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%I #48 Sep 08 2022 08:45:34
%S 1,3,12,28,51,81,118,162,213,271,336,408,487,573,666,766,873,987,1108,
%T 1236,1371,1513,1662,1818,1981,2151,2328,2512,2703,2901,3106,3318,
%U 3537,3763,3996,4236,4483,4737,4998,5266,5541,5823,6112,6408,6711,7021,7338
%N a(n) = (7*n^2 - 17*n + 12)/2.
%C Binomial transform of [1, 2, 7, 0, 0, 0, ...].
%C This sequence and 1, 6, 18, 37, 63, 96, ... with signature (3,-3,1) [not yet in OEIS] together contain all numbers k, so that 56*k - 47 is a square. - _Klaus Purath_, Oct 21 2021
%H G. C. Greubel, <a href="/A140065/b140065.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A007318 * [1, 2, 7, 0, 0, 0, ...].
%F a(n) = A000217(n) + 6*A000217(n-2) = (A140064(n) + A140066(n))/2. - _R. J. Mathar_, May 06 2008
%F O.g.f.: x*(1+6*x^2)/(1-x)^3. - _Alexander R. Povolotsky_, May 06 2008
%F a(n) = 7*n + a(n-1) - 12 for n>1, a(1)=1. - _Vincenzo Librandi_, Jul 08 2010
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 4. - _Klaus Purath_, Oct 21 2021
%e a(4) = 28 = (1, 3, 3, 1) * (1, 2, 7, 0) = (1 + 6 + 21 + 0).
%p seq((12-17*n+7*n^2)*1/2, n=1..40); # _Emeric Deutsch_, May 07 2008
%t Table[(7 n^2 - 17 n + 12)/2, {n, 1, 50}] (* _Bruno Berselli_, Mar 12 2015 *)
%t LinearRecurrence[{3,-3,1},{1,3,12},50] (* _Harvey P. Dale_, May 28 2017 *)
%o (PARI) x = 'x + O('x^50); Vec(x*(1+6*x^2)/(1-x)^3) \\ _G. C. Greubel_, Feb 23 2017
%o (Magma) [(7*n^2 - 17*n + 12)/2 : n in [1..60]]; // _Wesley Ivan Hurt_, Oct 10 2021
%Y Cf. A000217, A140064.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, May 03 2008
%E More terms from _R. J. Mathar_ and _Emeric Deutsch_, May 06 2008
%E More terms from _Vladimir Joseph Stephan Orlovsky_, Oct 25 2008
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