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Fifth diagonal (m=4) of triangle A084938; a(n) = A084938(n+4,n) = (n^4 + 18*n^3 + 131*n^2 + 426*n)/24.
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%I #17 Sep 08 2022 08:45:12

%S 0,24,64,126,217,345,519,749,1046,1422,1890,2464,3159,3991,4977,6135,

%T 7484,9044,10836,12882,15205,17829,20779,24081,27762,31850,36374,

%U 41364,46851,52867,59445,66619,74424,82896,92072,101990,112689

%N Fifth diagonal (m=4) of triangle A084938; a(n) = A084938(n+4,n) = (n^4 + 18*n^3 + 131*n^2 + 426*n)/24.

%H Vincenzo Librandi, <a href="/A090386/b090386.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5, -10, 10, -5, 1).

%F a(n) = A084938(n+4, n) = Sum_{k=0..4} A090238(4, k)*binomial(n, k).

%F a(0)=0, a(1)=24, a(2)=64, a(3)=126, a(4)=217, a(n)=5*a(n-1)- 10*a(n-2)+ 10*a(n-3)-5*a(n-4)+a(n-5). - _Harvey P. Dale_, Feb 23 2014

%t Table[(n^4+18n^3+131n^2+426n)/24,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,24,64,126,217},40] (* _Harvey P. Dale_, Feb 23 2014 *)

%o (Magma) [(n^4+18*n^3+131*n^2+426*n)/24: n in [0..40]]; // _Vincenzo Librandi_, Feb 24 2014

%Y Cf. A084938 A090238.

%K easy,nonn

%O 0,2

%A _Philippe Deléham_, Jan 30 2004

%E Corrected by _T. D. Noe_, Nov 08 2006