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).

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

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

Table of n, a(n) for n=1..25.

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