Talk:Pentagonal number theorem

does anybody know what a pentagonal number is?

Iam 11 years old and doing a project on the number 22 and need to know what a pentagonal number is so if any body would be so kind as to tell me I would be very thankful. Katey5000 (talk) 22:47, 30 September 2009 (UTC)

See pentagonal number. 22 is the 4th pentagonal number. Gandalf61 (talk) 08:58, 1 October 2009 (UTC)

Re. rv - was better before

Far enough. The addition of a second example in 12 makes the ideas clearer but floating across my mind is that infinite Euler product. How do we know the coefficient (of 12) is the sign we think it should be if there are an infinite number of terms (of'-1') that make up that distribution? I can see that the 5 and 12 will always have different signs but the actual sign we give to the first is arbitrary no? —Preceding unsigned comment added by 89.243.193.14 (talk) 15:55, 6 February 2011 (UTC)

Although the Euler product is infinite, only a finite number of its terms contribute anything (other than a factor of 1) to the term in x¹² on the right hand side. In other words, the coefficient of x¹² in the expansion of the infinite Euler product is the same as the coefficient of x¹² in the truncated finite product

$\prod _{n=1}^{12}(1-x^{n})$

There are 15 partitions of 12 into distinct parts:

1 partition into one part: 12

5 partitions into two distinct parts: 1+11, 2+10, 3+9, 4+8, 5+7

7 partitions into three distinct parts: 1+2+9, 1+3+8, 1+4+7, 1+5+6, 2+3+7, 2+4+6, 3+4+5

2 partitions into four distinct parts: 1+2+3+6, 1+2+4+5

so we can expand the term in x¹² as follows:

$-x^{12}$

$+x^{1}x^{11}+x^{2}x^{10}+x^{3}x^{9}+x^{4}x^{8}+x^{5}x^{7}$

$-x^{1}x^{2}x^{9}-x^{1}x^{3}x^{8}-x^{1}x^{4}x^{7}-x^{1}x^{5}x^{6}-x^{2}x^{3}x^{7}-x^{2}x^{4}x^{6}-x^{3}x^{4}x^{5}$

$+x^{1}x^{2}x^{3}x^{6}+x^{1}x^{2}x^{4}x^{5}$

$=-x^{12}$

Gandalf61 (talk) 18:07, 6 February 2011 (UTC)

I have a program in BASIC that uses this theorem to calculate Partition numbers.

10 cls

20 input "number of items = ";n

30 dim p(n)

40 let p(0) = 1

50 let p(1) = 1

60 let i = 2

70 print "n","p(n)"

80 print 1,1

90 rem begin loop

100 for k = 1 to i

110 let m = (-1)^(k-1)

120 let gk = k*(3*k-1)/2

130 let gj = k*(3*k+1)/2

140 if gk > i then goto 170

150 let t = i-gk

160 let p(i) = p(i)+m*p(t)

170 if gj > i then goto 210

180 let s = i-gj

190 let p(i) = p(i)+m*p(s)

200 next k

210 print i,p(i)

220 let i = i+1

230 if i > n then end

240 goto 90

250 end

146.90.100.129 (talk) 23:08, 10 February 2013 (UTC)

Incorrect signs in summation

As written, the summation for recursive computation of partition numbers does not give the correct sequence of signs. The summation has alternating signs. The explicit sum above it does not. I wonder what the correct expression for the power of (-1) is. -- 66.103.112.157 (talk) 20:28, 10 October 2015 (UTC)

No, I think it is correct. The alternating sum is over the entire integers, but it "folds" to the sum shown. Try computing a handful of terms of the summation (both positive and negative k). --JBL (talk) 21:00, 10 October 2015 (UTC)

That implies to me that the power of the $(-1)$ term should be not $k-1$, but instead g_k-1. So, instead of the current :$p(n)=\sum _{k\neq 0}(-1)^{k-1}p(n-g_{k})$ it should be :$p(n)=\sum _{k\neq 0}(-1)^{{g_{k}}-1}p(n-g_{k})$. But even this doesn't work precisely because the term for $g_{1}$ should be positive. 98.11.243.148 (talk) 15:44, 12 November 2024 (UTC)

No, it doesn't imply that. (I would say something more substantive but you don't say enough for me to understand the nature of your error.) --JBL (talk) 00:10, 13 November 2024 (UTC)

Case 2 in Franklin's bijective proof

The second exception in Franklin's bijective proof (m=s+1) corresponds to k=1-m (rather than k=m-1 as currently stated). — Preceding unsigned comment added by 89.158.32.144 (talk) 20:37, 23 October 2022 (UTC)

Yes unfortunately it was changed in June by an incompetent editor. I have corrected it; thank you for pointing it out. --JBL (talk) 22:29, 23 October 2022 (UTC)

The “pentagonal number theorem” is not only for the partition function (sequence A000041 in the OEIS), but also for the sum-of-divisors function (sequence A000203 in the OEIS)

The “pentagonal number theorem” is not only for the partition function (sequence A000041 in the OEIS), but also for the sum-of-divisors function (sequence A000203 in the OEIS), the only difference is when the last term is $\sigma (0)$, then change it to $n$ instead of 1 or 0, see the articles [1] and [2] and [3]. 218.187.64.145 (talk) 18:32, 9 June 2023 (UTC)