Conjecture on triangular numbers T(n)

In summary: Secondly, I proved the following: Theorem: If a triangular number T(n) is factored into the product A*B, then the difference between the product of the ith row and the (i+1)th row is one of the even factors of T(n).For example, the triangular number 6 (written as n*(n+1)/2) is factored into the product 12*24 (written as A*B) with A less or equal to B. The difference between the product of the 3rd row and the 5th row is 1, which is one of the even factors of 6. This theorem can be generalized to any triangular number.
  • #1
ramsey2879
841
3
Let a triangular number T(n) = n*(n+1)/2 be factored into the product A*B with A less or equal to B.
Let gcd(x,y) be the greatest common divisor of x and y
For each of pair (A,B) define C,D,E,F as follows

C = (gcd(A,n+1))^2,
D = 2*(gcd(B,n))^2,
E = 2*(gcd(A,n))^2,
F = (gcd(B,n+1))^2.


As an example.
Let n = 36 then T(n) = 666 which can be factored into 6 distinct pairs A,B. The
six sets (A,B,C,D,E,F) are as follows

1. (1,666,1,648, 1369,2)
2. (2,333,1,162,1369,8)
3. (3,222,1,72,1369,18)
4. (6,111,1,18,1369,72)
5. (9,74,1,8,1369,162)
6. (18,37,1,2,1369,648)

My conjecture is that the products (A+Ct)*(B+Dt) and (A+Et)*(B+Ft) are each triangular numbers (= T(r) = r(r+1))/2) for all integer t. Also, the above example gave a interesting result in that when the 12 results for r (= the integer part of the square root of 2T(r)) are sorted by value the difference between adjacent r values when t = 1 was one of the even factors of 666. It would seem to me that the easiest way to prove this conjecture would be to derive and prove a formula for r. I will give the corresponding r values below for t = 1
1. 72,110
2. 54,184
3. 48,258
4. 42,480
5. 40,702
6. 38,1368

Also I found that r follows an arithmetic series as t varies, so it is important to find the pattern for the above values, since the difference between them and 36 is the difference value of the arithmetic series. Anyone see a pattern?
 
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  • #2
just a silly question related with you are posting here: is there triangular prime numbers?

I think should be but in a quick search I didn't find an example
 
  • #3
ramsey2879 said:
It would seem to me that the easiest way to prove this conjecture would be to derive and prove a formula for r.

A formula can be derived from the (positive) solution of r^2 + r - 2*T(r) = 0, which is
[tex]r = {{\sqrt {1 + 8 T(r)} - 1} \over 2}}[/tex]
But I don't see how this helps to prove your conjecture.
 
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  • #4
al-mahed said:
just a silly question related with you are posting here: is there triangular prime numbers?

I think should be but in a quick search I didn't find an example

ramsey2879 said:
Let a triangular number T(n) = n*(n+1)/2

As stated in the original post, any triangular number can be written in the form n(n+1)/2. Since one of either n or n+1 must be even, that is the product of positive integers. The only way it could be prime is if one of those numbers is 1: if n= 1, then 1(2)/2= 1 which is not a prime number. If (n+1)/2= 1, then n+1= 2 so n= 1 again.

No, there are no prime triangular numbers.
 
  • #5
HallsofIvy said:
As stated in the original post, any triangular number can be written in the form n(n+1)/2. Since one of either n or n+1 must be even, that is the product of positive integers. The only way it could be prime is if one of those numbers is 1: if n= 1, then 1(2)/2= 1 which is not a prime number. If (n+1)/2= 1, then n+1= 2 so n= 1 again.

No, there are no prime triangular numbers.


Yes, this is obvious, I didn't see the most obvious thing !
 
  • #6
Actually, T(2)=3 is prime... but the reasoning holds for higher n.
 
  • #7
Progress report

ramsey2879 said:
Let a triangular number T(n) = n*(n+1)/2 be factored into the product A*B with A less or equal to B.
Let gcd(x,y) be the greatest common divisor of x and y
For each of pair (A,B) define C,D,E,F as follows

C = (gcd(A,n+1))^2,
D = 2*(gcd(B,n))^2,
E = 2*(gcd(A,n))^2,
F = (gcd(B,n+1))^2.

My conjecture is that the products (A+Ct)*(B+Dt) and (A+Et)*(B+Ft) are each triangular numbers (= T(r) = r(r+1))/2) for all integer t. Also, the above example gave a interesting result in that when the 12 results for r (= the integer part of the square root of 2T(r)) are sorted by value the difference between adjacent r values when t = 1 was one of the even factors of 666. It would seem to me that the easiest way to prove this conjecture would be to derive and prove a formula for r.

While only one prime, 3, can be represented as a triangular number, all primes other than 3 can be represented as n(n+1)/2 -i where i is a positive integer less than n in one and only one way, and as another posted noted triangular numbers and square numbers are related since 8T(n) + 1 = (2n+1)^2.

Progress report: my conjecture does indeed hold where n is even and A is a factor of n other than n. First of all I forgot to note that my formula for C,D,E,F was meant only for the case where n is even, if n is odd then make the substitution n' = -(n+1) and the formula also works. I still am working on the general conjecture.

Proof of my simplified conjecture follows:

The new values of A,B,C,D,E, and F can now be expressed as

1A. A,B,1,2*(n/2A)^2,2*A^2,(n+1)^2

The second conjectured result follows directly from the relation

[tex]T_{(n + 2A(n+1)t)} = (A +2t*(A^2))*(B+ t*(n+1)^2)[/tex]

The first conjectured result (using C and D) follows directly from the relation

[tex]T_{(n + nt/A)} = (A + t)*(B + 2t*(n/2A)^2)[/tex]


The above relations follow directly by expansion of both sides and comparison of like terms after making the substitution n(n+1)/2 = AB

Although the above is presumably dealing with integers only, I think the same relationship hold for n and A being within the field of rationals or within the complex number field as long as A,B,C,D,E,and F are defined as in line 1A. In this case the conditions that n is odd or that A is a factor of n are not necessary as long as T(n) is defined as n(n+1)/2 regardless of the nature of n.

I will concentrate on A = a factor of n+1 next
 
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  • #8
You conjecture is true for A = n/2 and B = n+1, when n = even number, for the expression (A+Dt)(B+Ct)

as gcd(n,n+1) = 1, then C = 1 and D = 2

[tex](A+Dt)(B+Ct) = (n/2 + t)(n+1 + 2t) = \frac{[n+2t]*[(n+2t)+1]}{2}[/tex]

But (A+Et)(B+Ft) is not a triangular number for all t

gcd(A,n) = n/2 ==> [tex]2[gcd(A,n)]^2 = \frac{n^2}{2}[/tex] ==>

==> [tex]A+Et = \frac{n + n^2t}{2}[/tex]

gcd(B,n+1) = n+1 ==> [tex]gcd(B,n+1)^2 = (n+1)^2[/tex] ==>

==> [tex]B+Ft = n + 1 + (n+1)^2t = n + n^2t + (2nt + t + 1)[/tex]

showing that (A+Et)(B+Ft) is triangular if and ony if t = 0
 
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  • #9
We have to test when A is not n/2, say A = kx and B = wy
 
  • #10
write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form
 
  • #11
al-mahed said:
You conjecture is true for A = n/2 and B = n+1, when n = even number, for the expression (A+Dt)(B+Ct)

as gcd(n,n+1) = 1, then C = 1 and D = 2

[tex](A+Dt)(B+Ct) = (n/2 + t)(n+1 + 2t) = \frac{[n+2t]*[(n+2t)+1]}{2}[/tex]

But (A+Et)(B+Ft) is not a triangular number for all t

gcd(A,n) = n/2 ==> [tex]2[gcd(A,n)]^2 = \frac{n^2}{2}[/tex] ==>

==> [tex]A+Et = \frac{n + n^2t}{2}[/tex]

gcd(B,n+1) = n+1 ==> [tex]gcd(B,n+1)^2 = (n+1)^2[/tex] ==>

==> [tex]B+Ft = n + 1 + (n+1)^2t = n + n^2t + (2nt + t + 1)[/tex]

showing that (A+Et)(B+Ft) is triangular if and ony if t = 0
You didn't look far enough!
[tex](A + Et)*(B+Ft) = \frac{n+n^2t}{2} *(n+n^2t +2nt + t + 1)[/tex]

[tex] = \frac{n+n^{2}t}{2}*((n+n^{2}t+ nt+1) + nt + t) [/tex]

[tex] = \frac{n +n^{2}t +nt}{2}*(n+n^{2}t + nt + 1)[/tex]

[tex] = T_{n + n^{2}t +nt}[/tex]

What's more important you can let A be any integer, rational or complex number and still the product
(A+Et)*(B+Ft) = m(m+1)/2 where T(n) = A*B is a triangular number and E and F are defined as in line 1A of my January 20 post and m = n + 2A(n+1)t
 
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  • #12
al-mahed said:
write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form

Thanks that is helpful, I will use these new terms in my next response.
 
  • #13
al-mahed said:
write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form
For (A+Ct)*(B + Dt); r = n+2xyt
by symmetry
For (A+Et)*(B+Ft); r = n+2wkt
 
  • #14
al-mahed said:
write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form

Now put k = (a^n)/2, y = 1/b^n, x = 1/b^n and w = c^n then my equations become

T(2ky + 2xyt) = (kx +tx^2)*(wy+2ty^2)

T(2ky +2kwt) = (ky +2tk^2)*(wy+tw^2)

Is it true that for n>2 this can have no solution except a = 0 since the requirement that 2ky + 1 = wx can not be met due to Fermat's theorm?
 
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  • #15
ramsey2879 said:
[tex] = \frac{n+n^{2}t}{2}*((n+n^{2}t+ nt+1) + nt + t) [/tex]

[tex] = \frac{n +n^{2}t +nt}{2}*(n+n^{2}t + nt + 1)[/tex]

This is not correct, you cannot put "nt" in the other side like this, and what about "t"?
 
  • #16
al-mahed said:
This is not correct, you cannot put "nt" in the other side like this, and what about "t"?

(n+n^2t)*(nt+t)=nt*(1+nt)*(n+1) = nt*(n + n^2t + nt + 1) so I was right to express it that way. Sorry for not showing all steps but I had trouble with my latex expressions.
 
  • #17
we have to write (A+Et)*(B+Ft) in the r(r+1)/2 form, right?

if [tex]A+Et = \frac{n + n^2t}{2}[/tex], then [tex]n + n^2t[/tex] can be called "r"

as[tex]B+Ft = n + (n+1)^2t + 1[/tex], and [tex]n + n^2t \neq n + (n+1)^2t[/tex], I can NOT write the product (A+Et)*(B+Ft) in the r(r+1)/2 form, since [tex] n + (n+1)^2t [/tex] is not "r",

try to find a concrete counter example proving that (A+Et)*(B+Ft) is triangular for t > 0, I may be mistaken, but I believe that there is not such example.
 
  • #18
al-mahed said:
we have to write (A+Et)*(B+Ft) in the r(r+1)/2 form, right?

if [tex]A+Et = \frac{n + n^2t}{2}[/tex], then [tex]n + n^2t[/tex] can be called "r"

as[tex]B+Ft = n + (n+1)^2t + 1[/tex], and [tex]n + n^2t \neq n + (n+1)^2t[/tex], I can NOT write the product (A+Et)*(B+Ft) in the r(r+1)/2 form, since [tex] n + (n+1)^2t [/tex] is not "r",

try to find a concrete counter example proving that (A+Et)*(B+Ft) is triangular for t > 0, I may be mistaken, but I believe that there is not such example.
I just finished showing that r = n + n^2t + nt . You are under the false impression that r must equal (A+Et) or (B+Ft) but all I said was that r(r+1)/2 = (A+Et)*(B+Ft)
In the example you chose r = n+n^2t +nt . Work it out and see for yourself.
 
  • #19
Sorry, my big mistake, pure laziness ! Of course you're right
 
  • #20
al-mahed said:
write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form
Let T(r) = r(r+1)/2
I have used this format as follows
Let n = 2ab and n+1 = cd then the following 4 equations hold for all t
T(n + 2act) = ac(b+ct)(d + 2at)
T(n + 2adt) = ad(b+dt)(c + 2at)
T(n + 2bct) = bc(a+ct)(d + 2bt)
T(n + 2bdt) = bd(a+dt)(c + 2bt)

Knowing that T(n) + T(n+1) = (n+1)^2, I tried letting a=c=1, b = x^n/y^n and c = z^n/y^n in the above formulas and got

(z^n/y^n)^2 = z^n(z^n + y^n + x^n)/2y^(2n) each time by choosing t = 1/2ac etc.
 

1. What is a triangular number?

A triangular number is a number that can be represented by a triangular arrangement of objects. It is the sum of the first n natural numbers, starting from 1. For example, the fourth triangular number is 10 because it can be represented by a triangle with 4 rows of objects, with 1, 2, 3, and 4 objects in each row.

2. What is the formula for calculating a triangular number?

The formula for calculating a triangular number is T(n) = (n * (n+1)) / 2. This means that to find the nth triangular number, you multiply n by the next number (n+1) and then divide the result by 2.

3. What is the significance of the Conjecture on triangular numbers T(n)?

The Conjecture on triangular numbers T(n) states that every positive integer can be expressed as the sum of at most three triangular numbers. This has been a topic of interest in number theory and has been proven to be true for all numbers up to 5 x 10^17.

4. How can the Conjecture on triangular numbers T(n) be applied in real life?

The Conjecture on triangular numbers T(n) has applications in combinatorics, geometry, and computer science. It can also be used in problem-solving and pattern recognition tasks.

5. Is there a proof for the Conjecture on triangular numbers T(n)?

No, there is currently no proof for the Conjecture on triangular numbers T(n). However, it has been verified to be true for all numbers up to a very large limit, and mathematicians continue to search for a proof of this conjecture.

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