[seqfan] Re: Inquality

[M. F. Hasler]

 Yasutoshi,
I think you cannot expect to get a valid result using "Mod 40" (in your
terms).
40 is missing the prime factor 3, which is important for the pattern of the
primes.
Try at least "Mod 120".

PS: My explanation for A068228 does not use a "Theorem" as you cite it,
but a concrete observation (number of gaps 12 up to a given limit).
You must be careful when trying to transpose a given explanation to a
different case.

Maximilian

On Thu, Jul 19, 2018 at 3:54 AM, <zbi74583_boat at yahoo.co.jp> wrote:

>         Hi  Seqfans    Once I posted the following sequence
>     a(n)=Prime(n)+Prime(n-M_3(Prime(n)))  Mod 4  Where if m=0,1,2 Mod 3
> then M_3(m)=0,1,-1
>
>
>        a(n) : 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
> 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
>
>     It has more easy four sequences as follows
>     1. a(m)=Prime(n-1)  Mod 4       where    b(m)=Prime(n)  b(i)=ith Prime
> such that 1  Mod 12       a(m) : 3,3,3,3,1,3,3,3,3,3,3,3     2.
> a(m)=Prime(n+1)  Mod 4       where    b(m)=Prime(n)  b(i))=ith Prime such
> that 5  Mod 12       b(m) : 3,3,3,3,3,1,3,3,3,3,3,3
>     3. 9a(m)=Prime(n-1)  Mod 4       where    b(m)=Prime(n)  b(i)=ith
> Prime such that 7  Mod 12       a(m) : 1,1,1,1,1,1,1,1,1,3,3,1
>     4. a(m)=Prime(n+1)  Mod 4       where    b(m)=Prime(n)  b(i)=ith Prime
> such that 11  Mod 12       a(m) : 1,1,1,1,1,1,1,1,1,1,1,3
>     The inequality of the number of terms of 3 and 1 has an easy
> explanation    Read Hasler's comment on A068228 and A040117
>     I summarize  them for sequences as follows and the four equations of
> the right side are for the sequence as follow    b(n)=Prime(n)+Prime(n+M_3(Prime(n)))
>      1 : 3 = 8 : 2,6 = 1 : 2          1 : 3 = 4 : 6,10 = 1 : 2       1 : 3
> = 8 : 2,6 = 1 : 2          1 : 3 = 4 : 6,10 = 1 : 2       3 : 1 = 8 : 2,6 =
> 1 : 2          3 : 1 = 4 : 6,10 = 1 : 2       3 : 1 = 8 : 2,6 = 1 : 2
>     3 : 1 = 4 : 6,10 = 1 : 2    For example  the first description
> represents that {number of terms of 1} : {number of terms of 3} = {gap is
> 8} : {gap is 2 or 6} = 1 : 2   The right side of the first line represents
> that {number of terms of 1} : {number of terms of 3} = {gap is 4} : {gap is
> 6 or 10} = 1 : 2
>
>     It is possible to explain A103271 using the same method
> a(n)=Prime(n)+Prime(n+1) Mod 4    a(n) : 0,0,2,0,2,0,2,0,0,0,2,0,2,0,0
> It has two sequences as follows
>     1. a(m)=Prime(n+1)  Mod 4         where  b(m)=Prime(n)  b(i)=ith Prime
> such that 1 Mod 4 a(m) : 3,1,3,3,1,3,3,3,3,1
>     2. a(m)=Prime(n+1)  Mod 4         where  b(m)=Prime(n)  b(i)=ith Prime
> such that 3 Mod 4 a(m) : 1,3,1,3,1,3,1,1,3,1
>     Explanation for sequence 1    1 Mod 4 means 1 or 5 Mod 12  hence the
> ratio of numbers of terms of 1 and 3 is the ratio of the right side of the
> first line on the summary and the left side of the second  line on the
> summary  They are 1 : 2
>     Explanation for sequence 2    3 Mod 4 means 7 or 11 Mod 12  hence the
> ratio of numbers of terms of 1 and 3 is the ratio  of the right side of the
> third line on the summary and the left side of the fourth line on the
> summary  They are 1 : 2
>     Difficulty 1    The phenomenon of A103271 and Oliver and
> Soundararajan's   result are the same in the essence  because the
> difference is only Mod 4 and Mod 10    I tried to explain it using the same
> method  and Mod 40 for example  but I didn't succeeded    Could anyone tell
> me how to do it?
>     Difficulty 2    For the sequence    b(n)=Prime(n)+Prime(n+M_3(Prime(n)))
> Mod 4   The explanation says that the ratio of 2 and 0 is 1 : 2    Indeed
> the sequence is    b(n) : 0,2,2,2,2,2,2,0,0,2,2,2,2,0,0,0,2,2,0,2,2,0,2,2
>   The ratio is 2 : 1  The explanation is not correct    Hasler's
> explanation depends on the following theorem    T.Gap  If Primes are small
> then gap is   also small    I think the theorem must be more exact
>     In Mathematics  if x is similar to y  then they are the same    Are
> S_0 and A103271 and Oliver and Soundararajan's result the same?
>
>
>     Yasutoshi
>
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