[seqfan] Re: help with a sequence
Bob Selcoe
rselcoe at entouchonline.net
Tue Aug 2 12:46:08 CEST 2016
Hi Jamie & Seqfans,
It's an interesting idea; I could certainly see adding, say, the sequence
for N as a set of prime numbers. But I'm not sure why you might want to
apply this particular construct to existing sequences. Why not, for
example, use (y-x)/3 instead?? Does Z have particular significance??
BTW - unless I'm mistaken, we also obtain quarter-squares if N is all
positive integers. I suppose that could be applied to Oppermann also, but
it seems a bit trivial to me.
Cheers,
Bob Selcoe
--------------------------------------------------
From: "Jamie Morken" <jmorken at shaw.ca>
Sent: Monday, August 01, 2016 11:34 PM
To: <seqfan at list.seqfan.eu>
Subject: [seqfan] help with a sequence
> Hi,
>
> Please excuse my lack of correct math terminology, I have a couple
> questions
> about some sequences that I found, and noticed they are already are in
> OEIS,
> but I don't think they are described with this formula, so if it makes
> sense please add it
> to OEIS, or please explain it more to me.
>
> For a partial set of the positive integers called N, for all combinations
> of pairs of integers x and y
> in the set where y is larger than x.
>
> Take the total count of unique formulas (y-x) / 2 = C where C is also in N
> and call this count Z.
>
>
> example with N as sets of prime numbers:
>
> 2,3 has Z=0..
> 2,3,5 has Z=0.. (+0)
> 2,3,5,7 has Z=1.. (+1)
> 2,3,5,7,11 has Z=3.. (+2)
> 2,3,5,7,11,13 has Z=5.. (+2)
> 2,3,5,7,11,13,17 has Z=9.. (+4)
> 2,3,5,7,11,13,17,19 has Z=11.. (+2)
> 2,3,5,7,11,13,17,19,23 has Z=14.. (+3)
> 2,3,5,7,11,13,17,19,23,29 has Z=18.. (+4)
> 2,3,5,7,11,13,17,19,23,29,31 has Z=20.. (+2)
> 2,3,5,7,11,13,17,19,23,29,31,37 has Z=24.. (+4)
> 2,3,5,7,11,13,17,19,23,29,31,37,41 has Z=29.. (+5)
> 2,3,5,7,11,13,17,19,23,29,31,37,41,43 has Z=33.. (+4)
>
> 1,3,5,9,11,14,18,20,24,29,33,37... sequence isn't in OEIS
>
> 1,2,2,4,2,3,4,2,4,5,4,4,... sequence is in OEIS http://oeis.org/A103274
> "Number of ways of writing prime(n) in the form 2*prime(i)+prime(j)"
>
>
> example with N as sets of odd numbers:
>
> 3,5,7 has Z=0..
> 3,5,7,9 has Z=1..(+1)
> 3,5,7,9,11 has Z=2..(+1)
> 3,5,7,9,11,13 has Z=4..(+2)
> 3,5,7,9,11,13,15 has Z=6..(+2)
> 3,5,7,9,11,13,15,17 has Z=9..(+3)
> 3,5,7,9,11,13,15,17,19 has Z=12..(+3)
> 3,5,7,9,11,13,15,17,19,21 has Z=16..(+4)
> 3,5,7,9,11,13,15,17,19,21,23 has Z=20..(+4)
> 3,5,7,9,11,13,15,17,19,21,23,25 has Z=25..(+5)
> 3,5,7,9,11,13,15,17,19,21,23,25,27 has Z=30..(+5)
> 3,5,7,9,11,13,15,17,19,21,23,25,27,29 has Z=36..(+6)
>
> 1,2,4,6,9,12,16,20,25,30,36, is in oeis:
>
> Quarter-squares: floor(n/2)*ceiling(n/2). Equivalently, floor(n^2/4)
> http://oeis.org/A002620
>
> primes mentioned regarding this sequeuce on that page:
> "
> Alternative statement of Oppermann's conjecture: For n>2, there is at
> least one prime between a(n) and a(n+1). - Ivan N. Ianakiev, May 23 2013.
> [This conjecture was mentioned in A220492, A222030. - Omar E. Pol, Oct 25
> 2013]
>
> For any given prime number, p, there are an infinite number of a(n)
> divisible by p, with those a(n) occurring in evenly spaced clusters of
> three
> as a(n), a(n+1), a(n+2) for a given p. The divisibility of all a(n) by p
> and the result are given by the following equations, where m >= 1 is the
> cluster number for that p: a(2m*p)/p = p*m^2 - m; a(2m*p + 1)/p = p*m^2;
> a(2m*p + 2)/p = p*m^2 + m. The number of a(n) instances between clusters
> is 2*p - 3. - Richard R. Forberg, Jun 09 2013
> "
>
> or also it is this sequence too:
> Numbers n such that ceil(sqrt(n)) divides n
> http://oeis.org/A087811
>
> Thanks.
>
> cheers,
> Jamie
>
>
> --
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>
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