[seqfan] Theorems on finite sequence of primes

[Vladimir Shevelev]

Dear SeqFans,

I proved the following (hoping new) result.
Theorem 1. Let q==2 (mod 3) be a fixed prime and for 
a prime p==1 (mod 3), A261029(2*q*p)>1. Then p<(2*q)^2. 

Proof. Recall that in A261029 we consider number of representations 
of nonneative numbers by the form: F(x,y,z)=x^3+y^3+z^3-3*x*y*z 
with the conditions 0<=x<=y<=z, z>=x+1. 

Note that F(x,y,z)=(x+y+z)*G(x,y,z), where G(x,y,z)=
x^2+y^2+z^2-x*y-x*z-y*z. It is easy to see that G==0 or 1(mod 3). 
Therefore, G differs from 2,2*p and q*p. In case, when G=1,
 x+y+z=2*q*p, we have the representation 2*p*q = F(k-1,k-1,k), 
where k=2(p*q+1)/3. 

Thus more representation one can obtain only in case G=p, x+y+z=2*q.
Denote by a=z-x>=1, then y=2*q-2*x-a. Then G=p yields 
9*x^2+(9a-12q)*x+(3*a^2-6*a*q+4q^2-p)=0. 
Solving this equation gives x=(4*q-3*a+-sqrt(4*p-3*a^2))/6. 
Since x>=0, the choice of minus is possible only if p<=
((4*q-3*a)^2-3*a^2)/4 or, since 1<=a<2*q, p<4q^2.
Now choose plus. Let b=sqrt(4*p-3a^2). The condition y>=x yields
b<=a. So p=(3*a^2+b^2)/4<=a^2=(z-x)^2<4q^2. 

It is left to show that the case G=2q, x+y+z=p is impossible.
Indeed, in this case x=(2*p-3*a+-sqrt(8*q-3*a^2))/6. 
Let c=sqrt(8*q-3*a^2), i.e., q=(3*a^2+c*2)/8. Since q is prime,
then it should be  2^3||3*a^2+c^2. But it is easy to show that 
it is impossible for any integers a,c. QED 
In the same way one can prove 
Theorem 2. If p,q == 1 (mod 3) are prime and A261029(2*q*p) > 2,
then sqrt(q)/2 < p < 4*q^2. 
These theorems are used for proof of the finiteness and completeness
of sequences that I submitted: A272381, A272382, A272384, A272404,
A272406, A272407, A272409.

Best regards,
Vladimir