OFFSET
1,2
COMMENTS
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.
LINKS
Robert Israel, Table of n, a(n) for n = 1..57
FORMULA
a(n) = Product_{i=1..n} A005061(i).
a(n) ~ c * 2^(n*(n+1)), where c = QPochhammer(3/4) = 0.015545038845451847... . - Vaclav Kotesovec, Oct 10 2016
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
MAPLE
seq(mul(4^i-3^i, i=1..n), n=0..20); # Robert Israel, Jun 01 2023
MATHEMATICA
Table[Product[4^i - 3^i, {i, n}], {n, 11}] (* Michael De Vlieger, Mar 07 2016 *)
FoldList[Times, Table[4^n-3^n, {n, 20}]] (* Harvey P. Dale, Jul 30 2018 *)
PROG
(PARI) a(n) = prod(k=1, n, 4^k-3^k); \\ Michel Marcus, Mar 05 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Bob Selcoe, Mar 02 2016
STATUS
approved