"A dream" of a series :-)
Gottfried Helms
Annette.Warlich at t-online.de
Thu Jun 5 15:24:00 CEST 2008
An analysis of the sequence with using more terms (currently 1024)
unfortunately exhibits some irregularites, so the pure sequence
is not such a dream as I thought in the beginning.
These irregularities may be corrected by correcting the
numerators, too - but I confess, that the charme of
the sequence has a certain loss.
What seems still unconditionally valid is, having A(n)
the denominator of the n'th term of the sequence of cofactors
C(n) at x^n in their most reduced form:
C(n) = N(n)/A(n) // where N(n) may be signed
1a) A(p) = p
1b) A(p^e*q^f*r^g) = p^E*q^F*r^G
meaning: A(n) has the same primefactors as n
-------------------------------------------------
2) A(p^e) = p^cy(e,p)
where cy(e,p) is the cyclotomic expression
cy(e,p) = (p^e-1)/(p-1)
Also it seems, that for n=p*q ( p and q different primes)
it is unconditionally
3) A(p*q) = p^q * q^p
= (p^(1/p) * q^(1/q))^n
--------------------------------------------------
For more primes there occur irregularites, but in most
of the checked cases we find
3b) A(p*q*r) = p^(q*r) * q^(p*r) * r*(p*q)
= (p^(1/p) * q^(1/q) * r(1/r))^n
To get regularities one may correct N(n)/A(n) by multiply
some powers of p,q or r to N(n) and A(n) simultaneously.
For A(p^e*q) there occur irregularities.
The general term seems, however, to be
4) A(p^e * q) = (p^cy(e,p))^q * q^(p^e)
Again the irrgularities may resolved by extending A(n)
and N(n) by some powers of p
5) A(p^e * q^f) = (p^cy(e,p))^q * q^(p^e)
p^(cy(e,p)*q^f )* q^(p^e*cy(f,q))
Again the irrgularities may resolved by extending A(n)
and N(n) by some powers of p. But note, that up to n=1024
I could check only very few examples.
-------------------------------------------------------
As I said, these irregularities reduce the charme of
the sequence much. I didn't analyze the sequence of
nurmerators (possibly including the correction factors)
yet. I think, only if the corrections lead to some
interesting composition-rule for the numerators, I would
give that sequence that honor, which I did, when I looked
at the n=1 to n=64 terms only...
We'll see...
It took 10 minutes to determine 1024 terms, with some
cubic or quartic growthrate in time. If someone had
an idea how to improve this significantly, please tell me.
Gottfried
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