[seqfan] Re: more strangeness (A000079 "Powers of 2" and many others)
arndt at jjj.de
Thu Mar 11 14:26:02 CET 2010
... no answer at all on this one.
Just edited approx 27 seqs, including
hopefully all with the keyword 'core'
and definitely all mentioned below
(will do more later).
I dearly hope that further submissions by this
gentleman are met with profound skepticism.
One thing struck me while editing:
there seems to be an inofficial contest
to produce the most awful code to compute
such seqs as '2n+1', '3n', 'n^2+1' etc.
(this is not by Vincenzo but mostly by
one other individual).
* Joerg Arndt <arndt at jjj.de> [Mar 04. 2010 14:51]:
> Dear Vincenzo,
> in seq A000079 you give:
> Except for the first term, number n such that if A=(7/8)*n^4;
> B=(7/16)*n^4; C=(17/16)*n^4; D=(5/4)*n^4; then A^3+B^3+C^3=D^3 [From
> Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Sep 08 2009]
> This is just an obfuscated way of saying that
> (7/8)^3 + (7/16)^3+ (17/16)^3 == (5/4)^3
> Further, the statement has nothing whatsover to do with the sequence
> at hand (plus the initial qualification is plain wrong).
> Kindly remove the comment including the corresponding example which
> For n=2, A=14, B=7, C=17, D=20, and 14^3+7^3+17^3=20^3 [From
> Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jun 25 2009]
> In addition, please do the following search in the database:
> librandi -author:librandi
> There seem to be more examples of relations that need
> to be removed, a quick glance suggests to firstly check:
> A005408 (see my last mail)
> A000668 (!)
> A004526 (! a(n)=floor(n*[sqrt(5)-sqrt(3)]) wrong!)
> A002145 (!)
> (and apparently many more, I just checked 2 of 255 pages)
> best regards, jj
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