A349948 Tetrahedral-sided isosceles Heron triangle pairs.
0, 10, 48, 190, 720, 2698, 10080, 37630, 140448, 524170, 1956240, 7300798, 27246960, 101687050, 379501248, 1416317950, 5285770560, 19726764298, 73621286640, 274758382270, 1025412242448, 3826890587530, 14282150107680, 53301709843198, 198924689265120
Offset: 1
Examples
10 is a term, so there exists one Heron isosceles triangle whose sides are the 10th, 11th, and 11th tetrahedral numbers (220, 286, 286) and another whose sides are the 15th, 15th, and 16th tetrahedral numbers (680, 680, 816). Those two triangles have areas 29040 and 221952, respectively. (See the n=2 row of the table below.) . Triangle sides Triangle sides k= ------------------ -------------------- n a(n) T(k) T(k+1) T(k+1) Area T(k+5) T(k+5) T(k+6) Area - ---- ---- ------ ------ ------ ------ ------ ------ ------ 1 0 0 1 1 0* 35 35 56 588 2 10 220 286 286 29040 680 680 816 221952 *(degenerate triangle)
Links
- Eric Weisstein's World of Mathematics, Tetrahedral Number
- Index entries for linear recurrences with constant coefficients, signature (5,-5,1).
Programs
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Mathematica
a[ n_] := 2*ChebyshevT[n, 2] - 4; (* Michael Somos, Mar 27 2022 *)
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PARI
Vec(2*x^2*(5 - x)/(1 - 5*x + 5*x^2 - x^3) + O(x^42))
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PARI
{a(n) = 2*polchebyshev(n,1, 2) - 4}; /* Michael Somos, Mar 27 2022 */
Formula
a(n+2) = 4*a(n+1) - a(n) + 8.
From Stefano Spezia, Mar 26 2022: (Start)
G.f.: 2*x^2*(5 - x)/((1-x)*(1 - 4*x +x^2)).
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 3.
a(n) = (2 + sqrt(3))^n + (2 - sqrt(3))^n - 4. (End)
a(n) = 2*A001075(n) - 4. - Michael Somos, Mar 27 2022
Comments