%I #42 Sep 08 2022 08:45:01

%S 101,402,903,1604,2505,3606,4907,6408,8109,10010,12111,14412,16913,

%T 19614,22515,25616,28917,32418,36119,40020,44121,48422,52923,57624,

%U 62525,67626,72927,78428,84129,90030,96131,102432,108933,115634,122535

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

%C The identity (200n+1)^2 - (100n^2+n)*20^2 = 1 can be written as A157956(n)^2 - a(n)*20^2 = 1 (see Barbeau's paper). Also, the identity (80000n^2 + 800n + 1)^2 - (100n^2 + n)*(8000n + 40)^2 = 1 can be written as A157664(n)^2 - a(n)*A157663(n)^2 = 1 (see the comment from _Bruno Berselli_ in A157664). - _Vincenzo Librandi_, Feb 04 2012

%H Vincenzo Librandi, <a href="/A055438/b055438.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 G.f.: x*(-101-99*x)/(x-1)^3. - _Vincenzo Librandi_, Feb 04 2012

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 04 2012

%t LinearRecurrence[{3, -3, 1}, {101, 402, 903}, 50] (* _Vincenzo Librandi_, Feb 04 2012 *)

%t Table[100n^2+n,{n,40}] (* _Harvey P. Dale_, May 15 2018 *)

%o (Magma) I:=[101, 402, 903]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 04 2012

%o (PARI) for(n=1, 50, print1(100*n^2+n", ")); \\ _Vincenzo Librandi_, Feb 04 2012

%Y Cf. A157956, A157663, A157664, A002378, A055437; a(n) = A055436(n) if 10 <= n < 100.

%Y Different from A031698.

%K nonn,easy

%O 1,1

%A _Henry Bottomley_, May 18 2000