OFFSET
1,2
COMMENTS
Generally, 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 up to and including the n-th iteration. Here, j=3 and k=2, so p=(2/3)^n and r = 1-a(n)/A047656(n+1). The limiting ratio of r ~ 0.9307279.
FORMULA
a(n) = Product_{i=1..n} A001047(i).
a(n) ~ c * 3^(n*(n+1)/2), where c = QPochhammer(2/3) = 0.0692720728018644... . - Vaclav Kotesovec, Oct 10 2016
MAPLE
MATHEMATICA
Table[Product[3^i - 2^i, {i, n}], {n, 15}] (* Wesley Ivan Hurt, Mar 02 2016 *)
FoldList[Times, Table[3^i-2^i, {i, 15}]] (* Harvey P. Dale, Feb 06 2017 *)
PROG
(Magma) [&*[ 3^k-2^k: k in [1..n] ]: n in [1..16]]; // Vincenzo Librandi, Mar 03 2016
(PARI) a(n) = prod(k=1, n, 3^k-2^k); \\ Michel Marcus, Mar 05 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bob Selcoe, Mar 02 2016
STATUS
approved