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 A227693 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). 5
 4, 25, 168, 1229, 9595, 78527, 664408, 5759130, 50833725, 455019102, 4118498801, 37616575907, 346165453783, 3205869110911, 29851888456753, 279286334215803, 2623780688311969, 24739953477533166, 234041108830344356, 2220562531262307905, 21124612016460745383, 201448482556532026684, 1925296277838503159171, 18437832696789559015711, 176901280909820032014422 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Coefficients are A= 0.1641239, B= 10.0861, C(0)=0 .9976796712309498, C(1)= 7.445960495e-02, C(2)= -6.73751166802e-03. This sequence gives a good approximation of pi(10^n) (A006880); see (A227694). 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. LINKS FORMULA a(n) = round((F[2n+1](Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i))))^2). EXAMPLE 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. MAPLE 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); CROSSREFS Cf. A006880, A227694. Sequence in context: A225137 A229255 A006880 * A175255 A081068 A163072 Adjacent sequences:  A227690 A227691 A227692 * A227694 A227695 A227696 KEYWORD nonn AUTHOR Vladimir Pletser, Jul 19 2013 STATUS approved

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Last modified April 23 14:15 EDT 2019. Contains 322386 sequences. (Running on oeis4.)