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A228063 Integer nearest to F[4n](S(n)), where F[4n](x) are Fibonacci polynomials and S(n) = Sum_{i=0..3} (C(i)*(log(log(A*(B+n^2))))^i) (see coefficients A, B, C(i) in comments).

%I A228063 #18 Feb 16 2025 08:33:20
%S A228063 4,21,143,1063,8371,68785,583436,5069633,44876757,403025174,
%T A228063 3660702622,33550877248,309726969451,2876065468123,26835315229835,
%U A228063 251389798269317,2362887262236150,22272676889496853,210455460654786509,1992806263723883464
%N A228063 Integer nearest to F[4n](S(n)), where F[4n](x) are Fibonacci polynomials and S(n) = Sum_{i=0..3} (C(i)*(log(log(A*(B+n^2))))^i) (see coefficients A, B, C(i) in comments).
%C A228063 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.
%C A228063 This sequence gives a good approximation of the number of primes with n digits (A006879); see A228064.
%C A228063 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.
%C A228063 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.
%D A228063 Jonathan Borwein, David H. Bailey, Mathematics by Experiment, A. K. Peters, 2004, p. 65 (Table 2.2).
%D A228063 John H. Conway and R. K. Guy, The Book of Numbers, Copernicus, an imprint of Springer-Verlag, NY, 1996, page 144.
%H A228063 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial.</a>
%F A228063 a(n) = round(F[4n](Sum_{i=0..3} (C(i)*(log(log(A*(B+n^2))))^i)) ).
%e A228063 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
%p A228063 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);
%Y A228063 Cf. A006879, A228064, A227693.
%K A228063 nonn
%O A228063 1,1
%A A228063 _Vladimir Pletser_, Aug 06 2013