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%I #53 Feb 19 2022 07:10:23
%S 1,2,16,416,33280,8053760,5863137280,12816818094080,84078326697164800,
%T 1654829626053597593600,97714379759212830706892800,
%U 17309711516825516108403231948800
%N a(n) = Product_{i=1..n} (3^i - 1).
%C 2*(10)^m|a(n) where 4*m <= n <= 4*m+3 for m >= 1. - _G. C. Greubel_, Nov 20 2015
%C Given probability p = 1/3^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1-a(n)/A047656(n+1) is the probability that the outcome has occurred at or before the n-th iteration. The limiting ratio is 1-A100220 ~ 0.4398739. - _Bob Selcoe_, Mar 01 2016
%H G. C. Greubel, <a href="/A027871/b027871.txt">Table of n, a(n) for n = 0..50</a>
%F a(n) ~ c * 3^(n*(n+1)/2), where c = A100220 = Product_{k>=1} (1-1/3^k) = 0.560126077927948944969792243314140014379736333798... . - _Vaclav Kotesovec_, Nov 21 2015
%F a(n) = 3^(binomial(n+1,2))*(1/3;1/3)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - _G. C. Greubel_, Dec 24 2015
%F a(n) = Product_{i=1..n} A024023(i). - _Michel Marcus_, Dec 27 2015
%F G.f.: Sum_{n>=0} 3^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 3^k*x). - _Ilya Gutkovskiy_, May 22 2017
%F From _Amiram Eldar_, Feb 19 2022: (Start)
%F Sum_{n>=0} 1/a(n) = A132324.
%F Sum_{n>=0} (-1)^n/a(n) = A100220. (End)
%p A027871 := proc(n)
%p mul( 3^i-1,i=1..n) ;
%p end proc:
%p seq(A027871(n),n=0..8) ; # _R. J. Mathar_, Jul 13 2017
%t Table[Product[(3^k-1),{k,1,n}],{n,0,20}] (* _Vaclav Kotesovec_, Jul 17 2015 *)
%t Abs@QPochhammer[3, 3, Range[0, 10]] (* _Vladimir Reshetnikov_, Nov 20 2015 *)
%o (PARI) a(n) = prod(i=1, n, 3^i-1); \\ _Michel Marcus_, Nov 21 2015
%o (Magma) [1] cat [&*[ 3^k-1: k in [1..n] ]: n in [1..11]]; // _Vincenzo Librandi_, Dec 24 2015
%Y Cf. A005329 (q=2), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
%Y Cf. A047656, A100220, A132324.
%K nonn
%O 0,2
%A _N. J. A. Sloane_
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