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 be a positive integer, and let denote the number of triplets consisting of three different positive integers the sum of which is exactly

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

if is divisible by 3, then there is one partition with three equal summands

If is even, then of the unordered partitions have two equal summands.

If is odd, then of the unordered partitions have two equal summands.

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

if

if

if

if

if

if

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

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

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.

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.

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.

Check out my post