A228111 - OEIS

%I #15 Dec 11 2019 09:50:11

%S 4,21,143,1063,8385,68929,584467,5074924,44885325,402777151,

%T 3656032622,33492393634,309106153431,2870123507479,26783122426197,

%U 250971797533095,2359952229466124,22256979400698116,210440626023838163,1994088284872617955,18931694933036811169

%N Integer nearest to (S(n)*F(4n)(S(n))), where F(4n)(x) are Fibonacci polynomials of multiple of 4 indices (4n) 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 Coefficients are A=16103485019141/2900449771918928, B=5262046568827901/29305205016290, C(0)=296261685121849/265642652464758, C(1)=38398556529727/750568780742436, C(2)=0, C(3)=-11594434149768/8254020049890781.

%C This sequence gives a good approximation of the number of primes with n digits (A006879); see (A228112).

%C 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.  However, when using Fibonacci polynomials, the exact relation is xF(4n)(x)=(F(2n+1)(x))^2- (F(2n-1)(x))^2.

%C To obtain this sequence, one switches to the product of x and multiple of 4 index Fibonacci polynomials F(4n)(x), and one obtains the sequence a(n) by computing x as a function of n such that (xF(4n)(x)) fit the values of pi(10^n)- pi(10^(n-1)) for 1 <= n <= 25, with pi(1)=0.

%D John H. Conway and R. K. Guy, The Book of Numbers, Copernicus, an imprint of Springer-Verlag, NY, 1996, page 144.

%H Vladimir Pletser, <a href="/A228111/b228111.txt">Table of n, a(n) for n = 1..500</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial.</a>

%F a(n) = round(S(n)*F(4n)(S(n))), where S(n)=Sum_{i=0..3} (C(i)*(log(log(A*(B+n^2))))^i)).

%e For n=1, xF(4)(x) = x^2*(x^2+2); replace x with Sum_{i=0..3} (C(i)*(log(log(A*(B+1))))^i)= 1.11173... to obtain a(1) = round((1.11173...)*F(4)(1.11173...)) = 4.

%e For n=2, xF(8)(x) = x^2*(x^2+2)*(x^4+4*x^2+2); replace x with Sum_{i=0..3} (C(i)*(log(log(A*(B+4))))^i)= 0.99998788... to obtain a(2) = round((0.99998788...)*F(8)(0.99998788...)) = 21.

%p with(combinat):A:=16103485019141/2900449771918928: B:=5262046568827901/29305205016290: C(0):=296261685121849/265642652464758: C(1):=38398556529727/750568780742436: C(2):=0: C(3):=-11594434149768/8254020049890781: b:=n->log(log(A*(B+n^2))): c:=n->sum(C(i)*(b(n))^i, i=0..3): seq(round(c(n)*fibonacci(4*n, c(n))), n=1..25);

%Y Cf. A006879, A228112, A228113, A228114, A228115, A228116.

%K nonn

%O 1,1

%A _Vladimir Pletser_, Aug 10 2013