1,871 reputation
1517
bio website go.helms-net.de
location Germany
age 61
visits member for 4 years, 1 month
seen 3 hours ago
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 factor-analysis and a matrix-orientated calculator MatMate). Around 2002 I came in contact with the internet-community in math-newsgroups and could improve my Collatz-discussion. Next subject was the Bernoulli-numbers, then integer matrices and since 2006 the problem of iterated exponentiation aka tetration. Serving half-term 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 profile-page.
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 $M=2^m-1$ 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.helms-net.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 Hamming-weight)
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-/totient-function four times iteratively -and of course such an iteration reduces the set of residue-classes 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