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.

A079555 Decimal expansion of Product_{k>=1} (1 + 1/2^k) = 2.384231029031371...

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Jan 25 2003

Keywords

Examples

			2.38423102903137172414989928867839723877161951650843345769...
		

Crossrefs

Programs

  • Mathematica
    digits = 105; NProduct[(1 + 1/2^k), {k, 1, Infinity}, WorkingPrecision -> digits+10, NProductFactors -> 200] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 14 2013 *)
    N[QPochhammer[-1/2,1/2]] (* G. C. Greubel, Dec 05 2015 *)
    1/N[QPochhammer[1/2, 1/4]] (* Gleb Koloskov, Apr 04 2021 *)
  • PARI
    prodinf(n=1,1+2.^-n) \\ Charles R Greathouse IV, May 27 2015
    
  • PARI
    1/prodinf(n=0, 1-2^(-2*n-1)) \\ Gleb Koloskov, Apr 04 2021

Formula

(1/2)*lim sup Product_{k=0..floor(log_2(n)), (1 + 1/floor(n/2^k))} for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
exp(sum{n>0, 2^(-n)*sum{k|n, -(-1)^k/k}})=exp(sum{n>0, A000593(n)/(n*2^n)}). - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n+1)/A132269(n)=2.3842310290313717241498992886... for n-->oo. - Hieronymus Fischer, Aug 20 2007
Equals (-1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 05 2015
2 + Sum_{k>1} 1/(Product_{i=2..k} (2^i-1)) = 2 + 1/3 + 1/(3*7) + 1/(3*7*15) + 1/(3*7*15*31) + 1/(3*7*15*31*63) + ... (conjecture). - Werner Schulte, Dec 22 2016
From Peter Bala, Dec 15 2020: (Start)
The above conjecture of Schulte follows by setting x = 1/2 and t = -1 in the identity Product_{k >= 1} (1 - t*x^k) = Sum_{n >= 0} (-1)^n*x^(n*(n+1)/2)*t^n/( Product_{k = 1..n} 1 - x^k ), due to Euler.
Constant C = 1 + Sum_{n >= 0} (1/2)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
C = 2 + Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*C = 7 + Sum_{n >= 0} (1/8)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*C = 50 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*15*C = 751 + Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
(End)
Equals 1/(1-P), where P is the Pell constant from A141848. - Gleb Koloskov, Apr 04 2021
Equals Sum_{k>=0} A000009(k)/2^k. - Vaclav Kotesovec, Sep 15 2021
From Amiram Eldar, Feb 19 2022: (Start)
Equals (sqrt(2)/2) * exp(log(2)/24 + Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(2))) (McIntosh, 1995).
Equals (1/2) * A081845.
Equals Sum_{n>=0} 1/A005329(n). (End)