200000*f(6)+ 583200000*f(1-6)=0$$ which is 4 times of your conjecture. $\endgroup$ – joro Oct 5 '15 at 7:43
  • $\begingroup$ Thanks Joro. Note that I started experimenting with the function $g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$, but that also induces a single complex pair of zeros off the critical line. The alternating version $f(s)$ doesn't seem to have any complex zeros off the line and i.m.o. had therefore more 'beauty'. Also tried to find a pattern linking integer values $g(n)$ and $g(1-n)$ and I believe it has exactly the same 'weights' ($b$) as you found for $f(s)$ however the initial offset values ($a$) are different: $1,9,139,36571,-468908713,136638677249$ for $n=1..6$ and $a+b*g(n)-b*g(1-n)$. $\endgroup$ – Agno Oct 5 '15 at 13:39
  • $\begingroup$ @Agno I consider significantly changing the question after a counterexample bad practice IMHO. $\endgroup$ – joro Oct 5 '15 at 14:50
  • $\begingroup$ Joro, I am sorry when I have upset you by making this change. No bad intentions here and I made explicit comments about the change in the edit reason as well as in the OP text. Having said that: are you absolutely sure the graph with the real zeros is correct? I have tried to replicate it and it keeps looking differently with no zeros. For the function $f(s)-f(1-s)$ you would expect zeros on the positive real axis if they show up on the negative axis as well (as in your graph). I get imaginary values (I guess Maple makes a certain branch cut here) for values of $s<-1$. $\endgroup$ – Agno Oct 5 '15 at 15:41

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