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3, 2, 6, 4, 1, 2, 1, 9, 2, 1, 2, 3, 2, 3, 5, 1, 2, 1, 1, 6, 1, 2, 5, 79, 6, 4, 5, 1, 1, 1, 1, 12, 1, 1
Decimal expansion of Product_{k >= 1} (1 - 1/2^k).
+10
99
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
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
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 354-361.
FORMULA
Lim inf Product_{k=0..floor(log_2(n))} floor(n/2^k)*2^k/n for n->oo. - Hieronymus Fischer, Aug 13 2007 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. - Hieronymus Fischer, Aug 13 2007 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 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)
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).
EXAMPLE
(1/2)*(3/4)*(7/8)*(15/16)*... = 0.288788095086602421278899721929230780088911904840685784114741...
MATHEMATICA
RealDigits[ Product[1 - 1/2^i, {i, 100}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
PROG
(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
CROSSREFS
Cf. A002884, A001318, A005327, A005329, A048652, A079555, A098844, A067080, A100220, A132019, A132020, A132026, A132038, A070933, A261584, A335764.
a(n) = Product_{i=1..n} (2^i - 1). Also called 2-factorial numbers.
(Formerly M3085)
+10
73
1, 1, 3, 21, 315, 9765, 615195, 78129765, 19923090075, 10180699028325, 10414855105976475, 21319208401933844325, 87302158405919092510875, 715091979502883286756577125, 11715351900195736886933003038875, 383876935713713710574133710574817125
COMMENTS
Conjecture: this sequence is the inverse binomial transform of A075272 or, equivalently, the inverse binomial transform of the BinomialMean transform of A075271. - John W. Layman, Sep 12 2002 To win a game, you must flip n+1 heads in a row, where n is the total number of tails flipped so far. Then the probability of winning for the first time after n tails is A005329 / A006125. The probability of having won before n+1 tails is A114604 / A006125. - Joshua Zucker, Dec 14 2005 Number of upper triangular n X n (0,1)-matrices with no zero rows. - Vladeta Jovovic, Mar 10 2008 Equals the q-Fibonacci series for q = (-2), and the series prefaced with a 1: (1, 1, 1, 3, 21, ...) dot (1, -2, 4, -8, ...) if n is even, and (-1, 2, -4, 8, ...) if n is odd. For example, a(3) = 21 = (1, 1, 1, 3) dot (-1, 2, -4, 8) = (-1, 2, -4, 24) and a(4) = 315 = (1, 1, 1, 3, 21) dot (1, -2, 4, -8 16) = (1, -2, 4, -24, 336). - Gary W. Adamson, Apr 17 2009 Number of chambers in an A_n(K) building where K=GF(2) is the field of two elements. This is also the number of maximal flags in an n-dimensional vector space over a field of two elements. - Marcos Spreafico, Mar 22 2012 Given probability p = 1/2^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, A114604(n)/ A006125(n+2) = 1-a(n)/ A006125(n+1) is the probability that the outcome has occurred up to and including the n-th iteration. The limiting ratio is 1- A048651 ~ 0.7112119. These observations are a more formal and generalized statement of Joshua Zucker's Dec 14, 2005 comment. - Bob Selcoe, Mar 02 2016 Also the number of dominating sets in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017 Empirical: Letting Q denote the Hall-Littlewood Q basis of the symmetric functions over the field of fractions of the univariate polynomial ring in t over the field of rational numbers, and letting h denote the complete homogeneous basis, a(n) is equal to the absolute value of 2^ A000292(n) times the coefficient of h_{1^(n*(n+1)/2)} in Q_{(n, n-1, ..., 1)} with t evaluated at 1/2. - John M. Campbell, Apr 30 2018 The series f(x) = Sum_{n>=0} x^(2^n-1)/a(n) satisfies f'(x) = f(x^2), f(0) = 1. - Lucas Larsen, Jan 05 2022
REFERENCES
Annie Cuyt, Vigdis Brevik Petersen, Brigitte Verdonk, Haakon Waadeland, and William B. Jones, Handbook of continued fractions for special functions, Springer, New York, 2008. (see 19.2.1)
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.
Mark Ronan, Lectures on Buildings (Perspectives in Mathematics; Vol. 7), Academic Press Inc., 1989.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n)/2^(n*(n+1)/2) -> c = 0.2887880950866024212788997219294585937270... (see A048651, A048652). G.f.: Sum_{n>=0} 2^(n*(n+1)/2) * x^n / (Product_{k=0..n} (1+2^k*x)).
Compare to: 1 = Sum_{n>=0} 2^(n*(n+1)/2) * x^n/(Product_{k=1..n+1} (1+2^k*x)). (End)
G.f. satisfies: A(x) = 1 + Sum_{n>=1} x^n/n! * d^n/dx^n x*A(x). - Paul D. Hanna, Apr 21 2012 a(n) = 2^(binomial(n+1,2))*(1/2; 1/2)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015 O.g.f. as a continued fraction of Stieltjes' type: A(x) = 1/(1 - x/(1 - 2*x/(1 - 6*x/(1 - 12*x/(1 - 28*x/(1 - 56*x/(1 - ... -(2^n - 2^floor(n/2))*x/(1 - ... )))))))) (follows from Heine's continued fraction for the ratio of two q-hypergeometric series at q = 2. See Cuyt et al. 19.2.1).
A(x) = 1/(1 + x - 2*x/(1 - (2 - 1)^2*x/(1 + x - 2^3*x/(1 - (2^2 - 1)^2*x/(1 + x - 2^5*x/(1 - (2^3 - 1)^2*x/(1 + x - 2^7*x/(1 - (2^4 - 1)^2*x/(1 + x - ... ))))))))). (End)
0 = a(n)*(a(n+1) - a(n+2)) + 2*a(n+1)^2 for all n>=0. - Michael Somos, Feb 23 2019 Sum_{n>=0} (-1)^n/a(n) = A048651. (End)
EXAMPLE
G.f. = 1 + x + 3*x^2 + 21*x^3 + 315*x^4 + 9765*x^5 + 615195*x^6 + 78129765*x^7 + ...
MAPLE
A005329 := proc(n) option remember; if n<=1 then 1 else (2^n-1)*procname(n-1); end if; end proc: seq( A005329(n), n=0..15);
MATHEMATICA
a[ n_] := If[ n < 0, 0, (-1)^n QPochhammer[ 2, 2, n]]; (* Michael Somos, Jan 28 2018 *)
PROG
(PARI) a(n)=polcoeff(sum(m=0, n, 2^(m*(m+1)/2)*x^m/prod(k=0, m, 1+2^k*x+x*O(x^n))), n) \\ Paul D. Hanna, Sep 17 2009 (PARI) Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D
a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1+sum(k=1, n, x^k/k!*Dx(k, x*A+x*O(x^n) ))); polcoeff(A, n) \\ Paul D. Hanna, Apr 21 2012 (PARI) {a(n) = if( n<0, 0, prod(k=1, n, 2^k - 1))}; /* Michael Somos, Jan 28 2018 */ (PARI) {a(n) = if( n<0, 0, (-1)^n * sum(k=0, n+1, (-1)^k * 2^(k*(k+1)/2) * prod(j=1, k, (2^(n+1-j) - 1) / (2^j - 1))))}; /* Michael Somos, Jan 28 2018 */ (Magma) [1] cat [&*[ 2^k-1: k in [1..n] ]: n in [1..16]]; // Vincenzo Librandi, Dec 24 2015 (GAP) List([0..15], n->Product([1..n], i->2^i-1)); # Muniru A Asiru, May 18 2018
CROSSREFS
Cf. A000225, A005321, A006125, A114604, A006088, A028362, A027871 (3-fac), A027872 (5-fac), A027873 (6-fac), A048651, A048652, A075271, A075272, A032085, A122746.
EXTENSIONS
Better definition from Leslie Ann Goldberg (leslie(AT)dcs.warwick.ac.uk), Dec 11 1999
Decimal expansion of Product_{k>=1} (1-1/2^k)^(-1).
+10
25
3, 4, 6, 2, 7, 4, 6, 6, 1, 9, 4, 5, 5, 0, 6, 3, 6, 1, 1, 5, 3, 7, 9, 5, 7, 3, 4, 2, 9, 2, 4, 4, 3, 1, 1, 6, 4, 5, 4, 0, 7, 5, 7, 9, 0, 2, 9, 0, 4, 4, 3, 8, 3, 9, 1, 3, 2, 9, 3, 5, 3, 0, 3, 1, 7, 5, 8, 9, 1, 5, 4, 3, 9, 7, 4, 0, 4, 2, 0, 6, 4, 5, 6, 8, 7, 9, 2, 7, 7, 4, 0, 2, 9, 4, 8, 4, 3, 3, 5, 3, 5, 0, 8, 8, 0
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 354-361.
FORMULA
Equals 1/QPochhammer(1/2, 1/2)_{infinity}. - G. C. Greubel, Jan 18 2018 Equals 1 + Sum_{n>=1} 2^(n*(n-1)/2)/((2-1)*(2^2-1)*...*(2^n-1)). - Robert FERREOL, Feb 22 2020
EXAMPLE
3.46274661945506361153795734292443116454075790290...
MAPLE
evalf(1+sum(2^(n*(n-1)/2)/product(2^k-1, k=1..n), n=1..infinity), 120); # Robert FERREOL, Feb 22 2020
MATHEMATICA
N[ Product[ 1/(1 - 1/2^k), {k, 1, Infinity} ], 500 ]
RealDigits[1/QPochhammer[1/2, 1/2], 10, 100][[1]] (* Vaclav Kotesovec, Jun 22 2014 *)
PROG
(PARI) { default(realprecision, 2080); x=prodinf(k=1, 1/(1 - 1/2^k)); for (n=1, 2000, d=floor(x); x=(x-d)*10; write("b065446.txt", n, " ", d)) } \\ Harry J. Smith, Oct 19 2009
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