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Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4*T(m) = 5*T(k) for some k.
1

%I #30 Sep 08 2022 08:46:00

%S 0,45,14535,4680270,1507032450,485259768675,156252138480945,

%T 50312703331095660,16200534220474321620,5216521706289400466025,

%U 1679703788890966475738475,540859403501184915787322970,174155048223592651917042257910,56077384668593332732371819724095

%N Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4*T(m) = 5*T(k) for some k.

%C For n > 1, a(n) = 322*a(n-1) - a(n-2) + 45. See A200994 for generalization.

%H Vincenzo Librandi, <a href="/A201004/b201004.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (323,-323,1).

%F G.f.: 45*x / ((1-x)*(x^2-322*x+1)). - _R. J. Mathar_, Aug 10 2014

%F From _Colin Barker_, Mar 02 2016: (Start)

%F a(n) = (-18 + (9-4*sqrt(5))*(161+72*sqrt(5))^(-n) + (9+4*sqrt(5))*(161+72*sqrt(5))^n)/128.

%F a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2. (End)

%e 4*0 = 5*0.

%e 4*45 = 5*36.

%e 4*14535 = 5*11628.

%e 4*4680270 = 5*3744216.

%t LinearRecurrence[{323, -323, 1}, {0, 45, 14535}, 20] (* _T. D. Noe_, Feb 15 2012 *)

%t CoefficientList[Series[-45 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 11 2014 *)

%o (PARI) concat(0, Vec(45*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ _Colin Barker_, Mar 02 2016

%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(45*x/((1-x)*(1-322*x+x^2)))); // _G. C. Greubel_, Jul 15 2018

%Y Cf. A001652, A029549, A053141, A075528, A200993-A201008.

%K nonn,easy

%O 0,2

%A _Charlie Marion_, Feb 15 2012

%E a(7) corrected by _R. J. Mathar_, Aug 10 2014