A297199 -id:A297199 - cp's OEIS frontend

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

Seiichi Manyama, Jan 14 2018

nonn

			   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.

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

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

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 *)

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)

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

A307609 Number of partitions of n^3 into consecutive positive cubes.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Ilya Gutkovskiy, Apr 18 2019

nonn

			20^3 = 11^3 + 12^3 + 13^3 + 14^3, so a(20) = 2.
2856^3 = 213^3 +...+ 555^3 = 273^3 +...+ 560^3, so a(2856) = 3. See also _Donovan Johnson_'s comment in A097811. - _Antti Karttunen_, Aug 22 2019

Antti Karttunen, Table of n, a(n) for n = 1..16861
Index entries for sequences related to sums of cubes

Cf. A000578, A030272, A097811, A131643, A217843, A259792, A297199.

A297199(n) = { my(s=0, k=1, c); while((c=k^3) <= n, my(u=n-c, i=k); while(u>0, i++; c = i^3; u=u-c); s += (!u); k++); (s); };
A307609(n) = A297199(n^3); \\ Antti Karttunen, Aug 22 2019

a(n) = [x^(n^3)] Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^3).
a(n) = A297199(A000578(n)).
a(n) >= 2 for n in A097811.

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Ruby

Formula