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a(n) = n*binomial(n+4, 4).
3

%I #29 Sep 08 2022 08:45:51

%S 0,5,30,105,280,630,1260,2310,3960,6435,10010,15015,21840,30940,42840,

%T 58140,77520,101745,131670,168245,212520,265650,328900,403650,491400,

%U 593775,712530,849555,1006880,1186680,1391280,1623160,1884960,2179485

%N a(n) = n*binomial(n+4, 4).

%C This sequence can be computed from Pascal's triangle. Find the fifth number in a row and multiply it by the second number of the next row. - _Alonso del Arte_, Jan 21 2018

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6, -15, 20, -15, 6, -1).

%F a(n) = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n) / 24.

%F For n > 0: a(n) = A003506(n+4, 5).

%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), with a(0)=0, a(1)=5, a(2)=30, a(3)=105, a(4)=280, a(5)=630. - _Harvey P. Dale_, Dec 03 2011

%F G.f.: 5*x/(1-x)^6. - _Colin Barker_, Mar 18 2012

%t Table[n Binomial[n + 4, 4], {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 5, 30, 105, 280, 630}, 40] (* _Harvey P. Dale_, Dec 03 2011 *)

%o (Magma) [ (n^5+10*n^4+35*n^3+50*n^2+24*n)/24: n in [0..40] ]; // _Vincenzo Librandi_, Dec 28 2010

%Y Cf. A033488, A027480, A003506.

%K nonn,easy

%O 0,2

%A _Reinhard Zumkeller_, Mar 05 2010, Mar 17 2010

%E Title switched with first Formula section entry, at the suggestion of _Alonso del Arte_, by _Jon E. Schoenfield_, Jan 28 2018