# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a027879 Showing 1-1 of 1 %I A027879 #42 May 07 2023 01:22:37 %S A027879 1,10,1200,1596000,23365440000,3763004112000000, %T A027879 6666387564654720000000,129909027758312519942400000000, %U A027879 27847153692160782464830528512000000000,65662131721505488121539650946349537280000000000 %N A027879 a(n) = Product_{i=1..n} (11^i - 1). %C A027879 It appears that the number of trailing zeros in a(n) is A191610(n). - _Robert Israel_, Nov 24 2015 %H A027879 Vincenzo Librandi, Table of n, a(n) for n = 0..43 %F A027879 10^n|a(n) for n>=0; 12*(10)^(n)|a(n) n>=2. - _G. C. Greubel_, Nov 21 2015 %F A027879 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 A027879 E.g.f. E(x) satisfies E'(x) = 11 E(11 x) - E(x). - _Robert Israel_, Nov 24 2015 %F A027879 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 A027879 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 A027879 Sum_{n>=0} (-1)^n/a(n) = A132267. - _Amiram Eldar_, May 07 2023 %p A027879 seq(mul(11^i-1,i=1..n),n=0..20; # _Robert Israel_, Nov 24 2015 %t A027879 FoldList[Times,1,11^Range[10]-1] (* _Harvey P. Dale_, Aug 13 2013 *) %t A027879 Abs@QPochhammer[11, 11, Range[0, 40]] (* _G. C. Greubel_, Nov 24 2015 *) %o A027879 (PARI) a(n)=prod(i=1,n,11^i-1) \\ _Anders Hellström_, Nov 21 2015 %o A027879 (Magma) [1] cat [&*[11^k-1: k in [1..n]]: n in [1..11]]; // _Vincenzo Librandi_, Dec 24 2015 %Y A027879 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 A027879 Cf. A132267, A191610. %K A027879 nonn %O A027879 0,2 %A A027879 _N. J. A. Sloane_ # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE