# Talk:Square pyramidal number

In all articles found the formula for the square pyramidal number is proofed by using the inductive methode. Isn't there a direct proof?
Jon van den Helder

I thought inductive was direct. An empirical proof (e.g., having a computer calculate the first thousand values using each formula and then having it compare that the lists are all the same) would convince more people, but symbological snobs would look down on it. Anton Mravcek 17:49, 14 July 2006 (UTC)
This article could use some diagrams to illustrate how the numbers are "built" as pyramids. — Gwalla | Talk 22:28, 10 February 2007 (UTC)

I tried making one at Image:Square pyramidal number.svg but my radial gradiants are not looking correct. I'll try to find out what's wrong with the svg coding, but it shouldn't be linked onto the actual article until it looks better. —David Eppstein 23:04, 10 February 2007 (UTC) Done! —David Eppstein 23:30, 10 February 2007 (UTC)
Nice work! Thanks! — Gwalla | Talk 18:56, 11 February 2007 (UTC)

what's wrong with pyramid number?  —Preceding unsigned comment added by 208.2.172.2 (talk) 01:05, 3 July 2009 (UTC) 

The infinite sum of Tetrahedral numbers reciprocals 
  
    
      
        
          
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              6
              
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    {\displaystyle \scriptstyle \sum _{n=1}^{\infty }{\frac {6}{n(n+1)(n+2)}}={\frac {3}{2}}}
  
.  What will be the infinite sum of Square pyramidal number? 46.115.58.82 (talk) 23:49, 12 October 2012 (UTC)

Try expanding using partial fractions. Ozob (talk) 14:45, 13 October 2012 (UTC)

I have an immense amount of experimentally discovered things I could add to "Squares in a Square" which involve different grid shapes and altogether different shapes such as triangles. So much so that I could very easily create a whole other Wikipedia article on it. I would enjoy doing so, but I thought I would get a second opinion. In addition, although my stuff is definitely true, as I said, I figured out this stuff experimentally, so I have nothing to reference but myself. I know that isn't a very good answer, but I could really improve the depth of understanding and knowledge of that topic. I hope you will comment. Frivolous Consultant (talk) 22:56, 17 October 2012 (UTC)

You need to find a published source for this material if you want to include it here; see Wikipedia:Verifiability and Wikipedia:No original research. In addition, if the solution doesn't involve the same sequence of numbers, it may be a bit off-topic for this article, because this article is not about square-in-square type problems, it's about the sequence of numbers. That doesn't mean it can't be mentioned at all (after all, we do mention the problem of rectangles in a square grid) but it means that it would probably be inappropriate to include much detail. —David Eppstein (talk) 23:01, 17 October 2012 (UTC)
I sort-of expected this answer, but I don't know what to do. People don't care enough about this topic, so no one makes anything about it. I've searched all over the Internet, but I haven't found anything. Although I haven't searched any books, I wouldn't know where to start, and besides, I would be exceptionally surprised if they had anything of value in them. If anyone can find anything, I'll defintely cite it, but as of now, I have nothing. I've thought about creating a formal website so that I have something to source, but I don't know how to do that. This isn't the first time something like this has happened to me before either. I could expand on that, but I'm already taking up too much space. I could add a formula to the "Squares in a square" part to improve it that wouldn't take up too much space and go off topic, but it still wouldn't have a reference. If you ask, I could figure out how to put it on here so you can look it over. That would be only a measely fraction of the unreferenceable knowledge I have on the topic. Also, I was suggesting creating a whole other page on Wikipedia for putting the rest and/or moving "Squares in a square" there, not adding a bunch to this article. I tried putting the suggestion on Wikipedia: Articles for creation today, but I couldn't get it to show up. I would appreciate help. Frivolous Consultant (talk) 00:20, 19 October 2012 (UTC)
Have you tried the Online Encyclopedia of Integer Sequences?  You should be able to generate some sequences of numbers from your results and plug them into OEIS.  I think it's pretty likely that you'll find some of your results there.  You may also find references, maybe enough to write an article.  Ozob (talk) 01:50, 21 October 2012 (UTC)
That website helped a lot. There is so much information on there about what I was looking for that it gave me a slight inferiority complex because of how many hours and hours I spent working on this stuff when it took me under a second to get a full sequence along with formulas. To my relief though, there was a few things that it didn't know, so I don't feel as bad, but it knows enough to give a good resorce. Be prepared for a new article on the subject, but there are two more things. First, I don't know what I should title it; I've recently been refering to the topic as tessellation conglomerates for lack of a better term, but that name is completely made up by me. Also, when I finish, it might be good to move "Squares in a square" to the page. (I realize what I've been typing takes up a lot of room. I won't be offended if you delete my previous entries.) Frivolous Consultant (talk) 23:21, 25 October 2012 (UTC)
If people "don't care enough about this topic" it may not be appropriate for wikipedia, meaning not notable--345Kai (talk) 03:12, 23 June 2015 (UTC)

I found that the "square pyramidal number" can be used to prove the Archimedes' theorem on the area of ​​parabolic segment. The proof, carried out without the use of "mathematical analysis", is readable at the following web adress:
https://drive.google.com/file/d/0B4iaQ-gBYTaJMDJFd2FFbkU2TU0/view?usp=sharing

See at: https://sites.google.com/site/leggendoarchimede  — Preceding unsigned comment added by Ancora Luciano (talk • contribs) 13:06, 27 June 2013 (UTC) 

We represent the square pyramidal number P6 = 91 with cubes of unit volume, as shown, and inscribe in building a pyramid (in red). Let V6 the volume of the inscribed pyramid. To obtain P6 you may add to V6 the excess external volume to the red pyramid. Such excess is: 2/3 for each cube placed on the central edge, and 1/2 for the cubes forming the steps of the building (enlarge for a better look of highlighted part). Then, calculating one has: 
P6 = V6+(2/3)*6+(1+2+3+4+5)
For the induction principle, will be:
Pn = Vn+(2n)/3+Σn(from 1 to n-1)n 
Pn = n3/3+2n/3+(n2+n)/2-n 
Pn = (2n3+3n2+n)/6
Geometric representation of the square pyramidal number using cubes, instead of sferes, is more useful.
George Polya, in his book "Mathematical discovery", Volume 2, 1968, presents this solution saying it "rained from the sky", as obtained algebraically with a trick, like a rabbit drawn out from the hat.
Returning to the introduction to this talk page, seems that this (realistic) geometrical derivation is a little more direct than the algebraic presented in the article.
In the article "Summation" was added the talk: "Sum of the first n cubes (geometrical proof)", please to see it.
--Ancora Luciano (talk) 06:15, 22 May 2013 (UTC)

Something seems amiss in the sentence starting "Now there". I suggest replacing "Now there" with "There are".
Bill01568 (talk) 16:29, 30 December 2013 (UTC)

Seems reasonable,  Done. LittleMountain5 22:35, 30 December 2013 (UTC)

It seems that current proof given in the article could be simplified by using 
  
    
      
        
          1
          
            2
          
        
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          2
          
            2
          
        
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        …
        +
        
          n
          
            2
          
        
        =
        1
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        2
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        3
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        3
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        …
        +
        n
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        n
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        …
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        n
      
    
    {\displaystyle 1^{2}+2^{2}+\ldots +n^{2}=1+2+2+3+3+3+\ldots +n+n+\ldots +n}
  
 instead of 
  
    
      
        
          1
          
            2
          
        
        +
        
          2
          
            2
          
        
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        …
        +
        
          n
          
            2
          
        
        =
        (
        2
        n
        −
        1
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        +
        (
        2
        n
        −
        3
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        +
        (
        2
        n
        −
        3
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        +
        …
        +
        1
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        1
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        …
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        1
      
    
    {\displaystyle 1^{2}+2^{2}+\ldots +n^{2}=(2n-1)+(2n-3)+(2n-3)+\ldots +1+1+\ldots +1}
  
, with sum of the column still being 
  
    
      
        2
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    {\displaystyle 2n+1}
  
. Unfortunately, I really can't be bothered to find citations for this proof. Fortunately, there are no citations for the current proof as well, so changing it won't make it worse. Opinions?
MYXOMOPbI4 (talk) 03:51, 7 September 2017 (UTC)

I'm not sure I see the point of long algebraic derivations at all, given that it's so straightforward to prove any such formula by induction. If you could replace this with a short conceptual visual (and sourced!) proof, that would be better, I think. —David Eppstein (talk) 04:06, 7 September 2017 (UTC)
The proof by induction is indeed not that hard, but you need to know the formula in the first place. This way you get the formula automatically, and the proof isn't long at all (it can be done with one picture, http://forumbgz.ru/user/upload/file580638.jpg (with some unimportant text in russian)). But I'm not sure if some random picture on the internet is considered a source. MYXOMOPbI4 (talk) 22:53, 7 September 2017 (UTC)
It's more or less obvious that the formula is a cubic polynomial and from that and the first four values you can immediately derive it by finite differences. And no, random pictures on the internet, especially from a site whose address syntax suggests that it's an open wiki, are not reliable sources. —David Eppstein (talk) 00:27, 8 September 2017 (UTC)

A link to Lucas Numbers should be provided for completeness. 199.209.255.246 (talk) 14:40, 10 September 2018 (UTC)

Only if the connection can be sourced. Otherwise the unsourced paragraph mentioning Lucas should be removed. —David Eppstein (talk) 16:33, 10 September 2018 (UTC)

I don't see any mention of Archimedes, who probably gave the first formula for the sum of squares of the first n natural numbers in his book 'On Spirals'.  It doesn't at first sight look like the modern formula, and the derivation is horribly complicated, but it is there after all.  See the discussion in Heath's edition of Archimedes, especially on page 109 in the Dover edition.2A00:23C8:7906:1301:A453:48F9:36D5:B594 (talk) 21:23, 7 March 2021 (UTC)

Added. I used the 1897 edition rather than the Dover edition, but the pagination is the same. —David Eppstein (talk) 00:43, 8 March 2021 (UTC)