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%I #23 Mar 03 2023 11:56:27
%S 4,25,168,1229,9592,78488,664356,5761311,50857532,455110791,
%T 4117706679,37598394076,345973354409,3204537723387,29847287869987,
%U 279317953220125,2624541016148480,24747919106286414,234089443816438414,2220530456953251916,21119025631088169139,201358809736398135352,1924434871799161020533,18434884359943473267194,176994218822287711757127
%N Integer nearest to (2^(n-1) + 3^(n-1))^(2*b(n)) where b(n) = (C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4) + C5)*(C6*n^C7 + Pi/2) (see coefficients in comments).
%C Coefficients are C1=27829/125000000, C2=-0.591561, C3=441/2500, C4=5, C5=19703973/31250000, C6=5.241804273*10^-3, C7=0.6246728093.
%C This sequence gives a good approximation of pi(10^n) (A006880); see (A229256).
%C To obtain this sequence, note first that the square roots of the first values of pi(10^n) (A006880) (see A221205) are equal or close to some values of A229194, i.e., A221205(n) = or ≈ A229194(2n+1) = round(2^(n-1) + 3^(n-1)) for 1 <= n <= 25. Then, values of pi(10^n), A006880(n) = or ≈ (A229194(2n+1))^2 = round((2^(n-1) + 3^(n-1)))^2 for 1 <= n <= 25. Finally, the fit is improved by multiplying the exponent 2 by the sequence b(n) which always has values close to one for 1 <= n <= 25, varying between 0.99382... and 1.01511....
%H Vladimir Pletser, <a href="/A229255/b229255.txt">Table of n, a(n) for n = 1..500</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeCountingFunction.html">Prime Counting Function</a>
%F a(n) = round((2^(n-1) + 3^(n-1))^(2*(C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4) + C5)*(C6*n^C7 + Pi/2))).
%e For n=1, b(1) = (C1*exp(C2)*cos(C3+C4) + C5)*(C6 + Pi/2))= 0.99382..., then a(1) = round(2^(2*0.99382...)) = round(3.96587...) = 4.
%p C1:=27829/125000000: C2:=-5.91561e-01: C3:=441/2500: C4:=5: C5:=19703973/31250000: C6:=5.241804273e-03: C7:=6.246728093e-01: b:=n-> (C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4)+C5)*(C6*n^C7+(Pi/2)): seq(round((2^(n-1)+3^(n-1))^(2*b(n))), n=1..25);
%Y Cf. A006880, A221205, A229194, A229256.
%K nonn,less
%O 1,1
%A _Vladimir Pletser_, Sep 17 2013
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