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A027878 a(n) = Product_{i=1..n} (10^i - 1).

%I A027878 #37 May 07 2023 01:22:55
%S A027878 1,9,891,890109,8900199891,890011088900109,890010198889020099891,
%T A027878 8900101098880002109889900109,890010100987899112108987901010099891,
%U A027878 890010100097889011121088788901111989989900109
%N A027878 a(n) = Product_{i=1..n} (10^i - 1).
%H A027878 G. C. Greubel, <a href="/A027878/b027878.txt">Table of n, a(n) for n = 0..50</a>
%F A027878 a(n) ~ c * 10^(n*(n+1)/2), where c = Product_{k>=1} (1-1/10^k) = A132038 = 0.890010099998999000000100009999999989999900000000... . - _Vaclav Kotesovec_, Nov 21 2015
%F A027878 3^n*(11)^(floor(n/2)) divides a(n) for n>=0. - _G. C. Greubel_, Nov 24 2015
%F A027878 Equals 10^(binomial(n+1,2))*(1/10;1/10)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - _G. C. Greubel_, Dec 24 2015
%F A027878 G.f.: Sum_{n>=0} 10^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 10^k*x). - _Ilya Gutkovskiy_, May 22 2017
%F A027878 From _Amiram Eldar_, May 07 2023: (Start)
%F A027878 Sum_{n>=0} 1/a(n) = A132326.
%F A027878 Sum_{n>=0} (-1)^n/a(n) = A132038. (End)
%t A027878 Table[Product[10^i-1,{i,n}],{n,0,10}] (* _Harvey P. Dale_, Aug 15 2011 *)
%t A027878 Abs@QPochhammer[10, 10, Range[0, 30]] (* _G. C. Greubel_, Nov 24 2015 *)
%o A027878 (PARI) a(n) = prod(k=1, n, 10^k - 1) \\ _Altug Alkan_, Nov 25 2015
%o A027878 (Magma) [1] cat [&*[10^k-1: k in [1..n]]: n in [1..11]]; // _Vincenzo Librandi_, Dec 24 2015
%Y A027878 Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027879 (q=11), A027880 (q=12).
%Y A027878 Cf. A132038, A132326.
%K A027878 nonn
%O A027878 0,2
%A A027878 _N. J. A. Sloane_