[seqfan] help with a sequence

[Jamie Morken]

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