A333700 a(n) = Sum_{k=1..n} pi(k) * pi(n-k).

0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718, 2863

Offset: 1

Author

Ilya Gutkovskiy, Apr 02 2020

Keywords

• nonn

Comments

Convolution of A000720 with itself.

Links

• Mathematics Stack Exchange, A curious equality of integrals involving the prime counting function?
• Eric Weisstein's World of Mathematics, Prime Counting Function

Crossrefs

Cf. A000720, A002815, A010051, A034387, A046992, A073610.

Programs

Mathematica

Table[Sum[PrimePi[k] PrimePi[n - k], {k, n}], {n, 50}]
nmax = 50; CoefficientList[Series[(1/(1 - x)^2) Sum[x^Prime[k], {k, 1, nmax}]^2, {x, 0, nmax}], x] // Rest

PARI

a(n) = sum(k=1, n, primepi(k)*primepi(n-k)); \\ Michel Marcus, Apr 03 2020

Formula

G.f.: (1/(1 - x)^2) * (Sum_{k>=1} x^prime(k))^2.

a(n) = Sum_{k=1..n} A046992(k) * A010051(n-k).

a(n) = Sum_{k=1..n} k * A073610(n-k+1).

From Jianing Song, Sep 27 2023: (Start)

a(n-1) = Integral_{0..n} pi(x) * pi(n-x) dx, since Integral_{0..n} pi(x) * pi(n-x) dx = Sum_{k=1..n} Integral_{k-1..k} pi(x) * pi(n-x) dx = Sum_{k=1..n} pi(k-1) * pi(n-k) = Sum_{k=0..n-1} pi(k) * pi(n-1-k) = a(n-1).

a(n) = (a(n-1) + a(n+1))/2 for n == 4 (mod 6) with n > 4, as shown in the Mathematics Stack Exchange link. (End)