

A181580


Smallest prime p such that p! ends with exactly n trailing 0's (or 0, if no such p exists).


1



2, 5, 11, 17, 23, 0, 29, 31, 37, 41, 47, 0, 53, 59, 61, 67, 71, 0, 79, 83, 89, 0, 97, 0, 101, 107, 113, 0, 0, 0, 0, 127, 131, 137, 0, 149, 0, 151, 157, 163, 167, 173, 0, 179, 181, 0, 191, 197, 0, 0, 0, 211, 0, 223, 0, 227, 233, 239, 241, 0, 0, 0, 251, 257, 263, 269, 271, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


LINKS



EXAMPLE

For any positive integer n, n! must have at least as many trailing 0's as does (n1)!; thus, other than the terms where a(n)=0, the sequence is strictly increasing.
2 is the first prime whose factorial (2! = 2) has no trailing 0's, so a(0)=2.
5 is the first prime whose factorial (5! = 120) has one trailing 0, so a(1)=5.
There is no number whose factorial has exactly 5 trailing 0's (since the factorials of 24 and 25 have 4 and 6 trailing 0's, respectively), so a(5)=0.


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



