OFFSET
0,2
COMMENTS
Given probability p = 1/5^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1 - a(n)/A109345(n+1) is the probability that the outcome has occurred at or before the n-th iteration. The limiting ratio is 1-A100222 ~ 0.2396672. - Bob Selcoe, Mar 01 2016
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..50
FORMULA
4^n|a(n) for n >= 1. - G. C. Greubel, Nov 21 2015
a(n) ~ c * 5^(n*(n+1)/2), where c = Product_{k>=1} (1-1/5^k) = A100222 . - Vaclav Kotesovec, Nov 21 2015
a(n) = 5^(binomial(n+1,2))*(1/5; 1/5)_{n}, where (a;q)_{n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015
a(n) = Product_{i=1..n} A024049(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 5^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 5^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A100222. - Amiram Eldar, May 07 2023
MAPLE
MATHEMATICA
Table[Product[(5^k-1), {k, 1, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2015 *)
Abs@QPochhammer[5, 5, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
Join[{1}, FoldList[Times, 5^Range[10]-1]] (* Harvey P. Dale, Dec 28 2021 *)
PROG
(PARI) a(n) = prod(i=1, n, 5^i-1); \\ Michel Marcus, Nov 21 2015
(Magma) [1] cat [&*[ 5^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved