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Nice question. Like other rational triangle questions it seems tied up with elliptic curves.

The sequence appears to start 5, 7, 10, 12, 15, 16, ... , which is not in OEIS.

Here are examples, with [a,b,c] denoting a triangle with sides a,b,c and both area and one side equal to a

5: [5, 13/6, 37/6], [5, 388489/115440, 351289/115440]
7: [7, 229/30, 61/30], [7, 363786649/97224240, 420740329/97224240]
10: [10, 241/60, 409/60], [10, 5858756089/1076373480, 16354950889/1076373480]
12: [12, 29/10, 101/10], [12, 4569289/2284620, 27772729/2284620]
15: [15, 1229/70, 221/70], [15, 350018295169/26465580960, 73496729569/26465580960]
16: [16, 34/15, 226/15], [16, 5547169/1498380, 19537249/1498380]

Because the triangles correspond to rational points on elliptic curves of positive rank, there are apparently infinitely many triangles for a given n, if there area any. But for some n there are no triangles. These correspond to the elliptic curves of rank 0.

There is at least a possibility I missed something, so others should double check.

@scythian: ah, yes there is [3, 5/2, 5/2]. I missed it because it is a torsion point on the curve. I guess I should check other rank 0 curves to see if a torsion point could give a triangle, but I doubt it will since the torsion group is of order 8 for the n=3 case and order 4 for the others. Anyway, I think the sequence will be an infinite sequence, but certainly have no proof of that. What I am saying is that except in the n=3 case when there is a solution for a particular value of n then there are infinitely many different triangles for that n.
For example, there are infinitely many different triangles with one side 16 and area 16 and the other two sides rational numbers.

Edit: Up to 34 the sequence goes as follow:

3, 5, 7, 10, 12, 15, 16, 18, 19, 23, 25, 26, 27, 28, 29, 30, 33, 34, ....

and I have triangle examples for each of those. For example, for n=34 I have this example: [a,b,c] =
[34, 8405427241/3424522920, 111771268729/3424522920]
That triangle has a side and its area both equal to 34.

I won't post the other examples, as this is getting long (unless someone really wants to see them).
Also, the missing values are associated with elliptic curves of rank 0 and don't have any triangles (unless I have overlooked something).

@Dragan K: interesting you have some parameterized subsequences -- I will look at this some more.

Edit2: Theorem (Rita the dog, 28 Jan 2012)

If S and T are any two rational numbers greater than 1, then

[S + 1/S, T + 1/T, S - 1/S + T - 1/T]

are the sides of a rational triangle and its area equals its third side, namely

S - 1/S + T - 1/T.

Conversely, if [a,b,c] is a rational triangle whose area is c, then there exist rational S,T larger than 1 such that [a,b,c] = [S + 1/S, T + 1/T, S - 1/S + T - 1/T].

Examples:

S=k^3+k ,T=k+1/k gives Dragan K's parameterized family [a,b,c] (see his additional details above).

S=81480/42029, T=334165/10248 gives the triangle
[8405427241/3424522920, 111771268729/3424522920, 34] which I mentioned above.

This still does not answer the original question of Dragan K, but it changes it into this;
For which rational S, T larger than 1 is S - 1/S + T - 1/T an integer, and which integers arise in this way.

Note: I have yet to try to write up a careful proof of this theorem, but I decided to share it now.

Edit3: A Pell subsequence.

Define recusively a(1) = 1, a(2) = 3, and for n>2, a(n) = 7*a(n-1) - a(n-2) -4. The sequence bigs:

1, 3, 16, 105, 715, 4896, 33553, 229971, 1576240, 10803705, ...

For n>1 it is a sub-sequence of Dragan K's desired integer sequence different from his parameterized sub-sequnece. For example, n=9 gives a(n) = 1576240 with associated triangle

[1576240, 3524578/1576239, 2484529385122/1576239]

which has 1576240 for both a side and its area.

Weitere Antworten(2)
Nice work so far by Rita & Scythian
Just to add in Dragan K's example the sequence does appear on OEIS
http://oeis.org/A054602

and ofcourse you can submit the sequence suggested by Rita, one of mine is already up there, which is still one of the open question for me
http://oeis.org/A153874

Weitere Antworten(3)
Well, immediately there's {3, 5, 16}. Any others are going to be hard to find, so I doubt if it'll be in OEIS if there isn't an infinite number of them. I'll keep looking.

Edit: Rita has found others as I have, 5, 7, 10, 12, 15, 16, but seems to be missing the most obvious one, 3. What I'm not sure about, is this sequence infinite? Just because the variables can be put into an elliptic equation doesn't seem to mean that there's an infinity of solutions, because of the requirement that both the area and one side be integers and the same.

Edit 2: Dragan K, but an infinity of DIFFERENT integers? Oftentimes I come up with something that while the generated sequence is infinite, that does't mean that there's that many distinct solutions. But I will have a look into this.

Edit 3: If that's so, that's interesting. Since Rita has a head start on this, I'll follow the conversation here. Unless I get some brilliant insight into this, but I doubt it.

Edit 4: I found some more values, matching a few of what Rita has, and including 72 and 135, seemingly with no end in sight. However, I have no idea of how to prove that there's an infinity of them, but certainly if some of the solutions are in parametric form, there is. I am as curious about this as Rita is too.

Edit 5: This problem is equivalent to finding all integer sums of :

(a² - b²) / ab + (c² - d²) / cd

where a, b, c, d are integers, both positive and negative. I agree now that the sequence is infinitely long, you've proved it with your parametrization. OEIS doens't have this series, but I think it should. Maybe you can pitch it to OEIS? I think it's a mathematically interesting and distinctive series.

Edit 6: Oh, that's interesting, Rita is saying the same thing. Finding all integer sums of:

(S - 1/S) + (T - 1/T)

where S and T are rational numbers. But proof isn't hard. Consider all triangles with an altitude of 1 with rational hypotenuses. Then ask what's the base? The base has to be the sum of 2 more rational numbers, it cannot be of the form p + r, q - r, where p, q are rational but r is not, no such solution exists for which this is true. Note that I said "altitude of 1", when 2 might be expected, but don't worry, work it out. Also, note that (a² - b²) / 2ab = 2mn / (m² - n²) using a simple linear substitution, so we don't have to consider two separate cases, it's all the same series.

Edit 7: Does anyone have an impossibility proof for c = 4? I think I have one, but it's very long and very shaky. I just wonder that the conditions for such impossibilities might lead to the dual series,

c = 1, 2, 4, 6, 8, 9, 11, 13, 14, 17, 20, 21, 22, etc.

which might be easier. But I'm out on a limb here.

Edit 8: This is the same as finding all the integer sums of the following, for integers a, b, c:

(a²b² - c²) / abc + (a² - b²c²) / abc

That is to say, S = ab / c, T = a / cb

I don't have the room to give the proof, nor I'm 100% sure of it. But it certainly makes for a faster search or confirmation of the numbers in the series. This is useful because then we know that both of the following have to be integers, their product being the sum of the above:

(1 + b²) / ac
(a² - c²) / b

so that if the sum is p, a prime number, then one of them = p, and the other = 1. We can use this to rule out some particular sums, such as 4. Anyway, I'm out of space now, Y!A is going to start truncating me.

Edit 9: Correction. Above analysis is a bit off, but it does lead to a superfast way of finding more terms. If for any given x, a²x² + 4a^4 + 8a² + 4 is a square for some a, then x is part of the sequence. It bugs me that it fails SOME of the time, i.e., can't find a to make it a square.

Edit 10: Eh, maybe this just needs an additional adjustment, hold on to see if I can improve on this.

Edit 11: Okay, final edit, out of room now. If for some positive integers a, b, c, d, all relatively prime with each other,

(a² - b²) / cd = p, an integer
(c² + d²) / ab = q, an integer

then pq is in the sequence, and conversely as well. This is one way to find out if a prime number is part of the sequence, for example, as in such cases, either p or q has to be 1. In general, S would be ac / bd while T would be ad / bc. As an example, S = 65*7 / 2*9 and T = 65*9 / 2*7 yields 67, which is a prime number, as 7² + 9² = 130 = 65*2, so that q here is 1, and p is 67.

Edit 12: I'm going nuts with this. Let N = xyz be a candidate for the sequence, where x, y, z are any integer factors, including 1. Then if there exists a positive integer m such that the following is rational:

√( (xy² - 4m²) / (xz²m² + 4) )

then N is part of the sequence. Note that given x, y, z, the search based on m is finite, so it can make a determination whether or not any number N is part of the sequence.

Edit 13: This needs just a tiny bit more work.