A081845 Decimal expansion of Product_{k>=0} (1 + 1/2^k).

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, 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, 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

Offset

1,1

Comments

Twice the product in A079555.

Links

G. C. Greubel, Table of n, a(n) for n = 1..2000

Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; citeseerx.

József Sándor, On Vandiver's arithmetical function - I, Notes on Number Theory and Discrete Mathematics, Vol. 27, No. 3 (2021), pp. 29-38. See p. 36, eq. 40.

Formula

lim sup Product_{k=0..floor(log_2(n))} (1 + 1/floor(n/2^k)) for n-->oo. - Hieronymus Fischer, Aug 20 2007

lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007

lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007

lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007

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

lim sup A132269(n+1)/A132269(n) = 4.76846205806274344... for n-->oo. - Hieronymus Fischer, Aug 20 2007

Sum_{k>=1} (-1)^(k+1) * 2^k / (k*(2^k-1)) = log(A081845) = 1.562023833218500307570359922772014353168080202860122... . - Vaclav Kotesovec, Dec 13 2015

Equals 2*(-1/2; 1/2)_{infinity}, where (a;q)_{infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015

Equals 1 + Sum_{n>=1} 2^n/((2-1)*(2^2-1)*...*(2^n-1)). - Robert FERREOL, Feb 21 2020

From Peter Bala, Jan 18 2021: (Start)

Constant C = 3*Sum_{n >= 0} (1/2)^n/Product_{k = 1..n} (2^k - 1).

Faster converging series:

C = (2*3*5)/(2^3)*Sum_{n >= 0} (1/4)^n/Product_{k = 1..n} (2^k - 1),

C = (2*3*5*9)/(2^6)*Sum_{n >= 0} (1/8)^n/Product_{k = 1..n} (2^k - 1),

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)

From Amiram Eldar, Mar 20 2022: (Start)

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).

Equals 1/A083864. (End)

Equals lim_{n->oo} A020696(2^n)/A006125(n+1) (Sándor, 2021). - Amiram Eldar, Jun 29 2022

Example

4.76846205806274344829979857....

Mathematica

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 *)
N[QPochhammer[-1, 1/2], 100] (* Vaclav Kotesovec, Dec 13 2015 *)
2*N[QPochhammer[-1/2, 1/2], 200] (* G. C. Greubel, Dec 20 2015 *)

Prog

(PARI) prodinf(k=0, 1/2^k, 1) \\ Hugo Pfoertner, Feb 21 2020

Crossrefs

Cf. A006125, A020696, A028361, A079555, A083864.

Cf. A000593, A048651, A100220, A098844, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132269, A132270.

Sequence in context: A292510 A284116 A200353 * A254339 A140877 A293904

Adjacent sequences: A081842 A081843 A081844 * A081846 A081847 A081848

Keyword

cons,nonn,easy

Author

Benoit Cloitre, Apr 09 2003

Status

approved