A299173 a(n) is the maximum number of squared consecutive positive integers into which the integer n can be partitioned.

1, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 2, 3, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 3, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 3, 0, 0, 0, 4, 5, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 2, 4, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset

1,5

Comments

a(k^2)>=1, the inequality being strict if k is in A097812.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

25 = 5^2 = 3^2 + 4^2 and no such partition is longer, so a(25) = 2.
30 = 1^2 + 2^2 + 3^2 + 4^2 and no such partition is longer, so a(30) = 4.
2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2 and no such partition is longer, so a(2018) = 12. (This special example is due to Seiichi Manyama.) - Jean-François Alcover, Feb 05 2018

Maple

N:= 200: # to get a(1)..a(N)
A:= Vector(N):
S:= n -> n*(n+1)*(2*n+1)/6:
M:= floor(sqrt(N)):
for d from 1 to M do
  for b from d to M do
    s:= S(b) - S(b-d);
    if s > N then break fi;
    A[s]:= d
od od:
convert(A, list); # Robert Israel, Feb 04 2018

Mathematica

terms = 100; jmax = Ceiling[Sqrt[terms]]; kmax = Ceiling[(3*terms)^(1/3)]; Clear[a]; a[_] = 0; Do[r = Range[j, j + k - 1]; n = r . r; If[k > a[n], a[n] = k], {j, jmax}, {k, kmax}]; Array[a, terms]

Crossrefs

Cf. A034705, A097812, A130052, A111044, A234304, A234311, A296338, A298467.

Sequence in context: A338690 A343221 A327169 * A351726 A285720 A269175

Adjacent sequences: A299170 A299171 A299172 * A299174 A299175 A299176

Keyword

nonn,look

Author

Jean-François Alcover, Feb 04 2018

Status

approved