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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
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OFFSET
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0,1
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COMMENTS
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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.
[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)))
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)
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)
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LINKS
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FORMULA
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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),
z(n) = a(3*n + 2) = A153893(n)^2 + A000079(n)^2 = 1 - 6*2^n + 10*2^(2*n).
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)
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EXAMPLE
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The three-column array pPT(n,k) begins:
n\k 0 1 2
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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
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MATHEMATICA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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