A248509 Length of longest sequence of distinct nonzero squares summing to n, or 0 if no such sequence exists.

1, 0, 0, 1, 2, 0, 0, 0, 1, 2, 0, 0, 2, 3, 0, 1, 2, 0, 0, 2, 3, 0, 0, 0, 2, 3, 0, 0, 3, 4, 0, 0, 0, 2, 3, 1, 2, 3, 4, 2, 3, 3, 0, 0, 3, 4, 0, 0, 3, 4, 4, 2, 3, 4, 5, 3, 4, 2, 3, 0, 3, 4, 4, 1, 4, 5, 0, 2, 3, 4, 4, 0, 2, 4, 5, 0, 3, 4, 5, 2, 4, 5, 3, 4

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

Cf. A129210.

Cf. A033461.

Sequence in context: A025891 A341000 A120630 * A281542 A331844 A191410

Adjacent sequences: A248506 A248507 A248508 * A248510 A248511 A248512

Keyword

nonn

Author

Robert Israel, Oct 07 2014

Status

approved