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So I have the closed formula of:

 (n(2n+1)(2n-1))/3

This closed form is the summation of all odd squares to n. I am now trying to figure out how to subtract something less than or equal to n.

I am trying to get the closed form of all odd squares up to n, so it should only add odd squares that are less than or equal to n. For example if n=9 then the summation should equal 10, because the perfect odd squares less than or equal to 9 are: 1 and 9, 1+9=10.

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  • $\begingroup$ It may not be readily clear what you're asking. You may consider clarifying and using Sigma Notation. (columbia.edu/itc/sipa/math/summation.html) $\endgroup$ – electronpusher May 15 '17 at 5:11
  • $\begingroup$ That is kind of the problem, I am not sure how to do a summation formula when I want to do a summation of the numbers, with the numbers not being greater than n itself. $\endgroup$ – There Is No Spoon May 15 '17 at 5:12
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    $\begingroup$ The first thing to do is, don't use the same letter to describe two different things. Either $n$ is the number of terms you add and you use some other letter to describe the upper bound (the number you want each term to be less than or equal to), or you use $n$ for the upper bound and some other letter to say how many terms to add. Then you may understand the answer that was already given. $\endgroup$ – David K May 15 '17 at 13:23
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First note that your formula can be interpreted as the nth term of a sequence:

$$\{n(2n+1)(2n-1)/3\}=1,10,35,84,165,...$$

You claim that your nth term formula gives the summation (more properly, the partial sums) of all squares of odd numbers, up to $n$. This implies that term $n=3$, for example, is the sum of the first three odd squares:

$$35=1+9+25=1^2+3^2+5^2$$

This appears to be correct, but I have not tested beyond the first four terms. I would be interested to see a proof of the claim that this sequence is the partial sums of the squares of odd numbers.

This is the nth partial sum of the squares of odd numbers:

$$S_n=\sum_{i=1}^{n}(2i-1)^2=1+9+25+...+(2n-1)^2$$

Note that odd numbers are produced by the moiety $\{2i+1\}$ or $\{2i-1\}$ and even numbers by $\{2i\}$; this may help you construct sequences and series in the future.

Let's examine some partial sums of this series:

$$S_1=\sum_{i=1}^{1}(2i-1)^2=1$$ $$S_2=\sum_{i=1}^{2}(2i-1)^2=1+9=10$$ $$S_3=\sum_{i=1}^{3}(2i-1)^2=1+9+25=35$$ $$S_4=\sum_{i=1}^{4}(2i-1)^2=1+9+25+49=84$$

Notice that the partial sums are indeed following your sequence (for the first four terms at least):

$$1,10,35,84,...=\{n(2n+1)(2n-1)/3\}$$

So, we may suspect an equation between the partial sums of the squares of odd numbers and the sequence of the partial sums:

$$\sum_{i=1}^{n}(2i-1)^2=\frac{n(2n+1)(2n-1)}{3}\equiv S_n$$

where "$\equiv$" means "define as", so your function $S_n$ can be calculated by either of the above expressions.

UPDATE

Given an arbitrary integer $N$, you seem to be seeking the sum of the squares of odd numbers up to but not exceeding $N$. That is, you desire the partial sum $S_n$ where none of the $n$ terms (all squares of odd numbers) exceeds $N$. (Correct me if I'm wrong).

I propose the following algorithm as a solution to your problem:

For arbitrary input integer $N$, define the integer index $n$:

$$n\equiv \Big\lfloor\frac{\sqrt N+1}{2}\Big\rfloor$$

where $\lfloor x\rfloor$ is the floor function, rounding down the input x to the nearest integer. I suppose I would show the derivation for the above expression if you request it, but I've already spent more time than I should have on this little puzzle.

When the appropriate parameter value $n$ is obtained from arbitrary input $N$, the value you seek is the $n$th partial sum of the squares of odd numbers:

$$S_n\equiv\sum_{i=1}^{n}(2i-1)^2 \\=\frac{n(2n+1)(2n-1)}{3} \\=\frac{1}{3}\Big(4n^3-n\Big)$$

where whichever form seems most convenient may be used for calculation.

TL;DR I think the function you desire is

$$f(N)\equiv\sum_{i=1}^{\lfloor(\sqrt N+1)/2\rfloor}(2i-1)^2 \\=\frac{1}{3}\Big(4\Big\lfloor\frac{\sqrt N+1}{2}\Big\rfloor^3-\Big\lfloor\frac{\sqrt N+1}{2}\Big\rfloor\Big)$$

I have listed two equivalent forms for convenience. This function is intended to return the sum of the squares of odd numbers, truncated so that no term (odd number square) is larger than the input $N$. I invite you to tabulate values of $f(N)$ for integer inputs $N$ and let me know if this function fulfills your your needs. If I may ask, what motivates your interest in such a function?

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  • $\begingroup$ So i understand this summation, and this is what I got originally as you see above with my: (n(2n+1)(2n-1))/3, however I am trying to exclude any odd square that is NOT <= n. therefore if n=9, the output should be 10, because the perfect odd squares that are <= 9 are 1 and 9, 1+9=10 $\endgroup$ – There Is No Spoon May 15 '17 at 6:20
  • $\begingroup$ How would I accomplish limiting the summation to exclude adding anything > n $\endgroup$ – There Is No Spoon May 15 '17 at 6:23
  • $\begingroup$ @There Is No Spoon See update. $\endgroup$ – electronpusher May 15 '17 at 13:11
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    $\begingroup$ This is a beautiful, elegant, and thorough answer. As a student, I appreciate gems like this. $\endgroup$ – gen-z ready to perish May 15 '17 at 16:24
  • $\begingroup$ @user56478 Glad you found it valuable, I know that feeling. :) $\endgroup$ – electronpusher May 15 '17 at 18:16

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