Re: Looking for a closed form for this!
- From: Dan <30pack@xxxxxxxxxxxxx>
- Date: Wed, 20 Jul 2005 13:59:21 EDT
Dan wrote:
>> Triangle #'s = 1,3,6,10,15,21...
>
>> Just observing odd composites of the form where p = prime
>> and where p > 5 and c = composite.
>> c == 0 (mod p)
>
>> Where every odd composite of the above form is just a
>> negation of a smaller triangle # from a larger one.
>> This also involves composites with more than 2 prime
>> factors where each prime factor >5.
>> e.g.
>
>> 55 = 10th triangle #
>> -6 = 3rd triangle #
>> ----
>> 49 = 7^2
>
<snip>
>>
>>
>> Please give a counter example if I am wrong!
>
>> If right, is there a closed form for any given
>> composite with the above composite restrictions
>> that will produce the two unknown tri. #'s?
>
>> In some cases there is more than 1 answer such as 77
>> (7*11)
>> 78 = 12th tri. #
>> -1 = 1st tri. #
>> ----
>> 77
>>
>> 105 = 14th tri. #
>> -28 = 7th tri. #
>> ----
>> 77
>>
>The n-th triangular number, T(n), is n(n + 1)/2. Your >observation is
>that for integers n > m > 0, we'll have
>T(n) - T(m) = n(n + 1)/2 - m(m + 1)/2
>= (n^2 + n)/2 - (m^2 + m)/2
>= ((n^2 - m^2) + (n - m)) / 2
>= ((n - m)(n + m) + (n - m)) / 2
>= (n - m)(n + m + 1) / 2
>Now, suppose we wish to express a number, AB, with A >= >B and B odd,
>as the difference of two triangular numbers. By the >above, we can
>set
>n - m = B [1]
>and
>(n + m + 1)/2 = A
>so
>n + m + 1 = 2A
>and hence
>n + m = 2A - 1 [2]
>Adding [1] and [2] gives us
>2n = B + 2A - 1
>so
>n = (2A + B - 1) / 2 [3]
>and from [1]
>m = n - B = (B + 2A - 1)/2 - B
>= (2A - B - 1)/ 2 [4]
>For example, with 216 we can write 216 = 24 * 9, so with
>A = 24 and B = 9 we have
>n = (2*24 + 9 - 1) / 2 = 28 from [3]
>and
>m = (2*24 - 9 - 1) / 2 = 19 from [4]
>and finally we see
>406 (= T(28))
>-190 (= -T(19)
>----
>216 tadaa!
>Regards,
>Rick
Neat closed form!
So there are multiple solutions not counting subtracting
two consecutive tri. #.
Using your closed form ---
6338580 = 3560th tri. #
-1987021 = 1993th tri. #
----------
4351559 = 1597*2777
On my OP this was represented as --
4367490 = 2955 tri. #
-15931 = 178th tri. #
---------
4351559 = 1567*2777
Also an interesting way to determine if a large integer
is a triangle # ---
n = a 20 digit integer
Then sqrt(n*2) and if the remainder is .4999999999xxxxxxxxx..x where x = any digit 0-9
In other words the .4999999999 will be 1/2 (10) the
length of the original integer. If integer is odd length
then the .499.. is half the length + 1. It is a triangle
number.
If the number shows < .49 then int (n) * int (n+1)/2
If the number shows >. 4999999999999999..9 then int (n+1) * int(n+2)/2
This will place these integers that are not triangle #'s
on the appropreate tri. # line.
Dan
.
- References:
- Looking for a closed form for this!
- From: Dan
- Looking for a closed form for this!
- Prev by Date: Re: set of a set etc.
- Next by Date: Re: set of a set etc.
- Previous by thread: Re: Looking for a closed form for this!
- Next by thread: Re: Looking for a closed form for this!
- Index(es):
Relevant Pages
|
