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A055401 Number of positive cubes needed to sum to n using the greedy algorithm. 6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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]

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = a(n-floor(n^(1/3))^3)+1 = a(A055400(n))+1 = a(n-A048762(n))+1 [with a(0) = 0]

EXAMPLE

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

PROG

(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 */

(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

CROSSREFS

Cf. A018888, A055400.

Cf. A002376 (least number of positive cubes needed to represent n).

Cf. A053610, A048766, A000573.

Sequence in context: A053843 A010886 A002376 * A053829 A033928 A194754

Adjacent sequences:  A055398 A055399 A055400 * A055402 A055403 A055404

KEYWORD

easy,nonn

AUTHOR

Henry Bottomley, May 16 2000

STATUS

approved

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Last modified December 20 06:03 EST 2014. Contains 252241 sequences.