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

0,9

Links

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

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.]

Example

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.

Prog

(Scheme, two variants)
(define (A261225 n) (- n (A055401 n)))
(definec (A261225 n) (if (zero? n) n (+ -1 (A048762 n) (A261225 (- n (A048762 n))))))

Crossrefs

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

Cf. also A260740.

Sequence in context: A186684 A255910 A108689 * A240830 A245402 A024583

Adjacent sequences: A261222 A261223 A261224 * A261226 A261227 A261228

Keyword

nonn

Author

Antti Karttunen, Aug 16 2015

Status

approved