%I #2 Mar 31 2012 10:30:54
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,2,2,2,2,2,1,1,1,
%U 1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3
%N The number of ways of minimal weight to make change for n cents using fairly valued United States coins (copper 1-cent coin, a nickel 5-cent coin, and silver 10-cent and 25-cent coins) assuming that silver is more valuable than nickel and that nickel is more valuable than copper.
%F G.f.: [1/(1-x^10)/(1-x^25)+x^5+x^15][1+x+x^2+x^3+x^4]
%e For n = 51 cents, the least weight is achieved with 50 cents in silver and 1 cent in copper. The 50 cents in silver can be achieved as two 25-cent coins or five 10-cent coins; thus there are a(51) = 2 ways to make 51 cents with minimal weight.
%Y Except for the values dependent upon nickel (i.e., a(5) through a(9) and a(15) through a(19)) this sequence can be constructed by repeating five times each term from sequence A008616.
%K easy,nonn
%O 0,51
%A _Lee A. Newberg_, May 14 2010