A261225 n minus the number of positive cubes needed to sum to n using the greedy algorithm: a(n) = n - A055401(n).

0, 0, 0, 0, 0, 0, 0, 0, 7, 7, 7, 7, 7, 7, 7, 7, 14, 14, 14, 14, 14, 14, 14, 14, 21, 21, 21, 26, 26, 26, 26, 26, 26, 26, 26, 33, 33, 33, 33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 47, 47, 47, 52, 52, 52, 52, 52, 52, 52, 52, 59, 59, 63, 63, 63, 63, 63, 63, 63, 63, 70, 70, 70, 70, 70, 70, 70, 70, 77, 77, 77, 77, 77, 77, 77, 77, 84, 84, 84, 89

Offset: 0

Author

Antti Karttunen, Aug 16 2015

Keywords

• nonn

Examples

			a(8) = 7, because when the greedy algorithm partitions 8 into cubes, it first finds 8 (= 2*2*2), thus A055401(8) = 1, and 8-1 = 7.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..12167

Crossrefs

Cf. A048762, A055401, A261226, A261227, A261228, A261229.

Cf. also A260740.

Formula

a(n) = n - A055401(n).

As a recurrence:

a(0) = 0; for n >= 1, a(n) = -1 + A048762(n) + a(n-A048762(n)). [Where A048762(n) gives the largest cube <= n.]