A132020 Decimal expansion of Product_{k>=0} (1 - 1/(2*4^k)).

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

Offset: 0

Author

Hieronymus Fischer, Aug 14 2007

Keywords

• nonn
• cons

Comments

This is the limiting probability that a large random symmetric binary matrix is nonsingular (cf. A086812, A048651). In other words, equals Lim_{n->oo} A086812(n)/A006125(n+1).- H. Tracy Hall, Sep 07 2024

Examples

			0.41942244179510759770995610770297425223395323439266674908044991663177...
		

Links

• John E. Cremona and R. W. K. Odoni, Some density results for negative Pell equations; an application of graph theory, Journal of the London Mathematical Society s2-39:1 (1989), pp. 16-28.

Crossrefs

Cf. A004171, A048651, A098844, A067080, A132019, A132026, A132028, A100221, A000217, A081845.

Programs

Maple

evalf(1+sum((-1)^n*2^(n*(n-1)/2)/product(2^k-1, k=1..n), n=1..infinity), 120); # Robert FERREOL, Feb 23 2020

Mathematica

RealDigits[ Product[1 - 1/(2*4^i), {i, 0, 175}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
RealDigits[QPochhammer[1/2, 1/4], 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)

PARI

prodinf(k=0,1-1.>>(2*k+1)) \\ Charles R Greathouse IV, Nov 16 2012

Formula

Equals lim inf_{n->oo} Product_{k=0..floor(log_4(n))} floor(n/4^k)*4^k/n.

Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^((1/2)*(1+floor(log_4(n)))*floor(log_4(n))).

Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^A000217(floor(log_4(n))).

Equals (1/2)*exp(-Sum_{n>0} (4^(-n)*(Sum_{k|n} 1/(k*2^k)))).

Equals lim inf_{n->oo} A132028(n)/A132028(n+1).

Equals Product_{k>0} (1-1/(2^k+1)). - Robert G. Wilson v, May 25 2011

From Robert FERREOL, Feb 23 2020: (Start)

Equals Product_{k>0} (1 + 1/2^k)^(-1) = 2/A081845.

Equals 1 + Sum_{n>=1} (-1)^n*2^(n*(n-1)/2)/((2-1)*(2^2-1)*...*(2^n-1)). (End)

From Peter Bala, Jan 15 2021: (Start)

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

Faster converging series:

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

(2^4)*C = 7*Sum_{n >= 0} 2^(-3*n)/Product_{k = 1..n} (1 - 4^k);

(2^9)*C = 7*31*Sum_{n >= 0} 2^(-5*n)/Product_{k = 1..n} (1 - 4^k), and so on.

Slower converging series:

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

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

7*31*C = -Sum_{n >= 0} 2^(7*n)/Product_{k = 1..n} (1 - 4^k), and so on. (End)

Equals Product_{n>=0} (1 - 1/A004171(n)). - Amiram Eldar, May 09 2023

Extensions

Name corrected by Charles R Greathouse IV, Nov 16 2012