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.

Showing 1-2 of 2 results.

A340533 Decimal expansion of log_2(4/Pi).

Original entry on oeis.org

3, 4, 8, 5, 0, 3, 8, 7, 0, 5, 2, 7, 6, 8, 1, 2, 0, 1, 9, 5, 6, 7, 2, 0, 7, 0, 4, 8, 9, 1, 9, 9, 2, 6, 6, 4, 9, 8, 1, 5, 2, 3, 0, 7, 3, 2, 3, 6, 9, 5, 8, 4, 7, 0, 5, 9, 3, 2, 1, 1, 4, 8, 4, 5, 1, 1, 8, 9, 7, 0, 3, 6, 4, 1, 5, 4, 5, 8, 5, 6, 1, 0, 3, 9, 7, 3, 5
Offset: 0

Views

Author

A.H.M. Smeets, Jan 10 2021

Keywords

Comments

Probability of a coefficient in the continued fraction being even, where the continued fraction coefficients satisfy the Gauss-Kuzmin distribution.

Examples

			0.348503870527681201956720704891992664981523...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Log[4/Pi]/Log[2], 10, 100][[1]] (* Amiram Eldar, Jan 10 2021 *)
  • PARI
    log(4/Pi)/log(2)

Formula

Equals 2 - A216582.
Equals log_2(A088538).
Equals -Sum_{k >= 1} log_2(1-1/(2*k+1)^2).
Equals 1-A340543.

A341641 Decimal expansion of the probability of two consecutive continued fraction coefficients being both even, when the continued fraction coefficients satisfy the Gauss-Kuzmin distribution.

Original entry on oeis.org

1, 1, 6, 9, 4, 0, 0, 0, 3, 5, 7, 8, 0, 6, 8, 0, 7, 6, 5, 6, 0, 5, 6, 0, 7, 5, 0, 9, 2, 0, 8, 5, 3, 4, 1, 0, 5, 7, 2, 6, 6, 5, 5, 6, 5, 8, 2, 1, 8, 6, 7, 0, 1, 5, 6, 8, 8, 1, 8, 1, 1, 5, 4, 4, 2, 7, 0, 7, 1, 9, 7, 0, 9, 4, 6, 6, 4, 4, 2, 8, 9, 5, 0, 6, 9, 0, 8
Offset: 0

Views

Author

A.H.M. Smeets, Feb 16 2021

Keywords

Examples

			0.1169400035780680765605607509208534105...
		

Crossrefs

Programs

  • PARI
    sumpos(j=1, log(gamma(1+1/(4*j+2))/gamma(1+(j+1)/(2*j+1))*gamma(1+(2*j+1)/4/j)/gamma(1+1/4/j)))/log(2)
    
  • PARI
    C = log(2)-1+(log(72*Pi)-4*log(gamma(1/4)))/log(2)
    C+sumpos(n=2, (-1)^n*(zeta(n)-1)/n*((2^(2-n)-2^(2-2*n)-1)*(zeta(n)-1)+(2^(n-1)-1)*2^(2-2*n)))/log(2)

Formula

Equals Sum_{j >= 1} log_2(Gamma(1+1/(4*j+2))/Gamma(1+(j+1)/(2*j+1))*Gamma(1+(2*j+1)/4/j)/Gamma(1+1/4/j)).
Showing 1-2 of 2 results.