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%I #61 Sep 08 2022 08:45:00
%S 1,5,45,441,4361,43165,427285,4229681,41869521,414465525,4102785725,
%T 40613391721,402031131481,3979697923085,39394948099365,
%U 389969783070561,3860302882606241,38213059042991845,378270287547312205
%N a(n)-th star number (A003154) is a square.
%C A two-way infinite sequence which is palindromic.
%C Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - _Colin Barker_, Jan 02 2015
%C Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - _Colin Barker_, Jan 02 2015
%H Colin Barker, <a href="/A054318/b054318.txt">Table of n, a(n) for n = 1..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 Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2016volume16/FG2016volume16.pdf#page=423">Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences</a>, Forum Geometricorum, Volume 16 (2016) 419-427.
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-11,1).
%F a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
%F a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
%F From _Michael Somos_, Mar 18 2003: (Start)
%F G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
%F 12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
%F a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
%F a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
%F a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
%F From _Peter Bala_, May 01 2012: (Start)
%F a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
%F a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
%F 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.
%F Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
%F A003154(a(n)) = A006061(n). - _Zak Seidov_, Oct 22 2012
%F a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - _Seiichi Manyama_, Aug 11 2016
%F 2*a(n) = 1+A072256(n). - _R. J. Mathar_, Feb 07 2022
%e a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.
%t CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* _Michael De Vlieger_, Aug 11 2016 *)
%t LinearRecurrence[{11,-11,1},{1,5,45},30] (* _Harvey P. Dale_, Nov 05 2016 *)
%o (PARI) a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)
%o (PARI) Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ _Colin Barker_, Jan 02 2015
%o (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
%o (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
%o (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
%Y A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
%Y Quintisection of column k=2 of A233427.
%Y Cf. A001844, A003215, A253475.
%K easy,nonn
%O 1,2
%A _Ignacio Larrosa Cañestro_, Feb 27 2000
%E More terms from _James A. Sellers_, Mar 01 2000
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