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A260688 a(n) = the least number of pieces of currency of denominations .01, .05, .10, .25, 1, 5, 10, 20, 50, 100 that the greedy algorithm uses to make n times .01 (n "cents") in change.

%I A260688 #26 Jul 19 2024 04:24:08
%S A260688 0,1,2,3,4,1,2,3,4,5,1,2,3,4,5,2,3,4,5,6,2,3,4,5,6,1,2,3,4,5,2,3,4,5,
%T A260688 6,2,3,4,5,6,3,4,5,6,7,3,4,5,6,7,2,3,4,5,6,3,4,5,6,7,3,4,5,6,7,4,5,6,
%U A260688 7,8,4,5,6,7,8,3,4,5,6,7,4,5,6,7,8,4,5,6,7,8,5,6,7,8,9,5,6,7,8,9
%N A260688 a(n) = the least number of pieces of currency of denominations .01, .05, .10, .25, 1, 5, 10, 20, 50, 100 that the greedy algorithm uses to make n times .01 (n "cents") in change.
%H A260688 US Treasury, <a href="https://www.treasury.gov/resource-center/faqs/Coins/Pages/denominations.aspx">Denominations of Coins</a>
%H A260688 US Treasury, <a href="https://www.treasury.gov/resource-center/faqs/Currency/Pages/denominations.aspx">Denominations of Paper Currency</a>
%o A260688 (Python)
%o A260688 def how_many(cents):
%o A260688     #d = denominations
%o A260688     d = ['$0.01', '$0.05', '$0.10', '$0.25',
%o A260688          '$1', '$5', '$10', '$20', '$50', '$100']
%o A260688     coins = {coin: 100*float(str(coin)[1:]) for coin in d}
%o A260688     how_many = {d[i]: 0 for i in range(10)}
%o A260688     while len(d) != 0:
%o A260688         how_many[d[-1]] = cents // coins[d[-1]]
%o A260688         cents %= coins[d[-1]]
%o A260688         d.pop()
%o A260688     return int(sum(how_many.values()))
%Y A260688 Cf. A053344, A067997, A080897, A108536.
%K A260688 nonn
%O A260688 0,3
%A A260688 _Edward Minnix III_, Nov 15 2015
%E A260688 Edited by _N. J. A. Sloane_, Apr 24 2016