|
%I #6 Sep 01 2013 22:35:40
%S 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,
%T 107,113,0,0,0,0,127,131,137,0,149,0,151,157,163,167,173,0,179,181,0,
%U 191,197,0,0,0,211,0,223,0,227,233,239,241,0,0,0,251,257,263,269,271,0
%N Smallest prime p such that p! ends with exactly n trailing 0's (or 0, if no such p exists).
%e For any positive integer n, n! must have at least as many trailing 0's as does (n-1)!; thus, other than the terms where a(n)=0, the sequence is strictly increasing.
%e 2 is the first prime whose factorial (2! = 2) has no trailing 0's, so a(0)=2.
%e 5 is the first prime whose factorial (5! = 120) has one trailing 0, so a(1)=5.
%e 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.
%Y Cf. A181579
%K nonn,base
%O 0,1
%A _Lekraj Beedassy_, Nov 02 2010
|
