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A050492 Thickened cube numbers: a(n) = n*(n^2 + (n-1)^2) + (n-1)*2*n*(n-1).

%I A050492 #48 Feb 17 2022 03:42:27
%S A050492 1,14,63,172,365,666,1099,1688,2457,3430,4631,6084,7813,9842,12195,
%T A050492 14896,17969,21438,25327,29660,34461,39754,45563,51912,58825,66326,
%U A050492 74439,83188,92597,102690,113491,125024,137313,150382,164255,178956
%N A050492 Thickened cube numbers: a(n) = n*(n^2 + (n-1)^2) + (n-1)*2*n*(n-1).
%C A050492 In other words, positive integers k such that 2*k - 1 is a perfect cube. - _Altug Alkan_, Apr 15 2016
%C A050492 a(n) represents the first term in a sum of (2*n - 1)^3 consecutive integers which equals (2*n - 1)^6. - _Patrick J. McNab_, Dec 24 2016
%H A050492 Vincenzo Librandi, <a href="/A050492/b050492.txt">Table of n, a(n) for n = 1..10000</a>
%H A050492 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A050492 a(n) = n*(4*n^2-6*n+3).
%F A050492 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=1, a(2)=14, a(3)=63, a(4)=172. - _Harvey P. Dale_, Oct 02 2011
%F A050492 G.f.: x*(1+10*x+13*x^2)/(1-4*x+6*x^2-4*x^3+x^4). - _Colin Barker_, Jan 04 2012
%F A050492 a(n) = ((2n-1)^3 + 1)/2. - _Dave Durgin_, May 07 2014
%F A050492 E.g.f.: x*(4*x^2 + 6*x + 1)*exp(x). - _G. C. Greubel_, Apr 15 2016
%e A050492        * *      *      * *
%e A050492 a(2) =  *   +  * *  +   *  = 14.
%e A050492        * *      *      * *
%t A050492 Table[n(n^2+(n-1)^2)+(n-1)2n(n-1),{n,40}] (* or *) LinearRecurrence[ {4,-6,4,-1},{1,14,63,172},40] (* _Harvey P. Dale_, Oct 02 2011 *)
%o A050492 (Magma) [n*(4*n^2-6*n+3): n in [1..40]]; // _Vincenzo Librandi_, Oct 03 2011
%o A050492 (PARI) a(n)=n*(4*n^2-6*n+3) \\ _Charles R Greathouse IV_, Nov 10 2015
%Y A050492 Cf. A001844, A046092, A050533.
%K A050492 nonn,easy,nice
%O A050492 1,2
%A A050492 Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 27 1999