A296338 - OEIS

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

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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