Interesting Seq ? Partition in groups with semiprime sums

Max relf at
Sat Jul 2 08:52:26 CEST 2005

zak seidov wrote:
> Dear seqfan gurus,
> i dare submit this sequence with hope
> that someone may wish to answer
> the Qs in it,
> thanks, zak
> %I A109411
> %S A109411 
> 3,1,4,1,1,5,2,3,1,1,13,3,1,3,2,2,2,1,4,6,2,1,6,1,2,2,1,14,4,1,1,1,
> %N A109411 Partition of sequence of natural numbers in
> the minimal 
> groups such that sum of terms in each group is a
> semiprime.
> %C A109411 The partiton starts as this: 
> {1-3},{4},{5-8},{9},{10},{11-15},{16-17},{18-20},{21},{22},{23-35},
> {36-38},{39},{40-42},{43-44},{45-46},{47-48},{49},{50-53},
> {54-59},{60-61},{62},{63-68},{69},{70-71},{72-73},{74},{75-88},
> {89-92},{93},{94},{95},{96-98},{99-103},{104-105}...
> The sequence A109411 gives the number of terms in each
> group. 
> The big question is: is the sequence finite? If a
> group begins with a 
> and ends with b then sum of terms is s=(a+b)(a-b+1)/2

Actually s=(a+b)(b-a+1)/2

> and it's not 
> evident that :
> a) there are a's such that it's impossible to find
> b>=a such that s is semiprime,  

Suppose that a is fixed. We are looking for such b that
either a+b=p and b-a+1=2q,
or a+b=2q and b-a+1=p
for some primes p,q.
p - 2q = 2a-1
2q - p = 2a-1.

Therefore, the question is equivalent to the following:
Given an odd integer n (=2a-1), can it be represented as p-2q or 2q-p where p,q are prime?
I believe the answer is "yes" but the problem may have the same complexity as Goldbach conjecture.


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