A221205 -id:A221205 - cp's OEIS frontend

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

Vladimir Pletser, Jul 19 2013

nonn

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.

			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.

Cf. A006880, A227694.

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);

a(n) = round((F[2n+1](Sum_{i=0..2} (C(i)*(log(log(A*(B+n^2))))^(2i))))^2).

A229255 Integer nearest to (2^(n-1) + 3^(n-1))^(2b(n)) where b(n) = (C1n^(Pi)exp(C2n)cos(C3n+C4) + C5)(C6n^C7 + Pi/2) (see coefficients in comments).

Original entry on oeis.org

4, 25, 168, 1229, 9592, 78488, 664356, 5761311, 50857532, 455110791, 4117706679, 37598394076, 345973354409, 3204537723387, 29847287869987, 279317953220125, 2624541016148480, 24747919106286414, 234089443816438414, 2220530456953251916, 21119025631088169139, 201358809736398135352, 1924434871799161020533, 18434884359943473267194, 176994218822287711757127

Offset: 1

Vladimir Pletser, Sep 17 2013

Coefficients are C1=27829/125000000, C2=-0.591561, C3=441/2500, C4=5, C5=19703973/31250000, C6=5.241804273*10^-3, C7=0.6246728093.
This sequence gives a good approximation of pi(10^n) (A006880); see (A229256).
To obtain this sequence, note first that the square roots of the first values of pi(10^n) (A006880) (see A221205) are equal or close to some values of A229194, i.e., A221205(n) = or ≈ A229194(2n+1) = round(2^(n-1) + 3^(n-1)) for 1 <= n <= 25. Then, values of pi(10^n), A006880(n) = or ≈ (A229194(2n+1))^2 = round((2^(n-1) + 3^(n-1)))^2 for 1 <= n <= 25. Finally, the fit is improved by multiplying the exponent 2 by the sequence b(n) which always has values close to one for 1 <= n <= 25, varying between 0.99382... and 1.01511....

			For n=1, b(1) = (C1*exp(C2)*cos(C3+C4) + C5)*(C6 + Pi/2) = 0.99382..., then a(1) = round(2^(2*0.99382...)) = round(3.96588...) = 4.

Vladimir Pletser, Table of n, a(n) for n = 1..500
Eric Weisstein's World of Mathematics, Prime Counting Function

Cf. A006880, A221205, A229194, A229256.

C1:=27829/125000000: C2:=-5.91561e-01: C3:=441/2500: C4:=5: C5:=19703973/31250000: C6:=5.241804273e-03: C7:=6.246728093e-01: b:=n-> (C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4)+C5)*(C6*n^C7+(Pi/2)):  seq(round((2^(n-1)+3^(n-1))^(2*b(n))), n=1..25);

a(n) = round((2^(n-1) + 3^(n-1))^(2*(C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4) + C5)*(C6*n^C7 + Pi/2))).

cp's OEIS Frontend

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).

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A229255 Integer nearest to (2^(n-1) + 3^(n-1))^(2b(n)) where b(n) = (C1n^(Pi)exp(C2n)cos(C3n+C4) + C5)(C6n^C7 + Pi/2) (see coefficients in comments).

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cp's OEIS Frontend

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).

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Maple

Formula

A229255 Integer nearest to (2^(n-1) + 3^(n-1))^(2*b(n)) where b(n) = (C1*n^(Pi)*exp(C2*n)*cos(C3*n+C4) + C5)*(C6*n^C7 + Pi/2) (see coefficients in comments).

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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).

A229255 Integer nearest to (2^(n-1) + 3^(n-1))^(2b(n)) where b(n) = (C1n^(Pi)exp(C2n)cos(C3n+C4) + C5)(C6n^C7 + Pi/2) (see coefficients in comments).