A272862 -id:A272862 - cp's OEIS frontend

A272860 Sums of two primes (in increasing order) when equal to the product of their prime-counting functions.

12, 18, 24, 96, 116, 120, 984, 990, 996, 8408, 23616, 23742, 23850, 24030, 24066, 24084, 480324, 480336, 481344, 3523814, 3523842, 3523884, 3524514, 9557160, 9558030, 9558240, 9558300, 25874592, 25874640, 70119798, 189960894, 189961344, 189962352, 189963594, 189963630, 189969102

Offset: 1

Giuseppe Coppoletta, Jun 19 2016

nonn

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)

			12 is a term because 12 = 5 + 7 = pi(5) * pi(7).

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.

Cf. A272861, A272862, A273286, A000040, A000720.

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 *)

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)

Numbers n = p+q = pi(p)*pi(q) for some primes p and q.
Equivalently, n = i*j = prime(i) + prime(j) for some i,j.
A272862 gives the corresponding terms pi(q) (with q>p). The terms pi(p) are given by A272860 / A272862

More terms from David A. Corneth, Jun 28 2016

A273286 Positive integers n such that n=p+q for some primes p,q with pi(p)*pi(q) = sigma(n).

Original entry on oeis.org

92, 130, 132, 136, 154, 270, 286, 384, 398, 456, 546, 608, 630, 636, 702, 934, 944, 2730, 4394, 4470, 4556, 5544, 12084, 14320, 17572, 22632, 27808, 27930, 31150, 31284, 32534, 36346, 41004, 41544, 42274, 56916, 58552, 61680, 66654, 74826, 86200

Offset: 1

Giuseppe Coppoletta, Jun 20 2016

nonn

Equivalently, integers n such that sigma(n) = i*j for some i,j with prime(i)+prime(j) = n. Each term is necessarily even, otherwise if n is odd n=2+q < sigma(2+q) = pi(2)*pi(q) = pi(q) < q which is absurd. Also p and q cannot be equal, otherwise sigma(2*p) = 3*(p+1) = pi(p)^2 with no solution.
Conjecture: the sequence is infinite and each term has only one decomposition into a sum of suitable primes p,q.
Using Rosser's theorem we can show that the primes p,q >= 19 and each of them can only occur for a finite number of terms n. - Robert Israel, Jun 30 2016

			92 = 19 + 73 with pi(19) * pi(73) = 8 * 21 = 168 = sigma(92).

Eric Weisstein's World of Mathematics, Rosser's Theorem

Cf. A000203, A000720, A272860, A272862.

N:= 10^6: # to use primes up to N
Primes:= select(isprime, [2,seq(i,i=3..N,2)]):
filter:= proc(n) local s,i,j;
  s:= numtheory:-sigma(n);
  for i in select(`>=`,numtheory:-divisors(s), ceil(sqrt(s))) minus {s} do
     if i > nops(Primes) then return FAIL
     elif Primes[i] + Primes[s/i] = n then return true fi
  od:
  false
end proc:
A:= NULL:
for n from 2 by 2 do
  v:= filter(n);
  if v = FAIL then break
  elif v then A:= A, n
  fi
od:
A; # Robert Israel, Jun 30 2016

Select[Range[10^3], Function[n, Length@ Select[Transpose@ {#, n - #} &@ Range[Floor[n/2]], And[Times @@ Boole@ PrimeQ@ {First@ #, Last@ #} == 1, DivisorSigma[1, First@ # + Last@ #] == PrimePi[First@ #] PrimePi[Last@ #]] &] > 0]] (* Michael De Vlieger, Jun 30 2016 *)

is(n) = if(n%2==1, return(0), my(x=n-1, y=1); while(x > y, if(ispseudoprime(x) && ispseudoprime(y) && sigma(x+y)==primepi(x)*primepi(y), return(1)); x--; y++); return(0)) \\ Felix Fröhlich, Jun 28 2016

is(n) = my( d=divisors(sigma(n))); for(i=1,ceil(#d/2), if(prime(d[i]) + prime(d[#d + 1-i]) == n, return(1))); return(0) \\ David A. Corneth, Jun 30 2016

def sol(n): return [j for j in divisors(sigma(n)) if j^2<= sigma(n) and is_prime(n-nth_prime(j)) and j * prime_pi(n-nth_prime(j))==sigma(n)]
v=[n for n in range(2,100000,2) if sol(n)]
print('list_n =',v)
w=[sigma(n) for n in v]; print('list_sigma(n) =',w)
list_pi(p)=flatten([sol(n) for n in range(2,100000,2) if sol(n)])
print('list_pi(p) =',list_pi(p))
list_pi(q)=[w[n]/list_pi[n] for n in range(len(v))]
print('list_pi(q) =',list_pi(q))

Integers n such that sigma(n) = pi(q) * pi(n-q) for some prime q.

cp's OEIS Frontend

A272860 Sums of two primes (in increasing order) when equal to the product of their prime-counting functions.

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