Matematické Fórum

stuart clark · 30. 04. 2019 16:45

Number of natural number $(a,b,c)$ in $a+b+c=n,$ where $1\leq a<b<c\leq n\in \mathbb{N}$

kerajs · 02. 05. 2019 01:46 — Editoval kerajs (02. 05. 2019 03:47)

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For $n \in \mathbb{N} \setminus \{ 0,1,2,3,4,5\}$ number triplets is: $\sum_{i=1}^{\lfloor \frac{n-3}{3} \rfloor} \lfloor \frac{n-(3i+1)}{2} \rfloor$

stuart clark · 02. 05. 2019 13:54

Thanks ↑ kerajs:. But i did not understand last line

please explain me, thanks

krakonoš

↑ stuart clark:
Hi Stuart Clarc
First I 'll take all the numbers $a,b,c\in \{0,1,2,...,n\}$ where $a+b+c=n$ and $n\ge 4$ .
Subtract the cases where the are two zeros, otherwise one zero.Finally, I will subtract the cases where the are the same digits.
1) n is odd number and not divisible three
$\frac{1}{6}\cdot \{(^{n+2}_{n})-3-3\cdot (n-1)-[\frac{n}{2}]\cdot 3\}$
2)n is even number and not divisible three
$\frac{1}{6}\cdot \{(^{n+2}_{n})-3-3\cdot (n-2)-[\frac{n}{2}]\cdot 3\}$
3) n is odd number and divisible three
$\frac{1}{6}\cdot \{(^{n+2}_{n})-3-3\cdot (n-1)-([\frac{n}{2}]-1)\cdot 3-1\}$
4)n is even number and divisible three
$\frac{1}{6}\cdot \{(^{n+2}_{n})-3-3\cdot (n-2)-([\frac{n}{2}]-1)\cdot 3-1\}$

$[floor function ]$