A253475 -id:A253475 - cp's OEIS frontend

A054318 a(n)-th star number (A003154) is a square.

1, 5, 45, 441, 4361, 43165, 427285, 4229681, 41869521, 414465525, 4102785725, 40613391721, 402031131481, 3979697923085, 39394948099365, 389969783070561, 3860302882606241, 38213059042991845, 378270287547312205

Offset: 1

Ignacio Larrosa Cañestro, Feb 27 2000

A two-way infinite sequence which is palindromic.
Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - Colin Barker, Jan 02 2015
Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - Colin Barker, Jan 02 2015

			a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.

Colin Barker, Table of n, a(n) for n = 1..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).
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
Quintisection of column k=2 of A233427.
Cf. A001844, A003215, A253475.

a:=[1,5,45];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* Michael De Vlieger, Aug 11 2016 *)
LinearRecurrence[{11,-11,1},{1,5,45},30] (* Harvey P. Dale, Nov 05 2016 *)

a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)

Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015

(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
From Michael Somos, Mar 18 2003: (Start)
G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
a(n+1) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(3) and T(n,x) the Chebyshev polynomial of the first kind.
Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
A003154(a(n)) = A006061(n). - Zak Seidov, Oct 22 2012
a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - Seiichi Manyama, Aug 11 2016
2*a(n) = 1+A072256(n). - R. J. Mathar, Feb 07 2022

More terms from James Sellers, Mar 01 2000

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

Original entry on oeis.org

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

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A054318 a(n)-th star number (A003154) is a square.

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

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

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