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%I #49 Mar 08 2022 20:32:42
%S 1,1225,1413721,1631432881,1882672131025,2172602007770041,
%T 2507180834294496361,2893284510173841030625,3338847817559778254844961,
%U 3853027488179473932250054441,4446390382511295358038307980025,5131130648390546663702275158894481
%N Squares (A000290) which are also hexagonal numbers (A000384).
%C Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - _Alexander Adamchuk_, Nov 06 2007
%C Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - _Richard Choulet_, May 01 2009
%C As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - _Ant King_ Nov 08 2011
%C Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - _Colin Barker_, Jan 16 2015
%C Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 25 2015
%D M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.
%H Colin Barker, <a href="/A046177/b046177.txt">Table of n, a(n) for n = 1..327</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexagonalSquareNumber.html">Hexagonal Square Number.</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SquareTriangularNumber.html">Square Triangular Number</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1155,-1155,1).
%F a(n) = A001110(2n-1). - _Alexander Adamchuk_, Nov 06 2007
%F a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - _Richard Choulet_, May 01 2009
%F a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - _Richard Choulet_, May 01 2009
%F From _Ant King_, Nov 07 2011: (Start)
%F a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
%F a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
%F a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
%F a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
%F a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
%F G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
%F (End)
%F a(n) = A096979(4*n - 3). - _Ivan N. Ianakiev_, Sep 05 2016
%F a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - _Ivan N. Ianakiev_, Sep 05 2016
%t LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* _Ant King_, Nov 08 2011 *)
%o (PARI) Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ _Colin Barker_, Jan 16 2015
%Y Cf. A008844, A046176, A253826.
%Y Cf. A001110 (Numbers that are both triangular and square).
%Y Cf. A000290, A000384, A016754, A253826.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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