## Number of partitions of n into exactly 3 parts

Determine the number of triplets consistinq of three different positive integers the sum of which is exactly n

In each of the following cases:

(1)   n   is even
(2)   n   is odd
(3)   n   is divisible by 3

*************************************

Let   $n$   be a positive integer, and let   $p(n)$   denote the number of triplets consisting of three different positive integers the sum of which is exactly   $n$

The number of ordered partitions of n into three positive summands is

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; = \; (n^2 \; - \; 3 n \; + \; 2)/2$

if   $n$   is divisible by 3, then there is one partition with three equal summands
If   $n$   is even, then   $(n - 2)/2$   of the unordered partitions have two equal summands.
If   $n$   is odd, then   $(n - 1)/2$   of the unordered partitions have two equal summands.

So the number of ordered partitions of   $n$   into distinct summands is:

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 2)/2 \; + \; 2 \; = \; (n^2 - 6 n + 12)/2$

$= \; 18 \, k^2 \; - \; 18 \, k \; + \; 6$     if   $n \; = \; 6 \, k$

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 1)/2 \; = \; (n^2 - 6 \, n + 5)/2$

$= \; 18 \, k^2 \; - \; 12 \, k$     if   $n \; = \; 6 \, k \; + \; 1$

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 2)/2 \; = \; (n^2 - 6 \, n + 8)/2$

$= \; 18 \, k^2 \; - \; 6 \, k$     if   $n \; = \; 6 \, k \; + \; 2$

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 1)/2 \; + \; 2 \; = \; (n^2 - 6 \, n + 9)/2$

$= \; 18 \, k^2$     if   $n \; = \; 6 \, k \; + \; 3$

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 2)/2 \; = \; (n^2 - 6 \, n + 8)$

$= \; 18 \, k^2 \; + \; 6 \, k$     if   $n \; = \; 6 \, k \; + \; 4$

$\dbinom{ \;n - 1 \;}{ \;2 \;} \; - \; 3(n - 1)/2 \; = \; (n^2 - 6 \, n + 5)$

$= \; 18 \, k^2 \; + \; 12 \, k$     if   $n \; = \; 6 \, k \; + \; 5$

To find the number of of unordered partitions, we divide the number of ordered 6.

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

### 5 Responses to Number of partitions of n into exactly 3 parts

1. paul says:

The format is

{–, –, –, –,..},{–, –, –, –,….} where the first bracket is the linear recurrence, the second bracket is the sequence generated, or the number of triplets.

For n is even

{2,-1,1,-2,1} , {1,2,4,7,10,14,19,24,30,37,44,52,61,70,80}

For n is odd

{2,-1,1,-2,1} , {1,3,5,8,12,16,21,27,33,40,48,56,65,75,85}

For n is divisible by 3

{2,0,-2,1} , {1,3,7,12,19,27,37,48,61,75,91,108,127,147,169}

this is how the sequence works, for the last example.

There are 4 terms to determine the sequence, so we take the first 4 terms in the sequence. (2 x 12) + (0 x 7) + (-2 x 3) + (1 x 1) = 24 + 0 – 6 + 1 = 19, or the next term. To find the next term you would take (2 x 19) + (0 x 12) + (-2 x 7) + (1 x 3) = 38 + 0 – 14 + 3 = 27. and so on. Using this method does require you to know the first few terms, 5 of them for the first 2 parts and 4 for the last one.

Paul

2. paul says:

I should have also stated at what number (n) each sequence started at,

for even it is 6, so for n = 6 there is 1 partition, for n = 8 there are 2 etc..
for n is odd starts at 7
for n is div by 3 starts at 6
Paul.

3. paul says:

Here are a few more sequences with different divisibility’s

n is divisible by 5, Start number is 10
Linear Recurrence {1,1,0,-1,-1,1}
Sequence {4,12,24,40,61,85,114,147,184,225,271,320,374,432,494}

n is divisible by 7, Start number is 7
Linear Recurrence {1,1,0,-1,-1,1}
Sequence {1,10,27,52,85,127,176,234,300,374,456,547,645,752,867}

n is divisible by 9, Start number is 9
Linear Recurrence {2,0,-2,1}
Sequence {3,19,48,91,147,217,300,397,507,631,768,919,1083,1261,1452}

n is divisible by 11, Start number is 11
Linear Recurrence {1,1,0,-1,-1,1}
Sequence {5,30,75,140,225,331,456,602,768,954,1160,1387,1633,1900,2187}

n is divisible by 12, Start number is 12
Linear Recurrence {3,-3,1}
Sequence {7,37,91,169,271,397,547,721,919,1141,1387,1657,1951,2269,2611}

Pal.

4. paul says:

I have determined there are only 6 recurrence sequences for all divisibility numbers, they are cyclic starting at 6, here they are

6, {3,-3,1}
7, {1,1,0,-1,-1,1}
8, {2,-1,1,-2,1}
9, {2,0,-2,1}
10, {2,-1,1,-2,1}
11, {1,1,0,-1,-1,1}
12, {3,-3,1}

etc.

Paul.