cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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).

Original entry on oeis.org

4, 21, 143, 1063, 8385, 68929, 584467, 5074924, 44885325, 402777151, 3656032622, 33492393634, 309106153431, 2870123507479, 26783122426197, 250971797533095, 2359952229466124, 22256979400698116, 210440626023838163, 1994088284872617955, 18931694933036811169
Offset: 1

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Author

Vladimir Pletser, Aug 10 2013

Keywords

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.

Examples

			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.
		

References

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

Crossrefs

Programs

  • 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);

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).