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A269576 a(n) = Product_{i=1..n} (4^i - 3^i).

%I A269576 #50 Jun 02 2023 02:00:00
%S A269576 1,7,259,45325,35398825,119187843775,1692109818073675,
%T A269576 99792176520894983125,24195710911432718503470625,
%U A269576 23942309231057283642583777144375,96180015123706384385790918441966041875
%N A269576 a(n) = Product_{i=1..n} (4^i - 3^i).
%C A269576 In general, for sequences of the form a(n) = Product_{i=1..n} j^i-k^i, where j>k>=1 and  n>=1: given probability p=(k/j)^n that an outcome will occur at the n-th stage of an infinite process, then r = 1 - a(n)/j^((n^2+n)/2) is the probability that the outcome has occurred at or before the n-th iteration. Here j=4 and k=3, so p=(3/4)^n and r = 1-a(n)/A053763(n+1). The limiting ratio of r is ~ 0.9844550.
%H A269576 Robert Israel, <a href="/A269576/b269576.txt">Table of n, a(n) for n = 1..57</a>
%F A269576 a(n) = Product_{i=1..n} A005061(i).
%F A269576 a(n) ~ c * 2^(n*(n+1)), where c = QPochhammer(3/4) = 0.015545038845451847... . - _Vaclav Kotesovec_, Oct 10 2016
%F A269576 a(n+3)/a(n+2) - 7 * a(n+2)/a(n+1) + 12 * a(n+1)/a(n) = 0. - _Robert Israel_, Jun 01 2023
%p A269576 seq(mul(4^i-3^i,i=1..n),n=0..20); # _Robert Israel_, Jun 01 2023
%t A269576 Table[Product[4^i - 3^i, {i, n}], {n, 11}] (* _Michael De Vlieger_, Mar 07 2016 *)
%t A269576 FoldList[Times,Table[4^n-3^n,{n,20}]] (* _Harvey P. Dale_, Jul 30 2018 *)
%o A269576 (PARI) a(n) = prod(k=1, n, 4^k-3^k); \\ _Michel Marcus_, Mar 05 2016
%Y A269576 Cf. A005061, A053763.
%Y A269576 Cf. sequences of the form Product_{i=1..n}(j^i - 1): A005329 (j=2), A027871 (j=3), A027637 (j=4), A027872 (j=5), A027873 (j=6), A027875 (j=7),A027876 (j=8), A027877 (j=9), A027878 (j=10), A027879 (j=11), A027880 (j=12).
%Y A269576 Cf. sequences of the form Product_{i=1..n}(j^i - k^1), k>1: A263394 (j=3, k=2), A269661 (j=5, k=4).
%K A269576 nonn
%O A269576 1,2
%A A269576 _Bob Selcoe_, Mar 02 2016