OFFSET
2,2
COMMENTS
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.
LINKS
Clark Kimberling, Table of n, a(n) for n = 2..2000
EXAMPLE
n ... a(n)
1 ... (undefined)
2 ... 1 = 2 - 1
3 ... 2 = 3 - 1
4 ... 3 = 4 - 1
5 ... 0 = 5 - 4 - 1
6 ... 1 = 6 - 4 - 1
7 ... 2 = 7 - 4 - 1
8 ... 3 = 8 - 4 - 1
9 ... 4 = 9 - 4 - 1
10 .. 0 = 10 - 9 - 1
11 .. 1 = 11 - 9 - 1
12 .. 2 = 12 - 9 - 1
13 .. 0 = 13 - 9 - 4
MATHEMATICA
z = 200; s = Table[n^2, {n, 1, z}]; s1 = Table[n, {n, 1, z}]; t = Table[{s1[[n]], #, Total[#] == s1[[n]]} &[DeleteCases[-Differences[FoldList[If[#1 - #2 >= 0, #1 - #2, #1] &, s1[[n]],
Reverse[Select[s, # < s1[[n]] &]]]], 0]], {n, z}]
r[n_] := s1[[n]] - Total[t[[n]][[2]]];
tr = Table[r[n], {n, 2, z}] (* A242305 *)
c = Table[Length[t[[n]][[2]]], {n, 2, z}] (* A242306 *)
f = 1 + Flatten[Position[tr, 0]] (* A242307 *) (* Peter J. C. Moses, May 06 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 11 2014
STATUS
approved