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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A119812 Decimal expansion of the constant defined by binary sums involving Beatty sequences: c = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n).

Original entry on oeis.org

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

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Author

Paul D. Hanna, May 26 2006

Keywords

Comments

Dual constant: A119809 = Sum_{n>=1} 1/2^A049472(n) = Sum_{n>=1} A001951(n)/2^n. The binary expansion of this constant is given by A080764 with offset n=1. Plouffe's Inverter describes approximations to this constant as "polylogarithms type of series with the floor function [ ]."

Examples

			c = 0.858267656461002055792260308433375148664905190083506778667684867..
Continued fraction (A119813):
c = [0;1,6,18,1032,16777344,288230376151842816,...]
where partial quotients are given by:
PQ[n] = 4^A000129(n-2) + 2^A001333(n-3) (n>2), with PQ[1]=0, PQ[2]=1.
The following are equivalent expressions for the constant:
(1) Sum_{n>=1} A049472(n)/2^n; A049472(n)=[n/sqrt(2)];
(2) Sum_{n>=1} 1/2^A001951(n); A001951(n)=[n*sqrt(2)];
(3) Sum_{n>=1} A080764(n)/2^n; A080764(n)=[(n+1)/sqrt(2)]-[n/sqrt(2)];
where [x] = floor(x).
These series illustrate the above expressions:
(1) c = 0/2^1 + 1/2^2 + 2/2^3 + 2/2^4 + 3/2^5 + 4/2^6 + 4/2^7 +...
(2) c = 1/2^1 + 1/2^2 + 1/2^4 + 1/2^5 + 1/2^7 + 1/2^8 + 1/2^9 +...
(3) c = 1/2^1 + 1/2^2 + 0/2^3 + 1/2^4 + 1/2^5 + 0/2^6 + 1/2^7 +...
		

Crossrefs

Cf. A119813 (continued fraction), A119814 (convergents); A119809 (dual constant); A000129 (Pell), A001333; Beatty sequences: A049472, A001951, A080764; variants: A014565 (rabbit constant), A073115.

Programs

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

Extensions

Removed leading zero and corrected offset R. J. Mathar, Feb 05 2009

A119813 Partial quotients of the continued fraction of the constant A119812 defined by binary sums involving Beatty sequences: c = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n).

Original entry on oeis.org

0, 1, 6, 18, 1032, 16777344, 288230376151842816, 1393796574908163946345982392042721617379328
Offset: 1

Views

Author

Paul D. Hanna, May 26 2006

Keywords

Comments

Convergents begin: [0/1, 1/1, 6/7, 109/127, 112494/131071,...], where the denominators of the convergents are equal to [2^A001333(n-1)-1], where A001333 are numerators of continued fraction convergents to sqrt(2). The number of digits in these partial quotients are (beginning at n=2): [1,1,2,4,8,18,43,102,246,594,1432,3457,8345,20146,48636,117417,...].

Examples

			c = 0.858267656461002055792260308433375148664905190083506778667684867..
The partial quotients start:
a(1) = 0; a(2) = 1; a(3) = 4^1 + 2^1; a(4) = 4^2 + 2^1;
a(5) = 4^5 + 2^3; a(6) = 4^12 + 2^7; a(7) = 4^29 + 2^17;
and continue as a(n) = 4^A000129(n-2) + 2^A001333(n-3) where
A000129(n) = ( (1+sqrt(2))^n - (1-sqrt(2))^n )/(2*sqrt(2));
A001333(n) = ( (1+sqrt(2))^n + (1-sqrt(2))^n )/2.
		

Crossrefs

Cf. A119812 (constant), A119814 (convergents); A119809 (dual constant).

Programs

  • PARI
    {a(n)=if(n==1,0,if(n==2,1, 4^round(((1+sqrt(2))^(n-2)+(1-sqrt(2))^(n-2))/(2*sqrt(2))) +if(n==3,2,2^round(((1+sqrt(2))^(n-3)-(1-sqrt(2))^(n-3))/2))))}

Formula

a(n) = 4^A000129(n-2) + 2^A001333(n-3) for n>2, with a(1)=0, a(2)=1.
Showing 1-2 of 2 results.