Triangular Number Conjecture: Recursive Series

[ramsey2879]

I have a new conjecture re triangular numbers that I think is fascinating.

Conjecture
For any two integers $$a$$ and $$b$$ such that $$ab$$ is a triangular number, then there is an integer $$c$$ such that $$a^2 + ac$$ and $$b^2 + bc$$ are both triangular numbers. Further, $$(6b-a+2c)*b$$ and $$(6b-a+2c)*(6b-a+3c)$$ are also triangular numbers so this property is recursive.

an interesting set of such recursive series is

0,1,6,35,204 ...(c = 0)
0,2,14,84,492...(c = 1)
0,3,22,133,780..(c = 2)
...
where the differences between any two sucessive terms of the $$i$$th columm form the recursive series $$0,1,8,49,288..(6*n_{(i-1)}-n_{(i-2)}+2)$$.

 

[matt grime]

Have you tried to prove this? My gut feeling is that it shouldn't be too hard.

 

[ramsey2879]

Have you tried to prove this? My gut feeling is that it shouldn't be too hard.

Well actually I proved various special cases in my earlier post and I believe that there are actually two values of $$c$$.
for any pair $$a$$ and $$b$$. They are



$$a+b \pm \sqrt{8ab + 1}$$

I still haven't completed the proof, but I am working on it. Got to sign off now.

 

[matt grime]

Well actually I proved various special cases in my earlier post

no you did not. you provided examples, you proved nothing.

 

[ramsey2879]

no you did not. you provided examples, you proved nothing.

I was referring to my earlier thread, sorry for not being specific.

Well actually I proved various special cases in my earlier post [thread] and I believe that there are actually two values of $$c$$.
for any pair $$a$$ and $$b$$. They are



$$c = a+b \pm \sqrt{8ab + 1}$$

Let $$2ab = n(n+1)$$
then $$c = a+b \pm (2n+1)$$

Case 1: c = a+b +2n + 1
$$(a^2 + ac) = 2a^2 + .5n^2 +.5n + 2na + a$$
$$\quad = .5(2a+n)(2a+n+1)$$

by simlarity of a and b $$(b^2 + bc) = .5(2b+n)(2b+n+1)$$

$$b*(6b-a+2c) = 6b^2 - ab + 2b(a + b + 2n + 1)$$
$$\quad = 8b^2 + .5n^2 + .5n + 4bn + 2b$$
$$\quad = .5(4b+n)(4b+n+1)$$

by similarity of a and b $$a*(6a-b+2c) = .5(4a+n)(4a+n+1)$$

$$(6b - a + 2c)*(6b - a + 3c) =$$
$$(8b + a +4n +2)*(9b+2a +6n +3)$$
$$72b^2 + 2a^2 + 24n^2 +6 +25ab +14an +84bn +42b +7a+24n$$
$$72b^2 + 2a^2 +24.5n^2 +6 +24ab +14an +84bn +42b +7a +24.5n$$
$$.5(12b+2a+7n+3)(12b+2a+7n+4)$$

by similarity of a and b
$$(6a-b +2c)*(6a-b+3c) = .5(12a +2b +7n +3)(12a +2b +7n +4)$$

Thus each of the above products are triangular numbers

Case 2 $$c = a + b - 2n - 1$$

$$(a^2 + ac) = 2a^2 + .5n^2 +.5n - 2na - a$$
$$\quad = .5(2a-n)(2a-n-1)$$

by simlarity $$(b^2 + bc) = .5(2b-n)(2b-n-1)$$

$$b*(6b-a+2c) = 6b^2 - ab + 2b(b + a - 2n - 1)$$
$$\quad = 8b^2 + .5n^2 + .5n - 4bn - 2b$$
$$\quad = .5(4b-n)(4b-n-1)$$

by similarity $$a*(6a-b+2c) = .5(4a-n)(4a-n-1)$$

$$(6b - a + 2c)*(6b - a + 3c) =$$
$$(8b + a -4n -2)*(9b+2a -6n -3)$$
$$72b^2 + 2a^2 + 24n^2 +6 +25ab -14an -84bn -42b -7a +24n$$
$$72b^2 + 2a^2 +24.5n^2 +24ab -14an -84bn -42b -7a +24.5n$$

$$.5(12b+2a-7n-3)(12b+2a-7n-4)$$

by similarity of a and b
$$(6a - b + 2c)*(6a-b+3c) = .5(12a+2b-7n-3)(12a+2b-7n-4)$$


Thus each of the above products are also triangular numbers

 

[ramsey2879]

I have a new conjecture re triangular numbers that I think is fascinating.

Conjecture
For any two integers $$a$$ and $$b$$ such that $$ab$$ is a triangular number, then there is an integer $$c$$ such that $$a^2 + ac$$ and $$b^2 + bc$$ are both triangular numbers. Further, $$(6b-a+2c)*b$$ and $$(6b-a+2c)*(6b-a+3c)$$ are also triangular numbers so this property is recursive.

an interesting set of such recursive series is

0,1,6,35,204 ...(c = 0)
0,2,14,84,492...(c = 1)
0,3,22,133,780..(c = 2)
...
where the differences between any two sucessive terms of the $$i$$th columm form the recursive series $$0,1,8,49,288..(6*n_{(i-1)}-n_{(i-2)}+2)$$.

Looking back at the above set of recursive series, there are some questions raised. I proved above that for each pair of numbers in the same columns, there are two possible values of c that lead to like recursive series, by letting $$c = a + b \pm 2n+1$$. While the the above choice use $$c = a+b - 2n -1$$. I still have to prove that the arithmetic series of numbers in each column will continue indefinitely in either direction and that when multiplied together by their neighbor on the same row will always equal a triangular number. I further believed that it should be simple to prove that if you take any two columns where $$c$$ was determined by one choice, i.e., adding the negative of $$2n+1$$ to the sum, $$a+b$$ as I did above and got one arithmetic series for neighboring columns, that you will get a different arithmetic series for the neighboring columns, and a corresponding new series of arithmetic differences if you take the other choice. However, the following set of recursive series and new series of differences is formed by using the first two columns with a, n=0, i.e. c = b + 1, rather than b-1:

0,1,10,63, ...(c=2)
0,2,18,112, ...(c=3)
0,3,26,161, ...(c=4)

Apparently, the series of differences is the same as before :0,1,8,49...

Now I tried using the second and third columns of the first set of recursive series.

1,6,63,...(c=14)
2,14,144,...(c=31)
3,22,225,...(c=48)
4,30,306,...(c=65)

A different series of differences is formed, i,e, 1,8,81,... instead of 1,8,49...!

I am now examining this further to determing the recursive relation of this series and will work on a proof of the present results.

 

[ramsey2879]

[tne following sets of series have the property that the product of the adjacent terms on any row is a triangular number and the columns are in arithmetic progression. How to find such sets was pointed out in previous posts]
0,1,10,63, ...(c=2)
0,2,18,112, ...(c=3)
0,3,26,161, ...(c=4)

Apparently, the series of differences is the same as before :0,1,8,49...

Now I tried using the second and third columns of the first set of recursive series.

1,6,63,...(c=14)
2,14,144,...(c=31)
3,22,225,...(c=48)
4,30,306,...(c=65)

A different series of differences is formed, i,e, 1,8,81,... instead of 1,8,49...!

I am now examining this further to determing the recursive relation of this series and will work on a proof of the present results.

I further determined that where $$ab$$ is a triangular number, there are exactly two pairs of numbers $$(c,d)$$ which are coprime, no others for a given pair $$[a,b]$$, such that $$(a + nc)(b + nd)$$ is always a triangular number for any integer $$n$$. Thus for the $$[a,b]$$ pair $$[1,6]$$ the two $$[c,d]$$ pairs are $$(1,8), (2,9)$$. I have not been able to give a proof of the general case but I have proven it for specific cases. Some cases are related to a hybird type fibonacci series where the odd terms equal the sum of the previous two terms as in the fibonacci series but the even terms equal a constant times the previous term plus the next previous term. Is there a simple explanation for this property?

 

[ramsey2879]

I further determined that where $$ab$$ is a triangular number, there are exactly two pairs of numbers $$(c,d)$$ which are coprime, no others for a given pair $$[a,b]$$, such that $$(a + nc)(b + nd)$$ is always a triangular number for any integer $$n$$. Thus for the $$[a,b]$$ pair $$[1,6]$$ the two $$[c,d]$$ pairs are $$(1,8), (2,9)$$. I have not been able to give a proof of the general case but I have proven it for specific cases. Some cases are related to a hybird type fibonacci series where the odd terms equal the sum of the previous two terms as in the fibonacci series but the even terms equal a constant times the previous term plus the next previous term. Is there a simple explanation for this property?

Interestingly enough, the determinant formed by the two pairs $$c_{1},d_{1}$$ and $$c_{2},d_{2}$$ is plus or minus the determinant of the two pairs of $$c's$$ and $$d's$$ for each pair of integer factors $$a'$$ and $$b'$$ that multiplied together equal $$ab$$ and can be made to be the same by switching $$a'$$,$$b'$$ if necessary.

 

[ramsey2879]

Interestingly enough, the determinant formed by the two pairs $$c_{1},d_{1}$$ and $$c_{2},d_{2}$$ is plus or minus the determinant of the two pairs of $$c's$$ and $$d's$$ for each pair of integer factors $$a'$$ and $$b'$$ that multiplied together equal $$ab$$ and can be made to be the same by switching $$a'$$,$$b'$$ if necessary.

For instance, the triangular number T(37) has 12 factors which yields the $$(a,b)$$ pairs 1,666; 2,333; 3,222; 6,111; 9,74 and 18,37. The respective sets of $$c,d$$ pairs in determinant format are

$$\left| \begin{smallmatrix}<br /> 1 & 648\\ 2 & 1369<br /> \end{smallmatrix}\right|$$

$$\left| \begin{smallmatrix}<br /> 1 & 162\\ 8 & 1369<br /> \end{smallmatrix}\right|$$

$$\left| \begin{smallmatrix}<br /> 1 & 72\\ 18 & 1369<br /> \end{smallmatrix}\right|$$

$$\left| \begin{smallmatrix}<br /> 1 & 18\\ 72 & 1369<br /> \end{smallmatrix}\right|$$

$$\left| \begin{smallmatrix}<br /> 1 & 8\\ 162 & 1369<br /> \end{smallmatrix}\right|$$

$$\left| \begin{smallmatrix}<br /> 1 & 2\\ 648 & 1369<br /> \end{smallmatrix}\right|$$

Each determinant equals $$37^2 - 36^2$$ and each of (1 + n)*(666+648n), (1 + 2n)*(666+1369n), (2+n)*(333+162n), ... (18+n)*(37+2n), (18+648n)*(37+1369n) are each triangular numbers for all integer n. In short if a given triangular number [a*b] has M factors, there are M different sets of products (a' + cn)(b' + dn) where a'b' = ab, c is prime to d and the products remain as triangular numbers for all integer n.