OFFSET
1,1
COMMENTS
Coefficients are C1=27829/125000000, C2=-0.591561, C3=441/2500, C4=5, C5=19703973/31250000, C6=5.241804273*10^-3, C7=0.6246728093.
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....
LINKS
Vladimir Pletser, Table of n, a(n) for n = 1..500
Eric Weisstein's World of Mathematics, Prime Counting Function
FORMULA
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))).
EXAMPLE
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.
MAPLE
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);
CROSSREFS
KEYWORD
nonn,less
AUTHOR
Vladimir Pletser, Sep 17 2013
STATUS
approved