A055401 Number of positive cubes needed to sum to n using the greedy algorithm.

0, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7

Offset: 0

Author

Henry Bottomley, May 16 2000

Keywords

• easy
• nonn

Comments

Define f(n) = n - k^3 where (k+1)^3 > n >= k^3; a(n) = number of steps such that f(f(...f(n)))= 0.

Also sum of digits when writing n in base where place values are positive cubes, cf. A000433. [Reinhard Zumkeller, May 08 2011]

Examples

			a(32)=6 because 32=27+1+1+1+1+1 (not 32=8+8+8+8).
a(33)=7 because 33=27+1+1+1+1+1+1 (not 33=8+8+8+8+1).
		

Links

• Antti Karttunen & Reinhard Zumkeller (terms 1-10000), Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A018888, A055400.

Cf. A002376 (least number of positive cubes needed to represent n; differs from this sequence for the first time at n=32, where a(32)=6, while A002376(32)=4).

Cf. A053610, A048766, A000578, A000433.

Cf. also A261225, A261226, A261227, A261228, A261229.

Programs

Haskell

a055401 n = s n $ reverse $ takeWhile (<= n) $ tail a000578_list where
  s _ []                 = 0
  s m (x:xs) | x > m     = s m xs
             | otherwise = m' + s r xs where (m',r) = divMod m x
-- Reinhard Zumkeller, May 08 2011
(Scheme, with memoization-macro definec)
(definec (A055401 n) (if (zero? n) n (+ 1 (A055401 (A055400 n)))))
;; Antti Karttunen, Aug 16 2015

Maple

f:= proc(n,k) local m, j;
if n = 0 then return 0 fi;
for j from k by -1 while j^3 > n do od:
m:= floor(n/j^3);
m + procname(n-m*j^3, j-1);
end proc:
seq(f(n,floor(n^(1/3))),n=0..100); # Robert Israel, Aug 17 2015

Mathematica

a[0] = 0; a[n_] := {n} //. {b___, c_ /; !IntegerQ[c^(1/3)], d___} :> {b, f = Floor[c^(1/3)]^3, c - f, d} // Length; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 17 2015 *)

PARI

F=vector(30,n,n^3); /* modify to get other sequences of "greedy representations" */ last_leq(v,F)=
{ /* Return last element <=v in sorted array F[] */
    local(j=1);
    while ( F[j]<=v, j+=1 );
    return( F[j-1] );
}
greedy(n,F)=
{
    local(v=n,ct=0);
    while ( v,  v-=last_leq(v,F); ct+=1; );
    return(ct);
}
vector(min(100,F[#F-1]),n,greedy(n,F)) /* show terms */
/* Joerg Arndt, Apr 08 2011 */

Formula

a(0) = 0; for n >= 1, a(n) = a(n-floor(n^(1/3))^3)+1 = a(A055400(n))+1 = a(n-A048762(n))+1.

Extensions

a(0) = 0 prepended by Antti Karttunen, Aug 16 2015