Talk:Square Numbers which are Sum of Sequence of Odd Cubes

The next one is $1^3 + 3^3 + \dots + 29^3 = 1189^2$.

The largest cube is A001653 cubed:
 * $1, 5, 29, 169, \dots$


 * No, the largest cube is $2 \times \text {A001653} - 1$ cubed.


 * So the next one is $\ds \sum_{j \mathop = 1}^{29} \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3$. See below.

The sum is A046177 squared:
 * $1, 35^2, 1189^2, 40391^2, \dots$

With square roots A046176.

The result can be obtained directly from Sum of Sequence of Odd Cubes. --RandomUndergrad (talk) 19:01, 30 August 2020 (UTC)


 * I've worked out where the mistake is.


 * I have calculated $1^3 + 3^3 + 5^3 + \dotsb + 27^3 + 29^3$ and got $101025$ which is not $1189^2$.


 * But then I calculated $29^2 \paren {2 \times 29^2 - 1} = 1189^2 = 1413721 = \ds \sum_{j \mathop = 1}^{29} \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3$. --prime mover (talk) 20:40, 30 August 2020 (UTC)