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A005329
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a(n) = Product_{i=1..n} (2^i - 1). Also called 2-factorial numbers.
(Formerly M3085)
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70
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1, 1, 3, 21, 315, 9765, 615195, 78129765, 19923090075, 10180699028325, 10414855105976475, 21319208401933844325, 87302158405919092510875, 715091979502883286756577125, 11715351900195736886933003038875, 383876935713713710574133710574817125
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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FORMULA
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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)
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EXAMPLE
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G.f. = 1 + x + 3*x^2 + 21*x^3 + 315*x^4 + 9765*x^5 + 615195*x^6 + 78129765*x^7 + ...
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MAPLE
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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);
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MATHEMATICA
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a[ n_] := If[ n < 0, 0, (-1)^n QPochhammer[ 2, 2, n]]; (* Michael Somos, Jan 28 2018 *)
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PROG
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(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
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CROSSREFS
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Cf. A000225, A005321, A006125, A114604, A006088, A028362, A027871 (3-fac), A027872 (5-fac), A027873 (6-fac), A048651, A048652, A075271, A075272, A032085, A122746.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Better definition from Leslie Ann Goldberg (leslie(AT)dcs.warwick.ac.uk), Dec 11 1999
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STATUS
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approved
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