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The two sequences involve areas of primitive Pythagorean triples and primorial products. Intersections are only considered once (no repeats). Conjecture: the sequence is infinite.

Conjecture: The next two entries are a(12) = 200560490130, a(13) = 401120980260.

From G. C. Greubel, Dec 29 2015: (Start)

6|a(n) for n>=1,

30|a(n) for n>=2,

a(n)/6 = {1, 5, 10, 30, 35, 385, 770, 10010, ...} is a subset of values found in A008706.

(End)

a(12) and a(13) confirmed. a(14) > 2*10^31, if it exists. - Giovanni Resta, Mar 31 2017

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. A given area often corresponds to more than one triangle; for example, a(9) = 60 for the triangles (a,b,c) = (6,25,29), (8,17,15), (13,13,10) and (13,13,24).

If only primitive integer triangles (that is, the lengths of the sides are coprime) are considered, then the possible areas are 6 times the terms in A083875. - T. D. Noe, Mar 23 2011

Number of terms < 10^k for increasing values of k: 1, 7, 34, 150, 636, 2536, 9757, 35987, 125350, 407538, ..., .

All terms are divisible by 6.

Also positive integers m with four (or more) different divisors (p, q, r, s) such that m = p*q = r*s and s = p+q+r. - Jose Aranda, Jun 28 2023

This sequence also gives Fibonacci's congruous numbers (or congrua) divided by 4 with multiplicities, not regarding leg exchange in the underlying primitive Pythagorean triangle. See A258150 and the example. - Wolfdieter Lang, Jun 14 2015

The squarefree part of an entry which is not squarefree is a primitive congruent number from A006991 belonging to a Pythagorean triangle with rational (not all integer) side lengths (and its companion obtained by exchanging the legs). See the W. Lang link. - Wolfdieter Lang, Oct 25 2016

All terms are divisible by 6.

Let k be even, k > 2, q = (k/2)^2 - 1, and b = (kq)/2. Then, for any k, b is a term of a(n). In other words, for any even k > 2, there is at least one such integer q > 2 that b = (kq)/2 and b is a term of a(n), while hypotenuse c = q + 2 (proved by Anton Mosunov). - Sergey Pavlov, Mar 02 2017

Let x be odd, x > 1, k == 0 (mod x), k > 0, y = (x-1)/2, q = ky + (ky/x), b = (kq)/2. Then b is a term of a(n), while hypotenuse c = q + k/x. As a special case of the above equation (k = x), for each odd k > 1 there exist such q and b that q = (k^2 - 1)/2, b = (kq)/2, and b is a term of a(n), while hypotenuse c = q + 1. - Sergey Pavlov, Mar 06 2017

Equivalently, integers of the form m*n*(m^2 - n^2) where m,n are positive integers with m > n. - James R. Buddenhagen, Aug 10 2008

The sequence giving the areas of all Pythagorean triangles is A009112 (sometimes called "Pythagorean numbers").

For example, the sequence does not contain 54, the area of the Pythagorean triangle with sides (9,12,15). - Robert Israel, Apr 03 2015

See also Theorem 2 of Mohanty and Mohanty. - T. D. Noe, Sep 24 2013

Among a(1) to a(30), only a(23) = 13123110 has multiplicity 3, the others have multiplicity 2. The three primitive Pythagorean triangles corresponding to a(23) are [4485, 5852, 7373], [3059, 8580, 9109] and [19019, 1380, 19069]. Leg exchange is not taken into account. - Wolfdieter Lang, Jun 15 2015

The area 13123110 of multiplicity three was discovered by C. L. Shedd in 1945, cf. Beiler, Gardner and Weisstein. - M. F. Hasler, Jan 20 2019

The problem is: given a square, find a positive integer that, whether added to or subtracted from that square, yields a square. That is, both x^2 + C = y^2 and x^2 - C = z^2. Equivalently: z^2 + C = x^2 and x^2 + C = y^2 (squares in arithmetic progression). This is treated in Fibonacci's 'The book of squares' (Liber Quadratorum (1225) but for rational x,y,z). See the Sigler reference, Proposition 14, pp. 53-74 (note that the formulation of this problem on p. 53 is not correct, 'from a square' should read 'from the same square'). See also van der Waerden, pp. 40-42, and A. Weil, pp. 13-14. The desired number C is called a congruum by Fibonacci (a congruous number in Sigler's translation).

For the history of this problem, see Dickson, pp. 459-472 (he uses the (misleading) term congruent number).

The following solution is based on primitive Pythagorean triangles. (Fibonacci's solution is based on sums of odd squares.) The triangle T(n, m) = 24*C(n, m) will have 0's for those (n, m) not leading to primitive Pythagorean triples.

Addition of the two equations, substitution of y = u + v > 0 and z = |u - v| and division by 2 leads to x^2 = u^2 + v^2. Consider primitive Pythagorean triples (u, v, x) with even v which are pairwise relatively prime. Then also GCD(u,v,x) = 1. A common factor f for u, v and x would lead to a multiplication by f^2 on both sides of the two equations. For primitive Pythagorean triples see A249866. One has u = n^2 - m^2, v = 2*n*m and x = n^2 + m^2 with GCD(n, m) = 1 , n > m >= 1, n + m odd. Then C = C(n, m) = 4*n*m*(n^2 - m^2) = 2*v(n, m)*u(n, m). This is four times the area of the Pythagorean triangle. C is divisible by 4! = 24 (see A020885). Define T(n, m) = C(n, m)/4!, for 2 <= m + 1 <= n. This is the area of the corresponding primitive Pythagorean triangle divided by 6.

The corresponding x = x(n, m), y = y(n, m) and z(n, m) number triangles are given in A222946, A225949 and A258149 respectively.

T(n, m) = n*m*(n^2 - m^2)/6, for m = 1, 2, ..., n-1, has for n >= 2 the minimum value at m = 1, and the next largest value appears for n >= 3 at m = n-1. Note all (n, m) pairs are considered here. The proof of the first part is easy. The proof of T(n, m) - T(n, n-1) > 0, for m = 2, 3, ..., n-2, and n >= 3, is equivalent to n^2*(m-2) + 3*n > m^3 +1 and this is easy to prove with n >= m+2 and m >= 2. Therefore the triangle T(n, m) with 0's attains for even n the smallest nonzero row entry at m = 1, and for odd n the smallest nonzero row entry appears at m = n-1 (last entry).

This allows us to find (after solution of two cubic equations for even and odd n, named ne = ne(N) and no = no(N)) a row number nmin(N) = max(ne(n), no(N)) such that N will not appear in any row n > nmin(N).

The original problem posed to Fibonacci by Giovanni di Palermo (Master John of Palermo) was to find a [rational] square that when increased or decreased by 5 gives a square. Fibonacci gave the solution in his Liber Quadratorum in Proposition 17 (see Sigler, pp. 77-81) as x^2 = (41/12)^2 = 1681/144, y^2 = (49/12)^2 = 2401/144 and z^2 = (31/12)^2 = 961/144. This corresponds to the integer quartet (C; x, y, z) = (720; 41, 49, 31) corresponding to the primitive Pythagorean triple [9, 40, 41]. See the examples for (n, m) = (5, 4).

The numbers without zeros, in nondecreasing order, are given in A020885 = A024406/6.

Comments from Eric Snyder, Feb 07 2023: (Start)

If m+n > 3 and not divisible by 3, then m+n | T(n,m).

Additionally, if 2n-1 > 3 and not divisible by 3, then 2n-1 = 6k+-1, and T(n,n-1) = (2n-1)*P(-+k), where P(-+k) is a generalized pentagonal number (A001318). For example, T(6,5) = 11*P(-2) = 11*5.

T(n,n-1) = A000330(n-1) for n>=2. (End)

Terms with even u or v form A024365. Squarefree terms form A147779.

Mohanty and Mohanty prove in Corollary 2.5 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if F(n) and F(n+1) have opposite parity. Every third number, starting at a(1) = 6, is not primitive Pythagorean.

Since a(n) = F(n+1)*F(n+2)*(F(n+2)^2 - F(n+1)^2), a(n) is in A073120. - Robert Israel, Apr 06 2015