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Integer nearest f(2^n), where f(x) = Sum of ( mu(k) * H(k)/k^(3/2) * Integral Log(x^(1/k)) ) for k = 1 to infinity, where H(k) is the harmonic number sum_{i=1..k} 1/i.
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%I #33 Feb 16 2025 08:33:16

%S 2,2,4,6,11,18,31,54,96,171,309,562,1029,1896,3514,6545,12247,23005,

%T 43371,82029,155598,295927,564164,1077892,2063545,3957761,7603593,

%U 14630713,28192867,54399529,105097590,203280493,393614506,762937782,1480207843,2874399615

%N Integer nearest f(2^n), where f(x) = Sum of ( mu(k) * H(k)/k^(3/2) * Integral Log(x^(1/k)) ) for k = 1 to infinity, where H(k) is the harmonic number sum_{i=1..k} 1/i.

%C This function gives a very good approximation to the number of primes less than or equal to n.

%C Also note that f(2^23) - pi(2^23) = 1, f(2^31) - pi(2^31) = 25, f(2^43) - pi(2^43) = 99, f(2^58) - pi(2^58) = -53540.

%H Chris K. Caldwell, <a href="https://t5k.org/howmany.shtml">How Many Primes Are There?</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PrimeCountingFunction.html">Prime Counting Function</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PrimeNumberTheorem.html">Prime Number Theorem</a>

%t f[n_Integer] := Sum[N[MoebiusMu[k]*HarmonicNumber[k]/k^(3/2)*LogIntegral[n^(1/k)], 50], {k, 1, 5!}]; Table[Round[f[2^n]], {n, 36}]

%Y Cf. A007053.

%K nonn

%O 1,1

%A _Arkadiusz Wesolowski_, Dec 30 2011