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A027876 a(n) = Product_{i=1..n} (8^i - 1).

%I A027876 #41 May 07 2023 01:23:40
%S A027876 1,7,441,225351,922812345,30237792108615,7926625536728661945,
%T A027876 16623330670976050126618695,278893192683059452825059069034425,
%U A027876 37432410397693271164043156886536608251975
%N A027876 a(n) = Product_{i=1..n} (8^i - 1).
%H A027876 G. C. Greubel, <a href="/A027876/b027876.txt">Table of n, a(n) for n = 0..50</a>
%F A027876 a(n) ~ c * 8^(n*(n+1)/2), where c = Product_{k>=1} (1-1/8^k) = A132036 = 0.859405994400702866200758580064418894909484979588... . - _Vaclav Kotesovec_, Nov 21 2015
%F A027876 7^n | a(n). - _G. C. Greubel_, Nov 21 2015
%F A027876 It appears that 7^m | a(n) iff 7^m | (7n)!. - _Robert Israel_, Dec 24 2015
%F A027876 a(n) = 8^(binomial(n+1,2))*(1/8;1/8)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - _G. C. Greubel_, Dec 24 2015
%F A027876 G.f. g(x) satisfies (1+x) g(x) = 1 + 8 x g(8x). - _Robert Israel_, Dec 24 2015
%F A027876 a(n) = Product_{i=1..n} A024088(i). - _Michel Marcus_, Dec 27 2015
%F A027876 G.f.: Sum_{n>=0} 8^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 8^k*x). - _Ilya Gutkovskiy_, May 22 2017
%F A027876 Sum_{n>=0} (-1)^n/a(n) = A132036. - _Amiram Eldar_, May 07 2023
%p A027876 seq(mul(8^i-1,i=1..n), n=0..20); # _Robert Israel_, Dec 24 2015
%t A027876 FoldList[Times,1,8^Range[10]-1] (* _Harvey P. Dale_, Dec 23 2011 *)
%o A027876 (PARI) a(n)=prod(i=1,n,8^i-1) \\ _Charles R Greathouse IV_, Nov 22 2015
%o A027876 (Magma) [1] cat [&*[ 8^k-1: k in [1..n] ]: n in [1..11]]; // _Vincenzo Librandi_, Dec 24 2015
%Y A027876 Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
%Y A027876 Cf. A132036.
%K A027876 nonn
%O A027876 0,2
%A A027876 _N. J. A. Sloane_