# [seqfan] Re: help with a sequence

Jamie Morken jmorken at shaw.ca
Thu Aug 4 13:44:52 CEST 2016

```Hi,

Yep you are right it is quarter-squares if N is all odd numbers, even numbers, or all positive integers too.

For your question about if Z has any significance, ie for the sequence N = 2,3,5,7,11

Z=3

(7-3)/2=2
(11-5)/2=3
(11-7)/2=2

The significance is related to something else I was working on, ie for each of those three
equations if you replace the minus sign with a plus sign then you get this:

(7+3)/2=5
(11+5)/2=8
(11+7)/2=9

So 5,8,9 are the centerpoints between 7,3 and 11,5 and 11,7 respectively.

Now if you ask the question how many of these equations does a given
centerpoint have, then that is what the significance would be since it shows
that there are peaks in the centerpoint count on numbers that have the most
unique prime factors ie primorials.

For example the centerpoint 8 has 2 valid equations using primes for N:
(11-5)/2=3
(13-3)/2=5

ie. here is the count of how many equations all the centerpoints from 0 to 216 have, and
centerpoint 210 a primorial has the most equations.

0    0
1    0
2    0
3    0
4    0
5    1
6    0
7    0
8    2
9    1
10    2
11    0
12    2
13    0
14    1
15    1
16    2
17    0
18    3
19    0
20    2
21    1
22    1
23    0
24    5
25    0
26    1
27    0
28    0
29    0
30    5
31    0
32    1
33    0
34    1
35    0
36    5
37    0
38    0
39    1
40    1
41    0
42    6
43    0
44    1
45    1
46    1
47    0
48    5
49    0
50    2
51    0
52    0
53    0
54    5
55    0
56    2
57    0
58    0
59    0
60    10
61    0
62    0
63    0
64    1
65    0
66    8
67    0
68    0
69    1
70    2
71    0
72    6
73    0
74    0
75    0
76    2
77    0
78    8
79    0
80    0
81    1
82    0
83    0
84    10
85    0
86    1
87    0
88    0
89    0
90    12
91    0
92    1
93    0
94    0
95    0
96    7
97    0
98    0
99    1
100    2
101    0
102    7
103    0
104    1
105    1
106    1
107    0
108    7
109    0
110    1
111    1
112    0
113    0
114    8
115    0
116    1
117    0
118    0
119    0
120    16
121    0
122    0
123    0
124    0
125    0
126    11
127    0
128    0
129    1
130    1
131    0
132    8
133    0
134    1
135    0
136    0
137    0
138    6
139    0
140    1
141    0
142    1
143    0
144    12
145    0
146    0
147    0
148    0
149    0
150    13
151    0
152    0
153    0
154    1
155    0
156    10
157    0
158    0
159    0
160    2
161    0
162    11
163    0
164    0
165    1
166    0
167    0
168    13
169    0
170    2
171    0
172    0
173    0
174    10
175    0
176    2
177    0
178    0
179    0
180    16
181    0
182    0
183    0
184    0
185    0
186    13
187    0
188    0
189    0
190    0
191    0
192    6
193    0
194    1
195    1
196    2
197    0
198    9
199    0
200    1
201    0
202    1
203    0
204    12
205    0
206    0
207    0
208    0
209    0
210    26
211    0
212    0
213    0
214    0
215    0
216    9

Here is a graph showing this too:

zoomed in: (shows the primorial peaks)
http://imgur.com/x7yDW1e

zoomed out: (shows the primorial multiple bands forming)
http://imgur.com/gcw7S39

cheers,
Jamie

----- Original Message -----
From: Bob Selcoe
rselcoe at entouchonline.net

Tue Aug 2 12:46:08 CEST 2016
Subject: [seqfan] Re: help with a sequenceHi 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
>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
```

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