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Thickened cube numbers: a(n) = n*(n^2 + (n-1)^2) + (n-1)*2*n*(n-1).
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%I #48 Feb 17 2022 03:42:27

%S 1,14,63,172,365,666,1099,1688,2457,3430,4631,6084,7813,9842,12195,

%T 14896,17969,21438,25327,29660,34461,39754,45563,51912,58825,66326,

%U 74439,83188,92597,102690,113491,125024,137313,150382,164255,178956

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

%C In other words, positive integers k such that 2*k - 1 is a perfect cube. - _Altug Alkan_, Apr 15 2016

%C 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 Vincenzo Librandi, <a href="/A050492/b050492.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = n*(4*n^2-6*n+3).

%F 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 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 a(n) = ((2n-1)^3 + 1)/2. - _Dave Durgin_, May 07 2014

%F E.g.f.: x*(4*x^2 + 6*x + 1)*exp(x). - _G. C. Greubel_, Apr 15 2016

%e * * * * *

%e a(2) = * + * * + * = 14.

%e * * * * *

%t 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 (Magma) [n*(4*n^2-6*n+3): n in [1..40]]; // _Vincenzo Librandi_, Oct 03 2011

%o (PARI) a(n)=n*(4*n^2-6*n+3) \\ _Charles R Greathouse IV_, Nov 10 2015

%Y Cf. A001844, A046092, A050533.

%K nonn,easy,nice

%O 1,2

%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 27 1999