OFFSET
1,1
COMMENTS
Each term is necessarily even and 3 < p < q in the formula n = p+q = pi(p)*pi(q). Indeed, assuming p<=q, if p=2 then n = 2+q = pi(2)*pi(q) = pi(q) < q. Inequality p > 3 easily follows from prime(k) > k*log(k) and if p=q then 2*p = pi(p)^2 with no solution.
Primes p,q can only occur for a finite number of terms n (see comments in A273286).
Conjecture: the sequence is infinite and each term has only one decomposition into a sum of suitable primes p,q.
From David A. Corneth, Jun 28 2016: (Start)
Pi(p) and pi(q) seem dependent on each other. Below is a small list of pi(p), the least corresponding pi(q) and the largest corresponding pi(q). If a value of pi(p) isn't listed, no terms are formed with it.
3, 4, 8
4, 24, 30
6, 164, 166
8, 1051, 1051
9, 2624, 2676
12, 40027, 40112
Can these bounds on pi(q) be expressed in terms of pi(p)? (End)
LINKS
Giuseppe Coppoletta, Table of n, a(n) for n = 1..43
Eric Weisstein's World of Mathematics, Rosser's Theorem
Pierre Dusart, Estimates of some functions over primes without R.H., arXiv:1002.0442 [math.NT], 2010.
FORMULA
EXAMPLE
12 is a term because 12 = 5 + 7 = pi(5) * pi(7).
MATHEMATICA
Select[Range[10^3], Function[n, MemberQ[Times @@ # & /@ PrimePi@ Select[Transpose@ {#, n - #} &@ Range[Floor[n/2]], Times @@ Boole@ PrimeQ@ {First@ #, Last@ #} == 1 &], n]]] (* Michael De Vlieger, Jun 29 2016 *)
PROG
(Sage)
def sol(n):
return [k for k in divisors(n) if k^2<= n and is_prime(n-nth_prime(k)) and k*prime_pi(n-nth_prime(k))==n]
N=25000
v=[n for n in range(2, N, 2) if len(sol(n))>0]
print('A272862 =', v)
list_pi=flatten([sol(n) for n in range(2, N, 2) if sol(n)])
print('list_pi(p) =', list_pi)
CROSSREFS
KEYWORD
nonn
AUTHOR
Giuseppe Coppoletta, Jun 19 2016
EXTENSIONS
More terms from David A. Corneth, Jun 28 2016
STATUS
approved