A369241 Number of representations of 2^n - 1 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 0, 2, 1, 3, 0, 2, 2, 6, 0, 13, 1, 13, 0, 15, 0, 24, 1, 49, 4, 47, 0, 156, 6, 129, 0, 441, 1, 616

Offset: 0

Author

Antti Karttunen, Jan 21 2024

Keywords

• nonn
• hard
• more

Comments

Any solutions for odd cases must have p = 3, with q and r > 3, because A000225(2n-1) == 1 (mod 3), while on even n, 2^n - 1 is a multiple of 3. This explains why the odd bisection grows much more sluggishly than the even bisection.

Question 2: Is there an infinite number of 0's in this sequence? See also comments in A369055.

Crossrefs

Cf. A000225, A369054, A369055, A369248.

Cf. also A369242, A369245.

Programs

PARI

A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
search_for_3k1_cases(n) = if(3!=(n%4), 0, my(p = 5, q, c=0); while(1, q = (n-(3*p)) / (3+p); if(q < p, return(c), if(1==denominator(q) && isprime(q), c++; write("b369241_by_solutions_of_odd_bisection_to.txt", n, " ", 3*p*q))); p = nextprime(1+p)));
A369241(n) = if(n%2, search_for_3k1_cases((2^n)-1), A369054((2^n)-1));

Formula

a(n) = A369054(A000225(n)).

For n >= 2, a(n) = A369055(2^(n-2)).