A006244 -id:A006244 - cp's OEIS frontend

A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

1, 7, 67, 661, 6541, 64747, 640927, 6344521, 62804281, 621698287, 6154178587, 60920087581, 603046697221, 5969546884627, 59092422149047, 584954674605841, 5790454323909361, 57319588564487767, 567405431320968307, 5616734724645195301, 55599941815130984701

Offset: 1

Colin Barker, Jan 08 2015

Also positive integers x in the solutions to 4*x^2-6*y^2-2*x+6*y-2 = 0, the corresponding values of y being A253475.

			7 is in the sequence because the 7th hexagonal number is 91, which is also the 6th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A003215, A253475, A006244, A350923.

LinearRecurrence[{11, -11, 1}, {1, 7, 67}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(-x*(x^2-4*x+1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: -x*(x^2-4*x+1) / ((x-1)*(x^2-10*x+1)).
a(n) = (2+(5-2*sqrt(6))^n*(3+sqrt(6))-(-3+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
4*a(n) = 1+3*A072256(n). - R. J. Mathar, Feb 07 2022
a(n) = A350923(n)/2. - Paolo Xausa, May 30 2025

A087125 Indices k of hex numbers H(k) that are also triangular.

Original entry on oeis.org

0, 5, 54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 0

Eric W. Weisstein, Aug 14 2003

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

Colin Barker, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A006244, A031138, A072256, A054318.

[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

CoefficientList[Series[(-x^2+5*x)/((1-x)*(1-10*x+x^2)), {x, 0, 25}], x] (* G. C. Greubel, Nov 04 2017 *)
LinearRecurrence[{11,-11,1},{0,5,54},30] (* Harvey P. Dale, Jun 14 2022 *)
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 *)

concat(0, Vec(x*(x-5)/((x-1)*(x^2-10*x+1)) + O(x^50))) \\ Colin Barker, Jun 23 2015

G.f.: (-x^2+5*x)/((1-x)*(1-10*x+x^2)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n > 2. - Colin Barker, Jun 23 2015
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
a(n) = 10*a(n-1) - a(n-2) + 4 for n > 1. - Charlie Marion, Feb 14 2023
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
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

cp's OEIS Frontend

A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

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