OFFSET
1,2
COMMENTS
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
LINKS
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 linear recurrences with constant coefficients, signature (11,-11,1).
FORMULA
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)
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
EXAMPLE
a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.
MATHEMATICA
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 *)
PROG
(PARI) a(n)=if(n<1, a(1-n), 1/2+subst(poltchebi(n)+poltchebi(n-1), x, 5)/12)
(PARI) Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015
(Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019
(Sage) (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
(GAP) 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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Ignacio Larrosa Cañestro, Feb 27 2000
EXTENSIONS
More terms from James A. Sellers, Mar 01 2000
STATUS
approved