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A243113
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Minimum of the cube root of the largest element over all partitions of n into at most 5 cubes.
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2
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0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 5, 3, 3, 3, 3, 3, 3, 4, 4, 5, 4, 3, 3, 3, 4, 4, 4, 4, 5, 4, 3, 4, 4, 3, 4, 4, 4, 4, 5
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OFFSET
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0,7
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COMMENTS
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It is known that every number can be written as the sum of at most 5 (positive or negative) cubes.
"Minimum of the cube root of the largest absolute element over all partitions of n into at most 5 cubes" gives a different sequence with differences at n=302, 509, 517, 518, 521, 581, 733, 860, 1076, 1228, 1642, 1733, 1741, 1885, 2012, ... . - Alois P. Heinz, Aug 26 2014
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LINKS
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EXAMPLE
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For n=5, a(n)=1. The partition of 5 into 1^3 + 1^3 + 1^3 + 1^3 + 1^3 has largest summand 1^3, while any other such partition, take 2^3 -1^3 -1^3 -1^3 for example, will have a larger largest part.
a(302) = 7: 7^3 +7^3 +4^3 +4^3 -8^3 = 302.
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MAPLE
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b:= proc(n, i, t) option remember; n=0 or (0<=i or n<=i^3)
and t>0 and (b(n, i-1, t) or b(n-i^3, i, t-1))
end:
a:= proc(n) local k; for k from 0
do if b(n, k, 5) then return k fi od
end:
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MATHEMATICA
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b[n_, i_, t_] := b[n, i, t] = n==0 || (0 <= i || n <= i^3) && t>0 && (b[n, i-1, t] || b[n-i^3, i, t-1]); a[n_] := For[k=0, True, k++, If[b[n, k, 5], Return[k]]]; Table[a[n], {n, 0, 120}] (* Jean-François Alcover, Feb 17 2017, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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