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%I #10 Jul 26 2013 15:18:32
%S 4,25,168,1229,9595,78527,664408,5759130,50833725,455019102,
%T 4118498801,37616575907,346165453783,3205869110911,29851888456753,
%U 279286334215803,2623780688311969,24739953477533166,234041108830344356,2220562531262307905,21124612016460745383,201448482556532026684,1925296277838503159171,18437832696789559015711,176901280909820032014422
%N Integer nearest to (F[2n+1](S(n)))^2, where F[2n+1](x) are Fibonacci polynomials of odd indices [2n+1] and S(n) = Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i)) (see coefficients A, B, C(i) in comments).
%C Coefficients are A= 0.1641239, B= 10.0861, C(0)=0 .9976796712309498, C(1)= 7.445960495e-02, C(2)= -6.73751166802e-03.
%C This sequence gives a good approximation of pi(10^n) (A006880); see (A227694).
%C To obtain this sequence, remark first that the square root of the first values of pi(10^n) (A006880) (see (A221205)) are close to odd indices Fibonacci numbers F[2n+1](1). Switching to odd indices Fibonacci polynomials F[2n+1](x), one obtains the sequence a(n) by computing x as a function of n such that (F[2n+1](x))^2 fit the values of pi(10^n) for 1<=n<=25.
%F a(n) = round((F[2n+1](Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i))))^2).
%e For n =1, F[3](x) = x^2+1; replace x by Sum_{i=0..2} (C(i)*(log(log(A*(B+1))))^(2i))= 1.016825… to obtain a(1)= round((F[3]( 1.016825…))^2)=4.
%p with(combinat, fibonacci): A:= 0.1641239: B:= 10.0861: C(0):= .9976796712309498: C(1):=7.445960495E-02: C(2):= -6.73751166802E-03: b:=n->log(log(A*(B+n^2))): c:=n->sum(C(i)*(b(n))^(2*i), i=0..2): seq(round(fibonacci(2*n+1, c(n))^2), n=1..25);
%Y Cf. A006880, A227694.
%K nonn
%O 1,1
%A _Vladimir Pletser_, Jul 19 2013
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