A296338 a(n) = number of partitions of n into consecutive positive squares.

1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1

Offset

1,25

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Formula

a(A034705(n)) >= 1 for n > 1.

G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^2). - Ilya Gutkovskiy, Apr 18 2019

Example

   1 = 1^2,                   so  a(1) = 1.
   4 = 2^2,                   so  a(4) = 1.
   5 = 1^2 + 2^2,             so  a(5) = 1.
   9 = 3^2,                   so  a(9) = 1.
  13 = 2^2 + 3^2,             so a(13) = 1.
  14 = 1^2 + 2^2 + 3^2,       so a(14) = 1.
  16 = 4^2,                   so a(16) = 1.
  25 = 3^2 + 4^2 = 5^2,       so a(25) = 2.
  29 = 2^2 + 3^2 + 4^2,       so a(29) = 1.
  30 = 1^2 + 2^2 + 3^2 + 4^2, so a(30) = 1.

Mathematica

nMax = 100; t = {0}; Do[k = n; s = 0; While[s = s + k^2; s <= nMax, AppendTo[t, s]; k++], {n, 1, nMax}]; tt = Tally[t]; a[_] = 0; Do[a[tt[[i, 1]]] = tt[[i, 2]], {i, 1, Length[tt]}]; Table[a[n], {n, 1, nMax}] (* Jean-François Alcover, Feb 04 2018, using T. D. Noe's program for A034705 *)

Prog

(Ruby)
def A296338(n)
  m = Math.sqrt(n).to_i
  ary = Array.new(n + 1, 0)
  (1..m).each{|i|
    sum = i * i
    ary[sum] += 1
    i += 1
    sum += i * i
    while sum <= n
      ary[sum] += 1
      i += 1
      sum += i * i
    end
  }
  ary[1..-1]
end
p A296338(100)

Crossrefs

Cf. A000290, A001227, A034705, A130052, A234304, A297199, A298467, A299173.

Sequence in context: A271231 A306798 A086079 * A374216 A133703 A073265

Adjacent sequences: A296335 A296336 A296337 * A296339 A296340 A296341

Keyword

nonn

Author

Seiichi Manyama, Jan 14 2018

Status

approved