OFFSET
1,5
COMMENTS
a(n) >= k for n > A129210(k).
a(n) > 0 iff A033461(n) > 0. - Reinhard Zumkeller, Oct 07 2014
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
P. T. Bateman, A. J. Hildebrand and G. B. Purdy, Sums of distinct squares, Acta Arithmetica 67 (1994), pp. 349-380.
Eryk Pawilan, Math Overflow question
EXAMPLE
1 = 1^2 so a(1) = 1.
2 and 3 are not sums of distinct squares, so a(2) = 0 and a(3) = 0.
4 = 2^2 so a(4) = 1.
5 = 1^2 + 2^2 so a(5) = 2.
MAPLE
N:= 100: # to get a(1) to a(N)
M:= floor(sqrt(N)):
A:= Array(0..N, 0..M):
sj:= unapply(sum(k^2, k=1..x), x):
for j from 1 to M do
for n from sj(j)+1 to N do A[n, j]:= -infinity od:
for n from 1 to j^2-1 do A[n, j]:= A[n, j-1] od:
for n from j^2 to min(sj(j), N) do A[n, j]:= max(A[n, j-1], 1+A[n-j^2, j-1]) od:
od:
subs(-infinity=0, [seq(A[n, M], n=1..N)]); # Robert Israel, Oct 07 2014
MATHEMATICA
Nt = 100 (* = number of terms *);
M = Floor[Sqrt[Nt]];
Clear[A]; A[_, _] = 0;
s[j_] := Range[j].Range[j];
For[j = 1, j <= M, j++,
For[n = s[j] + 1, n <= Nt, n++, A[n, j] = -Infinity];
For[n = 1, n <= j^2 - 1, n++, A[n, j] = A[n, j - 1]];
For[n = j^2, n <= Min[s[j], Nt], n++, A[n, j] = Max[A[n, j-1], 1+A[nj^2, j-1]]]
];
Table[A[n, M] /. DirectedInfinity[-1] -> 0, {n, 1, Nt}] (* Jean-François Alcover, Mar 04 2019, after Robert Israel *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Israel, Oct 07 2014
STATUS
approved