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%I #39 Sep 16 2022 21:25:00
%S 29,338,985,1970,3293,4954,6953,9290,11965,14978,18329,22018,26045,
%T 30410,35113,40154,45533,51250,57305,63698,70429,77498,84905,92650,
%U 100733,109154,117913,127010,136445,146218,156329,166778
%N a(n) = 169*n^2 + 140*n + 29.
%C The identity (57122*n^2 +47320*n +9801)^2 - (169*n^2 +140*n +29)*(4394*n +1820)^2 = 1 can be written as A156735(n)^2 - a(n)*A156636(n)^2 = 1.
%C The continued fraction expansion of sqrt(a(n)) is [13n+5; {2, 1, 1, 2, 26n+10}]. - _Magus K. Chu_, Sep 15 2022
%H Vincenzo Librandi, <a href="/A156640/b156640.txt">Table of n, a(n) for n = 0..10000</a>
%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) for n>2.
%F G.f.: (29 + 251*x + 58*x^2)/(1-x)^3. - _Vincenzo Librandi_, May 03 2014
%F E.g.f.: (29 +309*x +169*x^2)*exp(x). - _G. C. Greubel_, Feb 28 2021
%p A156640:= n-> 169*n^2 + 140*n + 29; seq(A156640(n), n=0..50); # _G. C. Greubel_, Feb 28 2021
%t LinearRecurrence[{3,-3,1},{29,338,985},50]
%t CoefficientList[Series[(29 +251x +58x^2)/(1-x)^3, {x, 0, 60}], x] (* _Vincenzo Librandi_, May 03 2014 *)
%o (Magma) I:=[29, 338, 985]; [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)=169*n^2+140*n+29 \\ _Charles R Greathouse IV_, Dec 23 2011
%o (Sage) [169*n^2 + 140*n + 29 for n in (0..50)] # _G. C. Greubel_, Feb 28 2021
%Y Cf. A156636, A156718, A156735.
%Y Cf. A154609 (13n+5).
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Feb 15 2009
%E Edited by _Charles R Greathouse IV_, Jul 25 2010
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