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According A180926, the elements of the set {$a:\exists m,n|60a=5n^2+5n=3m^2+3m$} satisfy the following recurrence relation:

$$a_{n}=\frac{62a_{n-1}+1+\sqrt{(48a_{n-1}+1)(80a_{n-1}+1)}}{2}$$ $$a_1=0$$

How this can be derived?

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    What exactly do you mean by characterize? Is the recurrence relation given in your link to A180926 not sufficient?2010-09-27
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    @Moron: I feel silly for asking; that's really what I wanted. Maybe someone can explain how to derive that?2010-09-27
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    @Charles: You can change your question by giving the recurrence relation here and asking how it can be dervied. I will go ahead and edit your question.2010-09-27
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    Please excuse me, but how does one conclude from "$6a(n)$ and $10a(n)$ are both triangular numbers" that $a=5n^{2}+5n=3m^{2}+3m$ ?2010-09-27
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    @Americo: It is _60_a = 5n^2 + 5n blah.2010-09-27

2 Answers 2

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Since we have

$$ 5(n^2 +n) = 3(m^2 +m)$$

multiplying by $4$ gives us

$$5(4n^2 + 4n + 1) - 5 = 3(4m^2 + 4m + 1) -3$$

if $x = 2n+1$ and $y = 2m+1$ we get

$$5x^2 - 3y^2 = 2$$

This is a generalized Pell's equation of the form $ax^2 - by^2 = c$ and I believe this is a solved problem.

For instance, see this: http://books.google.com/books?id=YD6UlfcRlGQC&pg=PA81

This should give us all the solutions, and so make the recurrence easy to verify.

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    It seems you're producing odd solutions $x$ and $y$, but the numbers in the sequence A180926 are not all odd...?2010-09-27
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    @Hans: The numbers in the sequence are of the from $(n^2 + n)/12$, where $x = 2n+1$. So it could be anything.2010-09-27
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    Of course... Sorry for the stupid comment.2010-09-27
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    Thank you, both for your answer and your patience with my question.2010-09-27
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    @Charles: You are welcome. Thank you for the nice problem.2010-09-27
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    @Charles: For full details of the solution see my answer here.2010-09-28
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    You can also feed this kind of equation to Dario Alpern's Generic Two integer variable equation solver, found at http://www.alpertron.com.ar/QUAD.HTM2010-10-16
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    @Ross: That is pretty neat! Thanks.2010-10-16
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Here's a complete solution, including a derivation of the recurrence relation in A180926. First, notice

$\rm\ 60\:a \:=\: 5\:(n^2+n)\ =\ 3\:(m^2+m)\ \ $ is easily transformed into the following Pell equation

$\rm\quad\quad\quad\quad\ \: 5\:x^2-3\:y^2 \:= \ 2\ \ \ $ for $\rm\ \ \ (x,\:y)\: =\: (2\:n+1,\:2\:m+1)$

Hence $\rm\ \ \ \: x^2 - 1\ =\ 4\: (n^2+n)\ =\ 48\:a,\quad\ y^2 - 1\ =\ 4\: (m^2+m)\ =\ 80\:a\quad\quad\quad\ \ (*)$

By standard Pell theory $\rm\ \: (x,y) \:\to\: (X,Y)\:=\: (4\:x+3\:y\ \ ,\ \ 5\:x+4\:y)\ $ is a solution too, since

$\rm\phantom{\quad\Rightarrow\quad}\ X+Y\omega \: =\: (4+5\:\omega)\:(x+y\:\omega) \: = \: (4\:x+3\:y + (5\:x+4\:y)\:\omega) \: ,\ \ \ \omega \:=\: \sqrt{3/5}$

$\rm\quad\Rightarrow\quad X- Y\omega \: =\: (4-5\:\omega)\:(x-y\:\omega)\quad\quad\:$ via conjugate prior equation

$\rm\quad\Rightarrow\quad 5\: X^2 - 3\: Y^2 \: =\ 5\: x^2 - 3\: y^2\: =\: 2\ \ $ via multiply the prior two equations, then scale by 5

This yields a recurrence to generate a new solution $\rm (X,Y) = (x_{n+1},\:y_{n+1})$ from a known solution $\rm\: (x,\:y) = (x_n,\:y_n)\:.\:$ We square this recurrence to obtain a recurrence for $\rm (A,\:a) = (a_{n+1},\:a_n)$

Namely $\rm\ \: (4\:x+3\:y)^2 =\ X^2$

$\rm\quad\quad\quad\quad \Rightarrow\quad\ 24\ x\:y\ =\ X^2-1 - 16\:(x^2-1) - 9\:(y^2-1) - 24$

$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 48\:A - 16\:(48\:a) - 9\:(80\:a) - 24\quad$ by $\ (*)$

$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 24\ (2\:A - 62\: a - 1)$

$\rm\quad\quad\quad\quad \Rightarrow\quad\ x^2\ y^2\ =\ (2\:A - 62\: a - 1)^2 $

But also $\rm\quad\quad\quad\: x^2\ y^2\ =\ (48\: a + 1)\ (80\: a + 1)\quad$ by $\ (*)$

Hence $\rm\ \ (2\ a_{n+1} - 62\ a_n - 1)^2 \ = \ (48\ a_n+1)\ (80\ a_n+1)\ \ $ as was to be proved.