OFFSET
1,1
COMMENTS
Coefficients are A=6.74100517717340111e-03, B=147.60482223254, C(0)=1.112640536670862472, C(1)=5.2280866355335360415e-02, C(2)=0, C(3)=-1.5569578292261924e-03.
This sequence gives a good approximation of the number of primes with n digits (A006879); see A228064.
As the squares of odd-indexed Fibonacci numbers F[2n+1](1) (see A227693) are equal or close to the first values of pi(10^n) (A006880), and as F[4n](1)=(F[2n+1](1))^2- (F[2n-1](1))^2, it is legitimate to ask whether the first values of the differences pi(10^n)- pi(10^(n-1)) (A006879) are also close or equal to multiple of 4 index Fibonacci numbers F[4n](1); e.g., for n=2, F[8](1)=21.
To obtain this sequence, one switches to multiple of 4 index Fibonacci polynomials F[4n](x), one obtains the sequence a(n) by computing x as a function of n such that F[4n](x) fit the values of pi(10^n)- pi(10^(n-1)) for 1 <= n <= 25, with pi(1)=0.
REFERENCES
Jonathan Borwein, David H. Bailey, Mathematics by Experiment, A. K. Peters, 2004, p. 65 (Table 2.2).
John H. Conway and R. K. Guy, The Book of Numbers, Copernicus, an imprint of Springer-Verlag, NY, 1996, page 144.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
FORMULA
a(n) = round(F[4n](Sum_{i=0..3} (C(i)*(log(log(A*(B+n^2))))^i)) ).
EXAMPLE
For n =1, F[4](x) = x^3+2x; replace x by Sum_{i=0..3} (C(i)*(log(log(A*(B+1))))^i)= 1.179499… to obtain a(1)= round(F[4]( 1.179499...))=4. For n=2, F[8](x) = x^7+6x^5+10x^3+4x; replace x by Sum_{i=0..3} (C(i)*(log(log(A*(B+4))))^i)= 0.999861... to obtain a(2)= round(F[8]( 0.999861…))=21
MAPLE
with(combinat):A:=6.74100517717340111e-03: B:=147.60482223254: C(0):=1.112640536670862472: C(1):=5.2280866355335360415e-02: C(2):=0: C(3):=-1.5569578292261924e-03: b:=n->log(log(A*(B+n^2))): c:=n->sum(C(i)*(b(n))^i, i=0..3): seq(round(fibonacci(4*n, c(n))), n=1..25);
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Pletser, Aug 06 2013
STATUS
approved