Seriously disagreement

Richard Mathar mathar at
Wed Sep 3 14:34:02 CEST 2008

shows that it counts primes generated from a list of 2x^2-1, x<=10^n that survive a
sieving procedure. The virgin starting list is A056220 = 2x^2-1 without the first term,
 = 2*10^2-1.
so 161/7=23 is the prime that is put into the list first but last before 199.
       46, 2262, 148933,...
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Date: Wed, 3 Sep 2008 13:33:51 -0400
From: "Alexander Povolotsky" <apovolot at>
To: "Richard Mathar" <mathar at>
Subject: Re: Seriously disagreement
Cc: seqfan at
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---------- Forwarded message ----------
From: Bernhard Helmes <bhelmes at>
Date: Sep 3, 2008 1:14 PM

I have added a description of the algorithms and the needed mathematical theory.

My mathematical explanation could be better formulate.
 If you have questions, ask me.

Nice Greetings from the primes
--         Development of Algorithmic Constructions      Jugglevideos         personal web page
Tel.: 0241 / 99 77 55 22 in Germany (0049)

There is a new page, SubmitC.html (linked from the usual
Submit new seq. or comment page) for making CHANGES
to a sequence.

But if you are just ADDING something, or changing the terms,
please use SubmitA.html instead - it is much easier to process.

Please use the new page very cautiously, and keep a private
copy of what you send - there will probably be bugs.

Both David Applegate and Russ Cox helped with the new page,
although they should not be blamed for any errors or for
my programming style.

The other major new page is SubmitN.html, for sending in
a new sequence (prenumbered or unnumbered - unnumbered sequences
automatically get assigned a number).  This was announced


pp> From seqfan-owner at  Tue Sep  2 11:08:44 2008
pp> Date: Tue, 02 Sep 2008 11:07:29 +0200
pp> From: Peter Pein <petsie at>
pp> To: Sequence Fans <seqfan at>
pp> Subject: Re: A109795 a(n)= n*(1+floor(n/10)) has arbitrary "10" and can generalize
pp> Seemingly the generating functions are (in Mathematica-syntax):
pp> gf[k_] := (Sum[x^i, {i, 1, k - 1}] + (k + 1) x^k) /
pp>           ((1 - x)^3*Sum[x^i, {i, 0, k - 1}]^2)
pp> Cheers,
pp> Peter

The generating function for floor(n/k) [which is following a recurrence
a(n)=a(n-k)+1] is x^k/[(1-x)(1-x^k)], as for example in A002264, A002265,
The generating function for 1+floor(n/k) is given from there by moving
the entire sequence k places to the left, which is equivalent to dropping
the leading k zeros and the factor x^k in the generating function, which yields
To multiply by n, the generating function must be differentiated with
respect to x (to move one "n" from the power in a(n)*x^n to the front)
and essentialy be post-multiplied by x to move from n*a(n)*x^(n-1) back
to n*a(n)*x^n. So the generating function of the k-th row in the
T(k,n) = n*(1+floor(n/k)) is 

or in fixed fonts ASCII art:


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