|
%I #31 Feb 17 2022 09:56:04
%S 0,5,22,59,124,225,370,567,824,1149,1550,2035,2612,3289,4074,4975,
%T 6000,7157,8454,9899,11500,13265,15202,17319,19624,22125,24830,27747,
%U 30884,34249,37850,41695,45792,50149,54774,59675,64860,70337,76114,82199
%N Thickened pyramidal numbers: a(n) = 2*(n+1)*n + Sum_{i=1..n} (4*i*(i-1) + 1).
%C This sequence is the partial sums of A053755. - _J. M. Bergot_, May 31 2012
%H Vincenzo Librandi, <a href="/A050533/b050533.txt">Table of n, a(n) for n = 0..1000</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) = (1/3)*n*(5 + 6*n + 4*n^2) = binomial(2*n+1, 3) + 2*(n+1)*n = A000447(n) + 4*A000217(n).
%F G.f.: x*(5+2*x+x^2)/(1-x)^4. - _Colin Barker_, Apr 16 2012
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Vincenzo Librandi_, Apr 27 2012
%t CoefficientList[Series[x*(5+2*x+x^2)/(1-x)^4,{x,0,40}],x] (* _Vincenzo Librandi_, Apr 27 2012 *)
%t LinearRecurrence[{4,-6,4,-1},{0,5,22,59},40] (* _Harvey P. Dale_, May 08 2012 *)
%o (PARI) a(n)=n*(4*n^2+6*n+5)/3 \\ _Charles R Greathouse IV_, Apr 16 2012
%o (Magma) I:=[0, 5, 22, 59]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Apr 27 2012
%Y Cf. A000292, A000447, A000217, A050492.
%K nonn,easy,nice
%O 0,2
%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999
|
