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a(n) = 25*n^2 - 14*n + 2.
8

%I #56 Sep 16 2022 21:24:21

%S 2,13,74,185,346,557,818,1129,1490,1901,2362,2873,3434,4045,4706,5417,

%T 6178,6989,7850,8761,9722,10733,11794,12905,14066,15277,16538,17849,

%U 19210,20621,22082,23593,25154,26765,28426,30137,31898,33709,35570,37481,39442,41453,43514

%N a(n) = 25*n^2 - 14*n + 2.

%C The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as A154359(n)^2 - a(n)*A154361(n)^2 = 1.

%C Numbers of the form (4*n-1)^2 + (3*n-1)^2. - _Bruno Berselli_, Dec 11 2011

%C From _Bruno Berselli_, Dec 13 2011: (Start)

%C More generally, considering together this sequence and A154355, A154358-A154361, for

%C r = (1/4)*(1250*(n-1)*(n-2) + 75*(2*n-3)(-1)^n + 321) with n>=0, i.e. the interleaving of A154358 and A154359 (649, 99, 99, 649, 2049, 3699,...)

%C s = (5/2)*(50*n+3*(-1)^n-75), the interleaving of A154360 and A154361 (-180, -70, 70, 180, 320, 430,...)

%C t = (1/8)*(50*(n-1)*(n-2) + 3*(2*n-3)*(-1)^n + 13), the interleaving of A154355 and A154357 (13, 2, 2, 13, 41, 74,...)

%C we verify that r^2 - t*s^2 = 1.

%C For n even we obtain (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1; for n odd we have the identity shown in the first comment. (End)

%C sqrt(A154357(n)) for n >= 1 has the continued fraction x; [1 1 1 1 2x] where x = 5n - 2 (the part in [] being repeated). - _Robert Israel_, May 26 2013

%C For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [10n-3; {4, 1, 5n-3, 1, 4, 20n-6}] - _Magus K. Chu_, Sep 16 2022

%H Vincenzo Librandi, <a href="/A154357/b154357.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 G.f.: (2 + 7*x + 41*x^2)/(1-x)^3. - _R. J. Mathar_, Jan 05 2011

%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - _Vincenzo Librandi_, Feb 08 2012

%F E.g.f.: (2 + 11*x + 25*x^2)*exp(x). - _G. C. Greubel_, Sep 14 2016

%t LinearRecurrence[{3, -3, 1}, {2, 13, 74}, 40] (* _Vincenzo Librandi_, Feb 08 2012 *)

%o (PARI) a(n)=25*n^2-14*n+2 \\ _Charles R Greathouse IV_, Dec 23 2011

%o (Magma) [25*n^2-14*n+2: n in [0..40]]; // _Bruno Berselli_, Sep 15 2016

%Y Cf. A154355, A154358, A154359, A154360, A154361.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Jan 08 2009

%E One entry and offset corrected by _R. J. Mathar_, Jan 05 2011

%E First comment rewritten by _Bruno Berselli_, Dec 11 2011