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%I #23 Sep 09 2022 11:08:01
%S 57,308,759,1410,2261,3312,4563,6014,7665,9516,11567,13818,16269,
%T 18920,21771,24822,28073,31524,35175,39026,43077,47328,51779,56430,
%U 61281,66332,71583,77034,82685,88536,94587,100838,107289,113940,120791,127842
%N a(n) = 100*n^2 - 49*n + 6.
%C The identity (80000*n^2 -39200*n +4801)^2 - (100*n^2 -49*n +6)*(8000*n -1960)^2 = 1 can be written as A157653(n)^2 - a(n)*A157652(n)^2 = 1.
%C The continued fraction expansion of sqrt(a(n)) is [10n-3; {1, 1, 4, 1, 1, 20n-6}]. - _Magus K. Chu_, Sep 09 2022
%H Vincenzo Librandi, <a href="/A157651/b157651.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&tstart=0">X^2-AY^2=1</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).
%F G.f.: x*(57 + 137*x + 6*x^2)/(1-x)^3.
%F E.g.f.: (6 + 51*x + 100*x^2)*exp(x) - 6. - _G. C. Greubel_, Nov 17 2018
%t LinearRecurrence[{3,-3,1},{57,308,759},40]
%o (Magma) I:=[57, 308, 759]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
%o (PARI) a(n) = 100*n^2 - 49*n + 6.
%o (Sage) [100*n^2-49*n+6 for n in (1..40)] # _G. C. Greubel_, Nov 17 2018
%o (GAP) List([1..40], n -> 100*n^2-49*n+6); # _G. C. Greubel_, Nov 17 2018
%Y Cf. A157652, A157653.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 03 2009
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