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A055401
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Number of positive cubes needed to sum to n using the greedy algorithm.
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6
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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
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OFFSET
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1,2
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COMMENTS
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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]
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LINKS
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_Reinhard Zumkeller_, Table of n, a(n) for n = 1..10000
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FORMULA
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a(n) = a(n-floor(n^(1/3))^3)+1 = a(A055400(n))+1 = a(n-A048762(n))+1 [with a(0) = 0]
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EXAMPLE
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a(33)=7 because 33=27+1+1+1+1+1+1 (not 33=8+8+8+8+1)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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easy,nonn
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AUTHOR
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Henry Bottomley, May 16 2000
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STATUS
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approved
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