bio  website  go.helmsnet.de 

location  Germany  
age  61  
visits  member for  4 years, 1 month 
seen  3 hours ago  
stats  profile views  1,148 
Mathematics is only a hobby  although I have done undergrad courses in the 70ties. But part of my job was doing statistics and this kept me near my favourite subject "linear algebra" (programmed factoranalysis and a matrixorientated calculator MatMate). Around 2002 I came in contact with the internetcommunity in mathnewsgroups and could improve my Collatzdiscussion. Next subject was the Bernoullinumbers, then integer matrices and since 2006 the problem of iterated exponentiation aka tetration. Serving halfterm jobs in teaching here at the university I found time to fiddle with that subjects in depth  and found love with the exploratory approach and impulse of the 18'th century numbertheory, namely L.Euler, "the master of us all"...
Due to lack of formal education I've to do my "research" widely on my own  but that's what I just like: to find structure, pattern, laws from the ground.
1d

comment 
Smoothness in Ecalle's method for fractional iterates
Everything is fine. Thank you very much! 
1d

comment 
Smoothness in Ecalle's method for fractional iterates
Please let me see your jpegs. I think my email occurs in my profilepage. 
Aug 25 
comment 
A family Mersenne composite numbers?
:) Nice approach to sort things out... 
Aug 24 
comment 
A family Mersenne composite numbers?
@Joro: true; for some small $t$ the factorizing looked a bit systematic, but in a second view for larger $t$ that simple pattern did not hold. 
Aug 24 
comment 
A family Mersenne composite numbers?
Hmm, so you mean $M=2^m1$ where $m=2^{2t+1}1+2t=4\cdot 2^w+w $. If $m$ is composite, then $M$ must also be composite. So it might be easier to prove that $m$ is composite? 
Aug 19 
accepted  An infinite set of identities using Stirling numbers 1st kind  are they all zero? 
Aug 19 
comment 
An infinite set of identities using Stirling numbers 1st kind  are they all zero?
Very nice, ideed. I suspected some separation into two sums in the manner as you did it with the difference of pairs of $s_1()$ but couldn't get the key entry. I'll have to go through your solution step by step and check the relevant identities with the $s_1$, although I think that this answers my question and it's ok to accept it now. Thank you very much, this has now been an open problem for 3 years... 
Aug 19 
revised 
An infinite set of identities using Stirling numbers 1st kind  are they all zero?
added two more images 
Aug 19 
revised 
An infinite set of identities using Stirling numbers 1st kind  are they all zero?
added images and enhanced the focus of he question 
Aug 19 
revised 
An infinite set of identities using Stirling numbers 1st kind  are they all zero?
added images and enhanced the focus of he question 
Aug 13 
comment 
Does the congruence $a^p \equiv 1 \pmod{b^p}$ with prime $p \ge 5$ force $b \le p$?
While you are preparing a refocused version of the question, perhaps is my older related treatize of interest: go.helmsnet.de/math/expdioph/fermatquotients.pdf 
Jul 30 
accepted  Cesaro(?)/Euler(?)  summation of the $s(p)=\sum_{k=0}^\infty (1)^{H(k)} (1+k)^p$ for $p=1,2,3,…$ (where $H(k)$ is the Hammingweight) 
Jul 17 
awarded  Yearling 
Jul 11 
comment 
Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
My first idea was to apply Euler's phi/totientfunction four times iteratively and of course such an iteration reduces the set of residueclasses much. But the intensity of the discussion around suggests I'm badly missing some other aspect by your question. What is the point that it is not so simple? 
Jul 9 
revised 
Error term for renewal function
added linkt to a more extensive study along my observations 
Jul 2 
awarded  Curious 
Jun 24 
revised 
Error term for renewal function
added 15 characters in body 
Jun 24 
revised 
Error term for renewal function
added 453 characters in body 
Jun 24 
revised 
Error term for renewal function
added 249 characters in body 
Jun 24 
answered  Error term for renewal function 