A048651 Decimal expansion of Product_{k >= 1} (1 - 1/2^k).

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• cons

Comments

This is the limiting probability that a large random binary matrix is nonsingular (cf. A002884).

This constant is very close to 2^(13/24) * sqrt(Pi/log(2)) / exp(Pi^2/(6*log(2))) = 0.288788095086602421278899775042039398383022429351580356839... - Vaclav Kotesovec, Aug 21 2018

This constant is irrational (see Penn link). - Paolo Xausa, Dec 09 2024

Examples

			(1/2)*(3/4)*(7/8)*(15/16)*... = 0.288788095086602421278899721929230780088911904840685784114741...
		

References

• Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 318, 354-361.

Links

• Harry J. Smith, Table of n, a(n) for n = 0..20000
• Steven R. Finch, Digital Search Tree Constants. [Broken link]
• Steven R. Finch, Digital Search Tree Constants. [From the Wayback machine]
• Marvin Geiselhart, Ahmed Elkelesh, Moustafa Ebada, Sebastian Cammerer, and Stephan ten Brink, On the Automorphism Group of Polar Codes, arXiv:2101.09679 [cs.IT], 2021.
• T. S. Jayram, Hellinger strikes back: a note on the multi-party information complexity of AND, LNCS 5687 (2009) 562-573.
• Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.
• Victor S. Miller, Counting Matrices that are Squares, arXiv:1606.09299 [math.GR], 2016.
• Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
• Michael Penn, A non-standard irrationality proof, YouTube video, 2024.
• V. Arvind Rameshwar, Shreyas Jain, and Navin Kashyap, Sampling-Based Estimates of the Sizes of Constrained Subcodes of Reed-Muller Codes, arXiv:2309.08907 [cs.IT], 2023.
• László Tóth, Alternating sums concerning multiplicative arithmetic functions, arXiv preprint arXiv:1608.00795 [math.NT], 2016.
• Eric Weisstein's World of Mathematics, Infinite Product.
• Eric Weisstein's World of Mathematics, Tree Searching.
• Wikipedia, Pentagonal number theorem.
• Wanchen Zhang, Yu Ning, Fei Shi, and Xiande Zhang, Extremal Maximal Entanglement, arXiv:2411.12208 [quant-ph], 2024. See p. 15.

Crossrefs

Cf. A002884, A001318, A005327, A005329, A048652, A065446, A079555, A098844, A067080, A100220, A132019, A132020, A132026, A132038, A070933, A261584, A335764.

Programs

Mathematica

RealDigits[ Product[1 - 1/2^i, {i, 100}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
RealDigits[QPochhammer[1/2], 10, 100][[1]] (* Jean-François Alcover, Nov 18 2015 *)

PARI

default(realprecision, 20080); x=prodinf(k=1, -1/2^k, 1); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b048651.txt", n, " ", d)); \\ Harry J. Smith, May 07 2009

Formula

exp(-Sum_{k>0} sigma_1(k)/k*2^(-k)) = exp(-Sum_{k>0} A000203(k)/k*2^(-k)). - Hieronymus Fischer, Jul 28 2007

From Hieronymus Fischer, Aug 13 2007: (Start)

Equals lim inf Product_{k=0..floor(log_2(n))} floor(n/2^k)*2^k/n for n->oo.

Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^(1/2*(1+floor(log_2(n)))*floor(log_2(n))) for n->oo.

Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^A000217(floor(log_2(n))) for n->oo.

Equals lim inf A098844(n)/(n+1)^((1+log_2(n+1))/2) for n->oo.

Equals (1/2)*exp(-Sum_{n>0} 2^(-n)*Sum_{k|n} 1/(k*2^k)). (End)

Limit of A177510(n)/A000079(n-1) as n->infinity (conjecture). - Mats Granvik, Mar 27 2011

Product_{k >= 1} (1-1/2^k) = (1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 27 2015

exp(Sum_{n>=1}(1/n/(1 - 2^n))) (according to Mathematica). - Mats Granvik, Sep 07 2016

(Sum_{k>0} (4^k-1)/(Product_{i=1..k} ((4^i-1)*(2*4^i-1))))*2 = 2/7 + 2/(3*7*31) + 2/(3*7*15*31*127)+2/(3*7*15*31*63*127*511) + ... (conjecture). - Werner Schulte, Dec 22 2016

Equals Sum_{k=-oo..oo} (-1)^k/2^((3*k+1)*k/2) (by Euler's pentagonal number theorem). - Amiram Eldar, Aug 13 2020

From Peter Bala, Dec 15 2020: (Start)

Constant C = Sum_{n >= 0} (-1)^n/( Product_{k = 1..n} (2^k - 1) ). The above conjectural result by Schulte follows by adding terms of this series in pairs.

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

C = (3/8)*Sum_{n >= 0} (-1/4)^n/( Product_{k = 1..n} (2^k - 1) ).

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

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

This latter identity generalizes as:

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

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

3*7*C = 6 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),

3*7*15*C = 91 - Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),

and so on, where the sequence [1, 0, 1, 6, 91, ...] is A005327.

(End)

From Amiram Eldar, Feb 19 2022: (Start)

Equals sqrt(2*Pi/log(2)) * exp(log(2)/24 - Pi^2/(6*log(2))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(2))) (McIntosh, 1995).

Equals Sum_{n>=0} (-1)^n/A005329(n).

Equals exp(-A335764). (End)

Equals 1/A065446. - Hugo Pfoertner, Nov 23 2024

Extensions

Corrected by Hieronymus Fischer, Jul 28 2007