%I #42 May 07 2023 01:22:37

%S 1,10,1200,1596000,23365440000,3763004112000000,

%T 6666387564654720000000,129909027758312519942400000000,

%U 27847153692160782464830528512000000000,65662131721505488121539650946349537280000000000

%N a(n) = Product_{i=1..n} (11^i - 1).

%C It appears that the number of trailing zeros in a(n) is A191610(n). - _Robert Israel_, Nov 24 2015

%H Vincenzo Librandi, <a href="/A027879/b027879.txt">Table of n, a(n) for n = 0..43</a>

%F 10^n|a(n) for n>=0; 12*(10)^(n)|a(n) n>=2. - _G. C. Greubel_, Nov 21 2015

%F a(n) ~ c * 11^(n*(n+1)/2), where c = Product_{k>=1} (1-1/11^k) = 0.900832706809715279949862694760647744762491192216... . - _Vaclav Kotesovec_, Nov 21 2015

%F E.g.f. E(x) satisfies E'(x) = 11 E(11 x) - E(x). - _Robert Israel_, Nov 24 2015

%F Equals 11^(binomial(n+1,2))*(1/11;1/11)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - _G. C. Greubel_, Dec 24 2015

%F G.f.: Sum_{n>=0} 11^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 11^k*x). - _Ilya Gutkovskiy_, May 22 2017

%F Sum_{n>=0} (-1)^n/a(n) = A132267. - _Amiram Eldar_, May 07 2023

%p seq(mul(11^i-1,i=1..n),n=0..20; # _Robert Israel_, Nov 24 2015

%t FoldList[Times,1,11^Range[10]-1] (* _Harvey P. Dale_, Aug 13 2013 *)

%t Abs@QPochhammer[11, 11, Range[0, 40]] (* _G. C. Greubel_, Nov 24 2015 *)

%o (PARI) a(n)=prod(i=1,n,11^i-1) \\ _Anders Hellström_, Nov 21 2015

%o (Magma) [1] cat [&*[11^k-1: k in [1..n]]: n in [1..11]]; // _Vincenzo Librandi_, Dec 24 2015

%Y 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), A027878 (q=10), A027880 (q=12).

%Y Cf. A132267, A191610.

%K nonn

%O 0,2

%A _N. J. A. Sloane_