A229255 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).

4, 25, 168, 1229, 9592, 78488, 664356, 5761311, 50857532, 455110791, 4117706679, 37598394076, 345973354409, 3204537723387, 29847287869987, 279317953220125, 2624541016148480, 24747919106286414, 234089443816438414, 2220530456953251916, 21119025631088169139, 201358809736398135352, 1924434871799161020533, 18434884359943473267194, 176994218822287711757127

Offset: 1

Author

Vladimir Pletser, Sep 17 2013

Keywords

• nonn
• less

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.

This sequence gives a good approximation of pi(10^n) (A006880); see (A229256).

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....

Examples

			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(3.96588...) = 4.
		

Links

• Vladimir Pletser, Table of n, a(n) for n = 1..500
• Eric Weisstein's World of Mathematics, Prime Counting Function

Crossrefs

Cf. A006880, A221205, A229194, A229256.

Programs

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);

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))).