[seqfan] Inquality

zbi74583_boat at yahoo.co.jp zbi74583_boat at yahoo.co.jp
Thu Jul 19 03:54:13 CEST 2018


        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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