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%I #7 Aug 31 2016 20:55:55
%S 0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6,4,5,1,2,5,6,7,8,6,7,8,9,7,8,9,10,8,9,
%T 10,11,2,3,4,5,3,4,5,6,4,5,6,7,5,6,7,8,6,7,3,4,7,8,9,10,8,9,10,11,9,
%U 10,11,12,10,11,12,13,4,5,6,7,5,6,7,8,6,7,8,9,7,8,9,10,8,9,5,6,9,10,11,12,1
%N a(n) = number of terms of A001563 needed to sum to n when always choosing the largest term that divides the remaining n, a(0) = 0.
%H Antti Karttunen, <a href="/A276332/b276332.txt">Table of n, a(n) for n = 0..4320</a>
%F a(0) = 0; for n >= 1, a(n) = 1 + a(n-A276330(n)) = 1 + a(A276331(n)).
%F Other identities and observations:
%F It seems that a(n) >= A276328(n) for all n.
%e For n=20, the largest term of A001563 that divides 20 is 4 and moreover, 4 is also the largest term of A001563 that divides 16, 12, 8 and 4, thus as 20 = 5*4, a(20) = 5.
%e For n=21, the largest term of A001563 that divides 21 is A001563(1) = 1, and case for 21 - 1 = 20 is shown above, thus a(21) = 1 + 5 = 6.
%o (Scheme, with memoization-macro definec)
%o (definec (A276332 n) (if (zero? n) n (+ 1 (A276332 (A276331 n)))))
%Y Cf. A001563, A276328, A276329, A276330, A276331.
%K nonn
%O 0,3
%A _Antti Karttunen_, Aug 30 2016
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