|
|
A252487
|
|
Smallest k such that n^7 = a_1^7 + ... + a_k^7 and all a_i are positive integers less than n.
|
|
1
|
|
|
128, 28, 66, 39, 28, 26, 21, 20, 18, 22, 22, 22, 20, 21, 14, 17, 14, 14, 17, 16, 17, 14, 16, 13, 15, 13, 12, 15, 13, 15, 13, 14, 13, 14, 13, 13, 14, 12, 12, 12, 13, 12, 12, 12, 11, 13, 13, 12, 12, 13, 12, 12, 11, 12, 11, 11, 12, 12, 11, 12, 9, 12, 11, 11, 11
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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, ... occur first at indices 568, 102, 62, ...
I conjecture that the sequence is bounded by the initial term a(2)=128. Probably even a(4)=66, a(5)=39, a(6)=28 and some more are followed only by smaller terms.
I've uploaded two scripts; one to compute the b-file and one to generate an IP file. For the first script, a parameter kmax can be set to gain a speedup but more memory is used. The other one (which also works with large integers now) should be used in case someone has a good IP-solver. Higher terms might be computable faster with a good IP solver. - Manfred Scheucher, Aug 14 2015
From results on Waring's problem, it is known that all a(n) <= A002804(7) = 143, and a(n) <= 33 for all sufficiently large n. - Robert Israel, Aug 16 2015
|
|
LINKS
|
|
|
MAPLE
|
M:= 10^8:
R:= Vector(M, 144, datatype=integer[4]):
for p from 1 to floor(M^(1/7)) do
p7:= p^7;
if p > 1 then A[p]:= R[p7] fi;
R[p7]:= 1;
for j from p7+1 to M do
R[j]:= min(R[j], 1+R[j - p7]);
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^7);
if lb > ub then return -1 fi;
bestyet:= ub;
for m from lb to 0 by -1 do
res:= procname(n-m*k^7, k-1, bestyet-m);
if res >= 0 then
bestyet:= res+m;
fi
od:
return bestyet
end proc:
for n from floor(M^(1/7))+1 to 50 do
A[n]:= F(n^7, n-1, 144)
od:
|
|
PROG
|
(PARI) a(n, verbose=0, m=7)={N=n^m; for(k=3, 999, forvec(v=vector(k-1, i, [1, n\sqrtn(k+1-i, m)]), ispower(N-sum(i=1, k-1, v[i]^m), m, &K)&&K>0&&!(verbose&&print1("/*"n" "v"*/"))&&return(k), 1))}
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|