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%I #18 Sep 28 2023 16:04:28
%S 0,0,0,1,4,8,14,22,32,45,58,73,90,110,132,158,184,214,246,282,320,363,
%T 406,455,506,562,618,678,738,804,872,944,1018,1099,1180,1269,1358,
%U 1450,1544,1644,1744,1852,1962,2078,2196,2321,2446,2581,2718,2863
%N a(n) = Sum_{k=1..n} pi(k) * pi(n-k).
%C Convolution of A000720 with itself.
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/q/3128367">A curious equality of integrals involving the prime counting function?</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeCountingFunction.html">Prime Counting Function</a>
%F G.f.: (1/(1 - x)^2) * (Sum_{k>=1} x^prime(k))^2.
%F a(n) = Sum_{k=1..n} A046992(k) * A010051(n-k).
%F a(n) = Sum_{k=1..n} k * A073610(n-k+1).
%F From _Jianing Song_, Sep 27 2023: (Start)
%F 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).
%F 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)
%t Table[Sum[PrimePi[k] PrimePi[n - k], {k, n}], {n, 50}]
%t nmax = 50; CoefficientList[Series[(1/(1 - x)^2) Sum[x^Prime[k], {k, 1, nmax}]^2, {x, 0, nmax}], x] // Rest
%o (PARI) a(n) = sum(k=1, n, primepi(k)*primepi(n-k)); \\ _Michel Marcus_, Apr 03 2020
%Y Cf. A000720, A002815, A010051, A034387, A046992, A073610.
%K nonn
%O 1,5
%A _Ilya Gutkovskiy_, Apr 02 2020
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