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%I #32 Mar 24 2023 09:46:07
%S 58,425,1130,2173,3554,5273,7330,9725,12458,15529,18938,22685,26770,
%T 31193,35954,41053,46490,52265,58378,64829,71618,78745,86210,94013,
%U 102154,110633,119450,128605,138098,147929,158098
%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 A156721(n)^2 - a(n)*A156627(n)^2 = 1.
%C The continued fraction expansion of sqrt(a(n)) is [13n-6; {1, 1, 1, 1, 1, 1, 26n-12}]. - _Magus K. Chu_, Sep 06 2022
%H Vincenzo Librandi, <a href="/A156639/b156639.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: x*(58 + 251*x + 29*x^2)/(1 - x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%t LinearRecurrence[{3, -3, 1}, {58, 425, 1130}, 40]
%t Table[169n^2-140n+29,{n,40}] (* _Harvey P. Dale_, Mar 24 2023 *)
%o (Magma) I:=[58, 425, 1130]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..40]];
%o (PARI) a(n)=169*n^2-140*n+29 \\ _Charles R Greathouse IV_, Dec 23 2011
%Y Cf. A156627, A156721.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Feb 15 2009
%E Edited by _Charles R Greathouse IV_, Jul 25 2010
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