# Triangular numbers facturable into n*(n+k)

1. May 14, 2010

### ramsey2879

Conjecture All triangular numbers T(i) in the recursive series i(0) = 0, i(1) = k*2+1 and following the recursive relationship i(j) = 6*i(j-1) - i(j-2) + 2 can be factured into the product n*(n+k) where n and k are integers. where k = 0 these are the square triangular numbers. I know that the series relationship for square triangular numbers is well known, but has anyone before proven or made the above conjecture?

2. May 15, 2010

### ramsey2879

Besides the recursive relation for the argments i of the triangular numbers the two factors differing by k each follow a closely related sequence the value n follows the sequence n(0) = 0, n(1) = k+1 and the following values are determined by the recursive relation n(i) = 6*n(i-1) - n(i-2) + 2k. For n = 2 the series is 0, 3, 22, 133, ... . Therefore the following products are triangular numbers 0*2, 3*5, 22*24, 133*135, ... k can take negative values also. E.G. for k = -2, the series is 0, -1, -10, -63, ... and the products 0*2, 1*3, 10*12, 63*65,... are also triangular numbers.
I am interested If k can take complex values also. For instance, if k can equal "i" then (1+i)*(1+2i) = (1+2i)*(2+2i)/2 which is in the form C * (C +1)/2 where C is a complex integer).

Last edited: May 15, 2010
3. May 19, 2010

### Martin Rattigan

EDIT: This is NOT a counterexample if conjecture is interpreted as OP intended. It is based on a misinterpretation. See sequel.

If $i_j$ satisfies $i_0=0,i_1=2k+1$ and $i_j=6i_{j-1}-i_{j-2}+2$, then for $k=28$, $i_5=68265=T_{369}=369.370/2$.

But $247(247+28)=67925<68265$ and $248(248+28)=68448>68265$, so $i_5$ is not expressible as $n(n+k)$.

Last edited: May 20, 2010
4. May 19, 2010

### ramsey2879

I believe I said t(i) can be expressed as n*(n + k)

t(i,k) = t(5,28) = 68265*(68265+1)/2 = 2330089245
n = 48257
n*(n+k) = 48257*(48257 + 28) = 2330089245

I am sorry that I did not make myself clear that t(i) meant to use i as the argument, not as the triangular number.

Last edited: May 19, 2010
5. May 19, 2010

### Martin Rattigan

Sorry I hadn't realised that the $i$ in $T(i)$ was one of the $i(n)$. I read it just as, "when an $i(j)$ is also a $T(i)$ for some $i$ then there is an $n$ such that $T(i)=i(j)=n(n+k)$ where $i(1)=2k+1$".

I thought there was probably something wrong because the cases where $i(j)=T(i)$ for some $i$ are quite rare.

So, back to the drawing board.

6. May 19, 2010

### ramsey2879

I found that this works for complex numbers also If you let the T(x) function take complex arguments. As an example, let k = 1+i and I = 2

then

I(0) = 0, I(1) = 3 + 2i, I(2) = (20 + 12i)

T(I) = (21 + 12i) * (10 + 6i) = 138 + 246i

n(0) = 0, n(1)= 2 + i , n(2) = 14 + 8i

n*(n + k) = (14 + 8i)*(15 + 9i) = 138 + 246i

7. May 19, 2010

### ramsey2879

Yeah, I need to read up on triangular numbers that are square to get a better start on proving this, but although I enjoy the diversion of looking for number patterns, especially with triangular numbers, I have just a little knowledge of number theory and thats it. If anyone can prove this for any k other than 0 or 1 which are known series, let alone proving it for all k, i would appreciate it. Also, I have arrived at a more general conjecture:

Let T(x) be defined as x*(x+1)/2. Let C be any complex integer and let the product
$$A_{0} * B_{0} = T(C)$$. Let $$B_{0} - A_{0} = K$$. Then there exists an infinite series of complex numbers $$D_{i}$$ such that $$T(C_i) = D_{i}*(D{i} + K)$$ where $$C_{i}) = 6C_{i-1) - C_{i-2} + 2$$ and $$D_{i} = 6D_{i-1} - D_{i-2} + 2K$$.

8. May 19, 2010

### ramsey2879

As an example with integers starting with K = 5, I = 3, A = 1, B = 6, we can set I(0) = -4,
I(1) = 3, D(0) = D(1) = 1. Then I(2), I(3), ... = 24, 143, ...; and D(2), D(3) ... = 15, 99, ...;

9. Jun 3, 2010

### ramsey2879

I think I found a proof of a more comprehensive theorm than my conjecture:

Theorm: For any given triangular number $$T(n_{0}) =n*(n+1)/2$$ and any given number $$N_{0}$$, the difference $$T(n_{i}) - N_{i}*(N_{i} + K)$$ is always constant for any constant K where the series $$N_{i}$$ is found by letting $$N_{1} = 3*N_{0} + K +1 +2*n_{0}$$ with the recursion $$N_{i} = 6*N_{i-1} - N_{i-2} + 2K$$ and letting $$n_{1} = 4*N_{0} +2K + 1 +3*n_{0}$$ for which the recursive relation is $$n_{i} = 6*n_{i-1} - n_{i-2} + 2$$.

I have the proof and will supply it latter.

Last edited: Jun 3, 2010
10. Jun 3, 2010

### ramsey2879

My proof is as follows:

I gave in my last post only the terms for $$N_{1}$$ and $$n_{1}$$ as function of $$N_{0}$$ and $$n_{0}$$ as well as the recursive formula for $$N_{i}$$ and $$n_{i}$$. Apart from the recursive formula, all of the above can be directly verified by doing the math and also one can simarly verify using alternative to N_1 and n_1 terms which follow N_0 and n_0 respectively, the terms preceding the given terms : $$N_{-1} = 3*N_{0} + K -1 -2*n_{0}$$ and $$n_{-1} = 1 +3*n_{0}-4*N_{0}-2K$$.

Once you have verified that for each of the three pairs of terms (N{-1}, n{-1}), (N{0}, n{0}) and (N{1},n{1}) that N{i}*(N{i}+K) - n{i}*(n{i} +1)/2 is the same and verified that each set of 3 terms follows the assigned recursive relation; the next step is to verify that the same formulas based upon N{0} and n{0} for the terms N{-1} and n{-1} will give N{0} and n{0} when based upon the pair (N{1},n{1}). This is shown as follows.

$$N_{0} = 3*N_{1} + K -1 - 2*n_{1}$$
$$= 3*(3*N_{0} + K + 1 + 2*n_{0}) + K - 1 -2*(4*N_{0} + 2K + 1 + 3*n_{0})$$
$$= (9-8)N_{0} + (3+1-4)K + (3-1-2) + (6 - 6)n_{0} = N_{0}$$

$$n_{0} = 1 + 3*n_{1}-4*N_{1} -2K$$
$$=-4(3N_{0} + K + 1 + 2n_{0}) -2K +1 +3(4N_{0} + 2K + 1 + 3n_{0})$$
$$= (12-12)N_{0} +(4+2-6)K +(-4 +1-2) + (-8+9)n_{0} = n_{0}$$

Does this prove my conjecture?

Last edited: Jun 4, 2010