OFFSET
0,9
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.
PROG
(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)));
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Antti Karttunen, Jan 21 2024
STATUS
approved