A157659 - OEIS

%I #40 Jul 25 2024 02:42:50

%S 99,398,897,1596,2495,3594,4893,6392,8091,9990,12089,14388,16887,

%T 19586,22485,25584,28883,32382,36081,39980,44079,48378,52877,57576,

%U 62475,67574,72873,78372,84071,89970,96069,102368,108867,115566,122465,129564

%N a(n) = 100*n^2 - n.

%C The identity (200*n - 1)^2 - (100*n^2 - n)*(20)^2 = 1 can be written as A157955(n)^2 - a(n)*(20)^2 = 1 (see Barbeau's paper).

%C Also, the identity (80000*n^2 - 800*n + 1)^2 - (100*n^2 - n)*(8000*n - 40)^2 = 1 can be written as A157661(n)^2 - a(n)*A157660(n)^2 = 1 (see also the second part of the comment at A157661). - _Vincenzo Librandi_, Jan 28 2012

%H Vincenzo Librandi, <a href="/A157659/b157659.txt">Table of n, a(n) for n = 1..10000</a>

%H E. J. Barbeau, <a href="http://www.math.toronto.edu/barbeau/home.html">Polynomial Excursions</a>, Chapter 10: <a href="http://www.math.toronto.edu/barbeau/hxpol10.pdf">Diophantine equations</a> (2010), pages 84-85 (row 14 in the first table at p. 85, case d(t) = t*(10^2*t-1)).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: x*(99 + 101*x)/(1-x)^3.

%F E.g.f.: x*(99 + 100*x)*exp(x). - _G. C. Greubel_, Nov 17 2018

%t LinearRecurrence[{3,-3,1},{99,398,897},50]

%o (Magma) I:=[99, 398, 897]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];

%o (PARI) a(n)=100*n^2-n \\ _Charles R Greathouse IV_, Dec 28 2011

%o (Sage) [100*n^2-n for n in (1..40)] # _G. C. Greubel_, Nov 17 2018

%o (GAP) List([1..40], n -> 100*n^2-n); # _G. C. Greubel_, Nov 17 2018

%Y Cf. A157660, A157661, A157955.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 04 2009