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A163521 a(n) = floor(Sum_{k = 2..10^n} k/log(k)). 2

%I #16 Mar 28 2021 19:32:40

%S 30,1255,78698,5762750,455059956,37607986470,3204942375900,

%T 279238346962895,24739954333817884

%N a(n) = floor(Sum_{k = 2..10^n} k/log(k)).

%C a(n) = Sum_{x=2..n} x/log(x) closely approximates the number of primes < n^2.

%C In fact, the sum is as good as Li(n^2) but summing a(n) is rather time consuming for large n.

%C For n = 10^9,

%C a(n) = 24739954333817884,

%C Pi(n^2) = 24739954287740860,

%C Li(n^2) = 24739954309690415,

%C R(n^2) = 24739954284239494,

%C where Li = Logarithmic integral approximation of Pi, and R = Riemann's approximation of Pi.

%C Now x/(log(x)-1) is a much better approximation of Pi(x) than x/log(x):

%C 10^18/(log(10^18)-1) = 24723998785919976,

%C 10^18/log(10^18) = 24127471216847323.

%C Ironically, though, a(n) = Sum_{x=2..n} x/(log(x)-1) is far from Pi(n^2).

%e For n = 9, floor(Sum_{x=2..10^n} x/log(x)) = 24739954333817884, the 9th term.

%t Table[Floor[Sum[j/Log[j], {j, 2, 10^n}]], {n, 1, 9}] (* _G. C. Greubel_, Jul 27 2017 *)

%o (PARI) nthsum(n) = for(j=1,n,print1(floor(sum(x=2,10^j,x/log(x)))","));

%K nonn

%O 1,1

%A _Cino Hilliard_, Jul 30 2009

%E Definition clarified by _R. J. Mathar_ and _Omar E. Pol_, Aug 01 2009

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