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.

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A119809 Decimal expansion of the constant defined by binary sums involving Beatty sequences: c = Sum_{n>=1} 1/2^A049472(n) = Sum_{n>=1} A001951(n)/2^n.

Original entry on oeis.org

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

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Author

Paul D. Hanna, May 26 2006

Keywords

Comments

Dual constant: A119812 = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n). Since this constant c = 2 + Sum_{n>=1} 1/2^A003151(n), where A003151(n) = n + floor(n*sqrt(2)), then the binary expansion of the fractional part of c has 1's only at positions given by Beatty sequence A003151(n) and zeros elsewhere. Plouffe's Inverter describes approximations to the fractional part of c as "polylogarithms type of series with the floor function [ ]."

Examples

			c = 2.32258852258806773012144068278798408011950250800432925665718...
Continued fraction (A119810):
c = [2;3,10,132,131104,2199023259648,633825300114114700748888473600,..]
where partial quotients are given by:
PQ(n) = 2^A001333(n-1) + 2^A000129(n-2) (n>1), with PQ(1)=2.
The following are equivalent expressions for the constant:
(1) Sum_{n>=1} 1/2^A049472(n); A049472(n)=[n/sqrt(2)];
(2) Sum_{n>=1} A001951(n)/2^n; A001951(n)=[n*sqrt(2)];
(3) Sum_{n>=1} 1/2^A003151(n) + 2; A003151(n)=[n*sqrt(2)]+n;
(4) Sum_{n>=1} 1/2^A097508(n) - 2; A097508(n)=[n*sqrt(2)]-n;
(5) Sum_{n>=1} A006337(n)/2^n + 1; A006337(n)=[(n+1)*sqrt(2)]-[n*sqrt(2)];
where [x] = floor(x).
These series illustrate the above expressions:
(1) c = 1/2^0 + 1/2^1 + 1/2^2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^4 +...
(2) c = 1/2^1 + 2/2^2 + 4/2^3 + 5/2^4 + 7/2^5 + 8/2^6 + 9/2^7 +...
(3) c = 2 + 1/2^2 + 1/2^4 + 1/2^7 + 1/2^9 + 1/2^12 + 1/2^14 +...
(4) c =-2 + 1/2^0 + 1/2^0 + 1/2^1 + 1/2^1 + 1/2^2 + 1/2^2 + 1/2^2 +...
(5) c = 1 + 1/2^1 + 2/2^2 + 1/2^3 + 2/2^4 + 1/2^5 + 1/2^6 + 2/2^7 +...
		

Crossrefs

Cf. A119810 (continued fraction), A119811 (convergents); A119812 (dual constant); A000129 (Pell), A001333; Beatty sequences: A049472, A001951, A003151, A097508, A006337; variants: A014565 (rabbit constant), A073115.

Programs

  • PARI
    {a(n)=local(t=sqrt(2),x=sum(m=1,10*n,floor(m*t)/2^m));floor(10^n*x)%10}

Formula

Equals Sum(1/(2^q-1)) with the summation extending over all pairs of integers gcd(p,q) = 1, 0 < p/q < sqrt(2) (O'Bryant, 2002). - Amiram Eldar, May 25 2023

A119810 Partial quotients of the continued fraction of the constant defined by binary sums involving Beatty sequences: c = Sum_{n>=1} 1/2^A049472(n) = Sum_{n>=1} A001951(n)/2^n.

Original entry on oeis.org

2, 3, 10, 132, 131104, 2199023259648, 633825300114114700748888473600, 883423532389192164791648750371459257913741948437810659652423818057613312
Offset: 1

Views

Author

Paul D. Hanna, May 26 2006

Keywords

Comments

Convergents A119811: [2/1,7/3,72/31,9511/4095,1246930216/536870911,...], where the denominators of the convergents are equal to [2^A000129(n-1)-1] and A000129 is the Pell numbers. The number of digits in these partial quotients are (beginning at n=1): [1,1,2,3,6,13,30,72,174,420,1013,2445,5901,14246,34391,83027,...].

Examples

			c = 2.32258852258806773012144068278798408011950250800432925665718...
The partial quotients start:
a(1) = 2^1; a(2) = 2^1 + 2^0; a(3) = 2^3 + 2^1;
a(4) = 2^7 + 2^2; a(5) = 2^17 + 2^5; a(6) = 2^41 + 2^12;
and continue as a(n) = 2^A001333(n-1) + 2^A000129(n-2) where
A001333(n) = ( (1+sqrt(2))^n + (1-sqrt(2))^n )/2;
A000129(n) = ( (1+sqrt(2))^n - (1-sqrt(2))^n )/(2*sqrt(2)).
		

Crossrefs

Cf. A119809 (decimal expansion), A119811 (convergents); A119812 (dual constant).

Programs

  • Mathematica
    (* b = A001333 *) b[0] = 1; b[1] = 1; b[n_] := b[n] = 2 b[n-1] + b[n-2]; a[1] = 2; a[n_] := 2^b[n-1] + 2^Fibonacci[n-2, 2]; Array[a, 10] (* Jean-François Alcover, May 04 2017 *)
  • PARI
    {a(n)=if(n==1,2,2^round(((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))/2) +2^round(((1+sqrt(2))^(n-2)-(1-sqrt(2))^(n-2))/(2*sqrt(2))))}

Formula

a(n) = 2^A001333(n-1) + 2^A000129(n-2) for n>1, with a(1)=2.
Showing 1-2 of 2 results.