A334638 Three-column array pPT read by rows: subsequence of primitive Pythagorean triples (x, y, z) with x = A153893^2 - A000079^2, y = 2*A153893*A000079, z = A153893^2 + A000079^2, ordered by increasing z.

3, 4, 5, 21, 20, 29, 105, 88, 137, 465, 368, 593, 1953, 1504, 2465, 8001, 6080, 10049, 32385, 24448, 40577, 130305, 98048, 163073, 522753, 392704, 653825, 2094081, 1571840, 2618369, 8382465, 6289408, 10479617, 33542145, 25161728, 41930753, 134193153, 100655104, 167747585, 536821761, 402636800, 671039489, 2147385345, 1610579968, 2684256257

Offset: 0

Author

Ralf Steiner, May 07 2020

Keywords

• nonn
• easy
• tabf

Comments

Let [h21] = {{1, 3}, {0, 2}} be the matrix [h_2]*[h_1] in Firstov's notation, from eqs. (24) and (39). Then primitive Pythagorean triples (pPT) (x(n), y(n), z(n)) = (u(n)^2 - v(n)^2, 2*u(n)*v(n), u(n)^2 + v(n)^2), with u(n) and v(n) of different parity, gcd(u(n), v(n)) = 1, and u(n) > v(n) > 0, are generated by (u(n), v(n))^T = [h21]^n*(2,1)^T (T for transpose).

For n > 0: (x(n), y(n), z(n)) = (1, 0, 1) (mod 4). Thus some z are Pythagorean primes (A002144).

The triples converge to the proportion (4:3:5) with:

lim_{n->infinity} x(n)/y(n) = 4/3, lim_{n->infinity} y(n)/z(n) = 3/5.

Altitude h(n) = x(n)*y(n)/z(n) is an irreducible fraction because of primitivity.

From Wolfdieter Lang, Jun 13 2020: (Start)

[h21]^n = sqrt(2)^n*(S(n, 3/sqrt(2))*[1_3] + S(n-1, 3/sqrt(2))*(1/sqrt(2))*([h21] - 3*[1_3])) with the Chebyshev S polynomials (A049310).

u(n) = sqrt(2)^n*(2*S(n, 3/sqrt(2)) - (1/sqrt(2))*S(n-1, 3/sqrt(2)))

= A153893(n),

v(n) = sqrt(2)^n*(S(n, 3/sqrt(2)) - (1/sqrt(2))*S(n-1, 3/sqrt(2)))

= A000079(n). Proof from the recurrence, using the Cayley-Hamilton theorem.

With the monic Chebyshev T polynomials, called R in A127672:

x(n)/3 = 2^(n+1)*(R(2*(n+1), 3/sqrt(2)) - (sqrt(2)/3)*R(2*n+1,3/sqrt(2)) - 1) = A171477(n),

y(n)/4 = 3*2^(n-1)*(sqrt(2)*R(2*n+1,3/sqrt(2)) - R(2*n,3/sqrt(2)) - 1/3)

= A010036(n),

z(n) = 3*2^(n+1)*((3/sqrt(2))*R(2*n+1, 3/sqrt(2)) - (4/3)*R(2*n,3/sqrt(2)) - 1).

Using 2^n*Rnx(2*n, 3/sqrt(2)) = A052539(n) = 2^(2*n) + 1, and

2^(n)*(sqrt(2)/3)*Rnx(2*n+1, 3/sqrt(2)) = A007583(n) = (2^(2*n + 1) + 1)/3,

produces the explicit formulas given by the author in the formula section.

G.f.s for {x(n)} G0(x) = 3/((1 - 4*x)*(1 - 2*x)*(1 - x)), for {y(n)} G1(x) = 4*(1-x)/((1 - 4*x)*(1 - 2*x)), and for {z(n)} = (5 - 6*x + 4*x^2)/((1 - 4*x)*(1 - 2*x)*(1 - x)). This produces the g.f. for the array, read as sequence {a(n)}: G(x) = G0(x^3) + x*G1(x^3) + x^2*G2(x^3) given in the formula section by Colin Barker.

(End)

Examples

			The three-column array pPT(n,k) begins:
n\k        0        1         2
-------------------------------
0:         3        4         5
1:        21       20        29
2:       105       88       137
3:       465      368       593
4:      1953     1504      2465
5:      8001     6080     10049
6:     32385    24448     40577
7:    130305    98048    163073
8:    522753   392704    653825
9:   2094081  1571840   2618369
10:  8382465  6289408  10479617
... - _Wolfdieter Lang_, Jun 13 2020
		

Links

• V. E. Firstov, A Special Matrix Transformation Semigroup of Primitive Pairs and the Genealogy of Pythagorean Triples; Mathematical Notes, volume 84, number 2, August 2008, pages 263-279; Link of the page (for the Russian article).
• Ralf Steiner, Weitere spezielle Folge primitiver pythagoraischer Dreiecke, ResearchGate.
• Wikipedia, Tree of primitive Pythagorean triples.
• Index entries for linear recurrences with constant coefficients, signature (0,0,7,0,0,-14,0,0,8).

Crossrefs

Cf. A000079, A010035, A049310, A083329, A093357, A010036, A134057, A153893, A171477.

Programs

Mathematica

h21={{1, 3}, {0, 2}}; l = {}; Do[v = MatrixPower[h21, n, {2, 1}]; p = v[[1]]; q = v[[2]];
a = p^2 - q^2; b = 2 p q; c = p^2 + q^2; l = AppendTo[l, {a, b, c}], {n, 0, 14}]; l // Flatten

PARI

Vec((3 + 4*x + 5*x^2 - 8*x^4 - 6*x^5 + 4*x^7 + 4*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 2*x^3)*(1 - 4*x^3)) + O(x^35)) \\ Colin Barker, Jun 12 2020

Formula

The three-column array PT(n, k) is for k = 0, 1, 2: x(n), y(n), z(n), for n >= 0, with

x(n) = a(3*n + 0) = A153893(n)^2 - A000079(n)^2 = 1 - 3*2^(n+1) + 2^(2*n+3) = binomial(2^(n+2) - 1, 2) = 3*A171477(n),

y(n) = a(3*n + 1) = 2*A153893(n)*A000079(n) = 2^(n+1)*(-1 + 3*2^n) = 4*A010036(n),

z(n) = a(3*n + 2) = A153893(n)^2 + A000079(n)^2 = 1 - 6*2^n + 10*2^(2*n).

From Colin Barker, May 08 2020: (Start)

G.f. (read as sequence {a(n)}): (3 + 4*x + 5*x^2 - 8*x^4 - 6*x^5 + 4*x^7 + 4*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 2*x^3)*(1 - 4*x^3)).

a(n) = 7*a(n-3) - 14*a(n-6) + 8*a(n-9), for n > 8.

(End)

Extensions

Edited, and corrected proportion by Wolfdieter Lang, Jun 13 2020

Minor grammatical edits. - N. J. A. Sloane, Sep 12 2020