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

 

Read:   Partition (number theory)

 
 

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

 
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.

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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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.

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