%I #53 Aug 06 2024 05:05:19
%S 4,7,6,8,4,6,2,0,5,8,0,6,2,7,4,3,4,4,8,2,9,9,7,9,8,5,7,7,3,5,6,7,9,4,
%T 4,7,7,5,4,3,2,3,9,0,3,3,0,1,6,8,6,6,9,1,5,3,8,4,2,0,3,0,1,5,9,7,8,3,
%U 6,2,5,8,6,0,7,2,0,7,4,5,1,0,3,7,3,0,7,0,4,2,0,7,3,1,3,6,1,0,4,0,0,0,5,3,7
%N Decimal expansion of Product_{k>=0} (1 + 1/2^k).
%C Twice the product in A079555.
%H G. C. Greubel, <a href="/A081845/b081845.txt">Table of n, a(n) for n = 1..2000</a>
%H Richard J. McIntosh, <a href="https://doi.org/10.1112/jlms/51.1.120">Some Asymptotic Formulae for q-Hypergeometric Series</a>, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; <a href="https://citeseerx.ist.psu.edu/pdf/4f03a5e304ec19f8a725774525aecd2a78f4ad81">citeseerx</a>.
%H József Sándor, <a href="https://nntdm.net/volume-27-2021/number-3/29-38/">On Vandiver's arithmetical function - I</a>, Notes on Number Theory and Discrete Mathematics, Vol. 27, No. 3 (2021), pp. 29-38. See p. 36, eq. 40.
%F lim sup Product_{k=0..floor(log_2(n))} (1 + 1/floor(n/2^k)) for n-->oo. - _Hieronymus Fischer_, Aug 20 2007
%F lim sup A132369(n)/A098844(n) for n-->oo. - _Hieronymus Fischer_, Aug 20 2007
%F lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - _Hieronymus Fischer_, Aug 20 2007
%F lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - _Hieronymus Fischer_, Aug 20 2007
%F 2*exp(Sum_{n>0} 2^(-n)*Sum_{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*2^n)). - _Hieronymus Fischer_, Aug 20 2007
%F lim sup A132269(n+1)/A132269(n) = 4.76846205806274344... for n-->oo. - _Hieronymus Fischer_, Aug 20 2007
%F Sum_{k>=1} (-1)^(k+1) * 2^k / (k*(2^k-1)) = log(A081845) = 1.562023833218500307570359922772014353168080202860122... . - _Vaclav Kotesovec_, Dec 13 2015
%F Equals 2*(-1/2; 1/2)_{infinity}, where (a;q)_{infinity} is the q-Pochhammer symbol. - _G. C. Greubel_, Dec 20 2015
%F Equals 1 + Sum_{n>=1} 2^n/((2-1)*(2^2-1)*...*(2^n-1)). - _Robert FERREOL_, Feb 21 2020
%F From _Peter Bala_, Jan 18 2021: (Start)
%F Constant C = 3*Sum_{n >= 0} (1/2)^n/Product_{k = 1..n} (2^k - 1).
%F Faster converging series:
%F C = (2*3*5)/(2^3)*Sum_{n >= 0} (1/4)^n/Product_{k = 1..n} (2^k - 1),
%F C = (2*3*5*9)/(2^6)*Sum_{n >= 0} (1/8)^n/Product_{k = 1..n} (2^k - 1),
%F C = (2*3*5*9*17)/(2^10)*Sum_{n >= 0} (1/16)^n/Product_{k = 1..n} (2^k - 1), and so on. The sequence [2,3,5,9,17,...] is A000051. (End)
%F From _Amiram Eldar_, Mar 20 2022: (Start)
%F Equals sqrt(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).
%F Equals 1/A083864. (End)
%F Equals lim_{n->oo} A020696(2^n)/A006125(n+1) (Sándor, 2021). - _Amiram Eldar_, Jun 29 2022
%e 4.76846205806274344829979857....
%t digits = 105; NProduct[1 + 1/2^k, {k, 0, Infinity}, WorkingPrecision -> digits+5, NProductFactors -> digits] // RealDigits[#, 10, digits]& // First (* _Jean-François Alcover_, Mar 04 2013 *)
%t N[QPochhammer[-1, 1/2], 100] (* _Vaclav Kotesovec_, Dec 13 2015 *)
%t 2*N[QPochhammer[-1/2, 1/2], 200] (* _G. C. Greubel_, Dec 20 2015 *)
%o (PARI) prodinf(k=0,1/2^k,1) \\ _Hugo Pfoertner_, Feb 21 2020
%Y Cf. A006125, A020696, A028361, A079555, A083864.
%Y Cf. A000593, A048651, A100220, A098844, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132269, A132270.
%K cons,nonn,easy
%O 1,1
%A _Benoit Cloitre_, Apr 09 2003