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Primitive reciprocal Pythagorean triangles 1/a^2 = 1/b^2 + 1/c^2 have a=fg, b=ef, c=eg where e^2 = f^2 + g^2; i.e., e,f,g represent the sides of primitive Pythagorean triangles. But the product of the two legs of primitive Pythagorean triangles are multiples of 12 and so the reciprocal of hypotenuses of reciprocal Pythagorean triangles are always multiples of 12 (A008594).

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A < B); sequence gives values of B, sorted.

Any term in this sequence is given by f(m,n) = 2*m*n or g(m,n) = m^2 - n^2 where m and n are any two positive integers, m > 1, n < m, the greatest common divisor of m and n is 1, m and n are not both odd; e.g., f(m,n) = f(2,1) = 2*2*1 = 4. - Agola Kisira Odero, Apr 29 2016

All terms are composite. - Thomas Ordowski, Mar 12 2017

a(1) is the only power of 2. - Torlach Rush, Nov 08 2019

The first term appearing twice is 420 = a(75) = a(76) = A024410(1). - Giovanni Resta, Nov 11 2019

From Bernard Schott, May 05 2021: (Start)

Also, ordered sides a of primitive triples (a, b, c) for integer-sided triangles where side a is the harmonic mean of the 2 other sides b and c, i.e., 2/a = 1/b + 1/c with b < a < c.

Example: a(2) = 12, because the second triple is (12, 10, 15) with side a = 12, satisfying 2/12 = 1/10 + 1/15 and 15-12 < 10 < 15+12.

The first term appearing twice 420 corresponds to triples (420, 310, 651) and (420, 406, 435), the second one is 572 = a(101) = a(102) = A024410(2) and corresponds to triples (572, 407, 962) and (572, 455, 770). The terms that appear more than once in this sequence are in A024410.

For the corresponding primitive triples and miscellaneous properties and references, see A343891. (End)

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives values of A, sorted.

Union of A081874 and A081925. - Lekraj Beedassy, Jul 28 2006

Each term in this sequence is given by f(m,n) = m^2 - n^2 or g(m,n) = 2mn where m and n are relatively prime positive integers with m > n, m and n not both odd. For example, a(1) = f(2,1) = 2^2 - 1^2 = 3 and a(4) = g(4,1) = 2*4*1 = 8. - Agola Kisira Odero, Apr 29 2016

All powers of 2 greater than 4 (2^2) are terms, and are generated by the function g(m,n) = 2mn. - Torlach Rush, Nov 08 2019

k is in this sequence iff A078926(k) > 0.

Also, ordered sides c of primitive triples (a, b, c) for integer-sided triangles where side a is the harmonic mean of the 2 other sides b and c, i.e., 2/a = 1/b + 1/c with b < a < c (A343893). - Bernard Schott, May 06 2021

a(n) are the ordered radii of inscribed circles in squares, from which the tangents to the circles are cut off by primitive Pythagorean triangles. - Alexander M. Domashenko, Oct 17 2024

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

It is an open question whether any two distinct Pythagorean triples can have the same product of their sides.

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)

See A020885 for this sequence with multiplicities. See A024365 for the areas with multiplicities.

This sequence gives also Fibonacci's congruous numbers divided by 24 without multiple entries. See A258150.

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