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%I #80 Jun 11 2023 12:09:50
%S 0,5,54,539,5340,52865,523314,5180279,51279480,507614525,5024865774,
%T 49741043219,492385566420,4874114620985,48248760643434,
%U 477613491813359,4727886157490160,46801248083088245,463284594673392294,4586044698650834699,45397162391834954700
%N Indices k of hex numbers H(k) that are also triangular.
%C From the law of cosines, the non-Pythagorean triple {a(n), a(n)+1=A253475(n+1), A072256(n+1)} forms a near-isosceles triangle with the angle bounded by the consecutive sides equal to the regular tetrahedron's central angle (see A156546 and A247412). This implies also that a(n) are those numbers k such that (16/3)*A000217(k)+1 is a perfect square. - _Federico Provvedi_, Apr 04 2023
%H Colin Barker, <a href="/A087125/b087125.txt">Table of n, a(n) for n = 0..1000</a>
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_1.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_2.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_3.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexNumber.html">Hex Number</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-11,1).
%F G.f.: (-x^2+5*x)/((1-x)*(1-10*x+x^2)).
%F a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n > 2. - _Colin Barker_, Jun 23 2015
%F a(n) = (-4 - (5-2*sqrt(6))^n*(-2 + sqrt(6)) + (2+sqrt(6))*(5+2*sqrt(6))^n)/8. - _Colin Barker_, Mar 05 2016
%F a(n) = 10*a(n-1) - a(n-2) + 4 for n > 1. - _Charlie Marion_, Feb 14 2023
%F a(n) = ((x^(n+1)+1)*(x^n-1))/(2*x^n*(x-1)), with x=5+2*sqrt(6). - _Federico Provvedi_, Apr 04 2023
%F a(n) = sqrt(3*A161680(A054318(n+1)) + 1/4) - 1/2 = floor(sqrt(3*A000217(A054318(n+1)-1) + 1/4). - _Federico Provvedi_, Apr 16 2023
%t CoefficientList[Series[(-x^2+5*x)/((1-x)*(1-10*x+x^2)), {x, 0, 25}], x] (* _G. C. Greubel_, Nov 04 2017 *)
%t LinearRecurrence[{11,-11,1},{0,5,54},30] (* _Harvey P. Dale_, Jun 14 2022 *)
%o (PARI) concat(0, Vec(x*(x-5)/((x-1)*(x^2-10*x+1)) + O(x^50))) \\ _Colin Barker_, Jun 23 2015
%o (Magma) [Round((-4-(5-2*Sqrt(6))^n*(-2+Sqrt(6)) + (2+Sqrt(6))*(5 + 2*Sqrt(6))^n)/8): n in [0..25]]; // _G. C. Greubel_, Nov 04 2017
%o Table[(x Sqrt[z^(2 n + 1) + z^-(2 n + 1) - 2] - 4) / 8 //. {x -> Sqrt[2], y -> Sqrt[3], z -> (5 + 2 x y)}, {n, 0, 100}] // Round (* _Federico Provvedi_, Apr 16 2023 *)
%Y Cf. A006244, A031138, A072256, A054318.
%K nonn,easy
%O 0,2
%A _Eric W. Weisstein_, Aug 14 2003
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