|
| |
|
|
A242293
|
|
Greedy residue sequence of cubes 2^3, 3^3, 4^3, ...
|
|
4
|
|
|
|
7, 18, 28, 25, 0, 1, 8, 0, 19, 15, 18, 0, 0, 19, 11, 15, 0, 0, 7, 9, 20, 27, 10, 0, 6, 0, 0, 15, 6, 11, 8, 9, 11, 6, 27, 10, 23, 0, 0, 0, 2, 2, 0, 9, 0, 9, 10, 0, 15, 27, 6, 17, 2, 21, 16, 0, 12, 5, 1, 17, 26, 6, 18, 6, 2, 0, 10, 1, 2, 14, 21, 10, 5, 17, 11
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,1
|
|
|
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. 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 A242293, s = (1^3, 2^3, 3^3, ...).
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
n ... n^3 ... a(n)
1 ... 1 .... (undefined)
2 ... 8 ..... 7 = 8 - 1
3 ... 27 .... 18 = 27 - 8 - 1
4 ... 64 .... 28 = 64 - 27 - 8 - 1
5 ... 125 ... 25 = 125 - 64 - 27 - 8 - 1
6 ... 216 ... 0 = 216 - 125 - 64 - 27
7 ... 343 ... 1 = 343 - 216 - 125 - 1
8 ... 512 ... 8 = 512 - 125 - 27 - 8 - 1
9 ... 729 ... 0 = 729 - 512 - 216 - 1
|
|
|
MATHEMATICA
|
z = 200; s = Table[n^3, {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}] (* A242293 *)
c = Table[Length[t[[n]][[2]]], {n, 2, z}] (* A242294 *)
f = 1 + Flatten[Position[tr, 0]] (* A242295*)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
