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March 6

Generalized pentagonal numbers

I understand pentagonal numbers and the image at the top of Pentagonal number is an easily understood graphical representation of pentagonal numbers. Given that, I can believe the given formula (though I haven't attempted to verify it to myself yet). However Generalized pentagonal numbers have no "meaning" to me; we take the pentagonal number formula and use it on a strange sequence of "n"s. Why should we consider that is a useful thing to do? Is there some way of visualising the sequence akin to the graphic mentioned earlier? The sequence starts 0, 1, 2, 5, 7, 12, 15 and I can see 0, 1, 5, 12 being pentagonal, but there is nothing apparently pentagonal about 2, 7, 15. So can anyone explain and/or provide some graphics for the sequence? -- SGBailey (talk) 11:58, 6 March 2025 (UTC)[reply]

For square and triangular numbers, n² and n(n+1)/2, you get the same set of numbers if you plug in negative n. This is not true for pentagonal numbers though. In the positive direction in goes 0, 1, 5, 12, 22, ..., and in the negative direction it goes ... 40, 26, 15, 7, 2, 0. (See (sequence A005449 in the OEIS).) I don't know if there's a natural geometric definition of the second set of numbers; the article has a section "Generalized pentagonal numbers and centered hexagonal numbers" which tells us that each centered hexagonal number is the sum of a (regular) pentagonal number and the corresponding (offset by one) negative pentagonal number. This section is unsourced though and I'm not convinced it's anything more than a mathematical coincidence. To me, the real use of generalized pentagonal numbers is Euler's Pentagonal number theorem which gives a relatively simple recurrence relation for the Partition numbers, see that article for details. Note that the definition of the partition numbers apparently has nothing to do with polygonal or geometric numbers of any kind, so it's really kind of an accident that numbers involved in the theorem were related to a sequence that was already well known. The pentagonal number theorem is important because it's a much easier (if more complex) way to compute these numbers than directly from the definition. (Finding an even easier asymptotic formula was of great interest in the early part of the 20th century, see the section "Approximation formulas" in the article.) The partition numbers have connections to other areas of mathematics such as representation theory. One could, I suppose, define generalized n-gonal numbers for any n in the same way, but afiak there isn't much in the way of applications for them. --RDBury (talk) 16:36, 6 March 2025 (UTC)[reply]

Now you have pointed out that it is p(n) and p(-n), the input sequence has become obvious - well it was before, I just didn't see it (!!!). I now observe that p(-n) = p(n) + n . This can be illustrated by drawing the p(n) pentagons and adding a duplicate row below the bottom edge. Thus

 *       *           *
 o     *   *       *   *
        * *      *  * *  *
        o o       *     *
                   * * *
                   o o o
 1,2    5,7        12,15

Thanks. -- SGBailey (talk) 11:06, 7 March 2025 (UTC)[reply]

Now that I've thought about it some, there are two "natural" geometric arrangements in which both p(n) and p(-n) show up in Franklin's bijective proof of the pentagonal number theorem. These are exactly the arrangements (i.e. Ferrers diagrams) that don't cancel themselves out, so they're the ones that turn up in the generating function. And if you ever want to waste some time with a bit of mathematical doodling I sure you can find many other pleasing geometrical ways to compose and decompose both pentagonal and negative pentagonal numbers. If t(n) = n(n+1)/2 is the nth triangular number, and s(n) = n² then p(-n) = s(n)+t(n) and p(n) = s(n)+t(n-1). These correspond to Ferrers diagrams in Franklin's proof.) Or p(n)+p(-n) = 3s(n), where the right hand side gives you a variation on the Centered hexagonal number where there's a small triangle in the center instead of a single dot:

         * * * *
        * * * * *
       * * * * * *
        o o o o o
         o o o o
          o o o

I don't know if such results are particularly significant, but if you're bored on a rainy spring afternoon... --RDBury (talk) 17:31, 7 March 2025 (UTC)[reply]