A201004 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.
0, 45, 14535, 4680270, 1507032450, 485259768675, 156252138480945, 50312703331095660, 16200534220474321620, 5216521706289400466025, 1679703788890966475738475, 540859403501184915787322970, 174155048223592651917042257910, 56077384668593332732371819724095
Offset: 0
Examples
4*0 = 5*0. 4*45 = 5*36. 4*14535 = 5*11628. 4*4680270 = 5*3744216.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Index entries for linear recurrences with constant coefficients, signature (323,-323,1).
Programs
-
Magma
m:=25; R
:=PowerSeriesRing(Integers(), m); Coefficients(R!(45*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018 -
Mathematica
LinearRecurrence[{323, -323, 1}, {0, 45, 14535}, 20] (* T. D. Noe, Feb 15 2012 *) CoefficientList[Series[-45 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *) -
PARI
concat(0, Vec(45*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016
Formula
For n > 1, a(n) = 322*a(n-1) - a(n-2) + 45. See A200994 for generalization.
G.f.: 45*x / ((1-x)*(x^2-322*x+1)). - R. J. Mathar, Aug 10 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-18 + (9-4*sqrt(5))*(161+72*sqrt(5))^(-n) + (9+4*sqrt(5))*(161+72*sqrt(5))^n)/128.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2. (End)
a(n) = 45*A298271(n). - Amiram Eldar, Dec 01 2024
Extensions
a(7) corrected by R. J. Mathar, Aug 10 2014