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.

A340543 Decimal expansion of log(Pi/2)/log(2).

Original entry on oeis.org

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

Views

Author

A.H.M. Smeets, Jan 11 2021

Keywords

Comments

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

Examples

			0.65149612947231879804327929510800733501847692676304...
		

Crossrefs

Cf. A019669. Essentially the same as A216582.
Cf. A340533.

Programs

  • Mathematica
    RealDigits[Log2[Pi/2], 10, 120][[1]] (* Amiram Eldar, May 29 2023 *)
  • PARI
    log(Pi/2)/log(2)

Formula

Equals A216582 - 1.
Equals log_2(A019669).
Equals Sum_{k >= 0} -log_2(1-1/(2*k+2)^2).
Equals 1-A340533.

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.