A228111 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).

4, 21, 143, 1063, 8385, 68929, 584467, 5074924, 44885325, 402777151, 3656032622, 33492393634, 309106153431, 2870123507479, 26783122426197, 250971797533095, 2359952229466124, 22256979400698116, 210440626023838163, 1994088284872617955, 18931694933036811169

Offset

1,1

Comments

Coefficients are A=16103485019141/2900449771918928, B=5262046568827901/29305205016290, C(0)=296261685121849/265642652464758, C(1)=38398556529727/750568780742436, C(2)=0, C(3)=-11594434149768/8254020049890781.

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

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.

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.

References

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

Links

Vladimir Pletser, Table of n, a(n) for n = 1..500

Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Formula

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)).

Example

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.
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.

Maple

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);

Crossrefs

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

Sequence in context: A158577 A006879 A228063 * A305986 A233481 A308337

Adjacent sequences: A228108 A228109 A228110 * A228112 A228113 A228114

Keyword

nonn

Author

Vladimir Pletser, Aug 10 2013

Status

approved