

A252486


Smallest k such that n^6 = a_1^6+...+a_k^6 where all the a_i are positive integers less than n.


4



64, 36, 15, 29, 22, 21, 15, 19, 15, 17, 15, 16, 14, 15, 13, 12, 11, 11, 13, 14, 12, 13, 13, 12, 12, 12, 12, 12, 11, 11, 11, 11, 11, 13, 11, 11, 11, 10, 11, 11, 11, 11, 11, 10, 11, 11, 11, 11, 11, 11, 11, 11, 9, 11, 10, 11, 11, 11, 9, 10, 11, 11, 11, 11, 10
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OFFSET

2,1


COMMENTS

Inspired by Fermat's Last Theorem: 2 never occurs in this sequence.
No n is known for which a(n)<7, according to the MathWorld page. The values 7, 8, 9, 10 and 11 occur first at indices 1141, 251, 54, 39, 18, cf. sequence A252476.
I conjecture that the sequence is bounded by the initial term. Probably even a(3)=36, a(5)=29, a(6)=22 and some more are followed only by smaller terms.
From results on Waring's problem, it is known that all a(n) <= A002804(6) = 73, and a(n) <= 24 for all sufficiently large n.  Robert Israel, Aug 17 2015


LINKS



MAPLE

M:= 10^8:
R:= Vector(M, 74, datatype=integer[4]):
for p from 1 to floor(M^(1/6)) do
p6:= p^6;
if p > 1 then A[p]:= R[p6] fi;
R[p6]:= 1;
for j from p6+1 to M do
R[j]:= min(R[j], 1+R[j  p6]);
od
od:
F:= proc(n, k, ub)
local lb, m, bestyet, res;
if ub <= 0 then return 1 fi;
if n <= M then
if n = 0 then return 0
elif R[n] > ub then return 1
else return R[n]
fi
fi;
lb:= floor(n/k^6);
if lb > ub then return 1 fi;
bestyet:= ub;
for m from lb to 0 by 1 do
res:= procname(nm*k^6, k1, bestyetm);
if res >= 0 then
bestyet:= res+m;
fi
od:
return bestyet
end proc:
for n from floor(M^(1/6))+1 to 50 do
A[n]:= F(n^6, n1, 73)
od:


MATHEMATICA

a[n_] := Module[{k}, For[k = 7, True, k++, If[IntegerPartitions[n^6, {k}, Range[n1]^6] != {}, Print[n, " ", k]; Return[k]]]];


PROG

(PARI) a(n, verbose=0, m=6)={N=n^m; for(k=3, 64, forvec(v=vector(k1, i, [1, n\sqrtn(k+1i, m)]), ispower(Nsum(i=1, k1, v[i]^m), m, &K)&&K>0&&!(verbose&&print1("/*"n" "v"*/"))&&return(k), 1))}


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



