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A051895 Partial sums of second pentagonal numbers with even index (A049453).

%I A051895 #33 Sep 08 2022 08:44:59
%S A051895 0,7,33,90,190,345,567,868,1260,1755,2365,3102,3978,5005,6195,7560,
%T A051895 9112,10863,12825,15010,17430,20097,23023,26220,29700,33475,37557,
%U A051895 41958,46690,51765,57195,62992,69168,75735,82705,90090,97902,106153,114855,124020,133660
%N A051895 Partial sums of second pentagonal numbers with even index (A049453).
%C A051895 For A049453(n+1), the corresponding formula would be a(n)=(n+1)*(6*n+7) and its partial sums would be given by a(n)=(n+1)*(n+2)*(4*n+7)/2.
%D A051895 A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
%H A051895 Vincenzo Librandi, <a href="/A051895/b051895.txt">Table of n, a(n) for n = 0..1000</a>
%H A051895 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A051895 a(n) = n*(n+1)*(4*n+3)/2.
%F A051895 G.f.: x*(7+5*x)/(1-x)^4. - _Colin Barker_, Jan 12 2012
%F A051895 a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - _Vincenzo Librandi_, Apr 27 2012
%F A051895 a(n) = A002492(n) + A016061(n). - _J. M. Bergot_, Apr 20 2018
%t A051895 Table[(n(4n-1)(n-1))/2,{n,40}]  (* _Harvey P. Dale_, Mar 11 2011 *)
%t A051895 CoefficientList[Series[x*(7+5*x)/(1-x)^4,{x,0,50}],x] (* _Vincenzo Librandi_, Apr 27 2012 *)
%o A051895 (Magma) I:=[0, 7, 33, 90]; [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..50]]; // _Vincenzo Librandi_, Apr 27 2012
%o A051895 (PARI) a(n) = n*(n+1)*(4*n+3)/2; \\ _Altug Alkan_, Apr 20 2018
%Y A051895 Cf. A002492, A016061, A017605, A049453.
%K A051895 nonn,easy
%O A051895 0,2
%A A051895 _Barry E. Williams_, Dec 17 1999