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%I #4 May 09 2014 17:39:43
%S 3,4,2,0,1,0,1,0,2,0,2,0,1,0,1,3,0,0,2,0,2,0,1,0,1,0,1,3,0,2,0,0,2,0,
%T 2,0,1,0,1,0,1,0,1,3,0,2,0,0,1,0,2,0,2,0,1,0,1,0,2,0,1,0,1,3,0,2,0,0,
%U 1,0,1,0,2,0,2,0,1,0,1,0,2,0,2,0,1,0
%N Greedy residue sequence of squares 2^2, 3^2, 4^2, ...
%C Suppose that s = (s(1), s(2), ... ) is a sequence of real numbers such that for every real number u, at most finitely many s(i) are < u, and suppose that x > min(s). We shall apply the greedy algorithm to x, using terms of s. Specifically, let i(1) be an index i such that s(i) = max{s(j) < x}, and put d(1) = x - s(i(1)). If d(1) < s(i) for all i, put r = x - s(i(1)). Otherwise, let i(2) be an index i such that s(i) = max{s(j) < x - s(i(1))}, and put d(2) = x - s(i(1)) - s(i(2)). If d(2) < s(i) for all i, put r = x - s(i(1)) - s(i(2)). Otherwise, let i(3) be an index i such that s(i) = max{s(j) < x - s(i(1)) - s(i(2))}, and put d(3) = x - s(i(1)) - s(i(2)) - s(i(3)). Continue until reaching k such that d(k) < s(i) for every i, and put r = x - s(i(1)) - ... - s(i(k)). Call r the s-greedy residue of x, and call s(i(1)) + ... + s(i(k)) the s-greedy sum for x. If r = 0, call x s-greedy summable. If s(1) = min(s) < s(2), then taking x = s(i) successively for i = 2, 3,... gives a residue r(i) for each i; call (r(i)) the greedy residue sequence for s. When s is understood from context, the prefix "s-" is omitted. For A241833, s = (1^2, 2^2, 3^2, 4^2, ... ).
%H Clark Kimberling, <a href="/A241833/b241833.txt">Table of n, a(n) for n = 2..2000</a>
%e n ... n^2 .. a(n)
%e 1 ... 1 .... (undefined)
%e 2 ... 4 .... 3 = 4 - 1
%e 3 ... 9 .... 4 = 9 - 4 - 1
%e 4 ... 16 ... 2 = 16 - 9 - 4 - 1
%e 5 ... 25 ... 0 = 25 - 16 - 9
%e 6 ... 36 ... 1 = 36 - 25 - 9 - 1
%e 7 ... 49 ... 0 = 49 - 36 - 9 - 4
%e 8 ... 64 ... 1 = 64 - 49 - 9 - 4 - 1
%t z = 200; s = Table[n^2, {n, 1, z}]; t = Table[{s[[n]], #, Total[#] == s[[n]]} &[ DeleteCases[-Differences[FoldList[If[#1 - #2 >= 0, #1 - #2, #1] &, s[[n]], Reverse[Select[s, # < s[[n]] &]]]], 0]], {n, z}]; r[n_] := s[[n]] - Total[t[[n]][[2]]]; tr = Table[r[n], {n, 2, z}] (* A241833 *)
%t c = Table[Length[t[[n]][[2]]], {n, 2, z}] (* A241834 *)
%t f = 1 + Flatten[Position[tr, 0]] (* A241835 *)
%t f^2 (* A241836 *)
%t (* _Peter J. C. Moses_, May 06 2014 *)
%Y Cf. A241834, A241835, A241836, A242252, A000290.
%K nonn,easy
%O 2,1
%A _Clark Kimberling_, May 09 2014
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