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A166942 One fifth of product plus sum of five consecutive nonnegative numbers.

%I A166942 #20 Sep 08 2022 08:45:48
%S A166942 2,27,148,509,1350,3031,6056,11097,19018,30899,48060,72085,104846,
%T A166942 148527,205648,279089,372114,488395,632036,807597,1020118,1275143,
%U A166942 1578744,1937545,2358746,2850147,3420172,4077893,4833054,5696095
%N A166942 One fifth of product plus sum of five consecutive nonnegative numbers.
%C A166942 a(n) = ((n*...*(n+4))+(n+...+(n+4)))/5, n >= 0.
%C A166942 Binomial transform of 2, 25, 96, 144, 96, 24, 0, 0, 0, 0, ....
%C A166942 Partial sums of A062938 where initial term 1 is replaced by 2.
%H A166942 Vincenzo Librandi, <a href="/A166942/b166942.txt">Table of n, a(n) for n = 0..1000</a>
%F A166942 a(n) = (n^5 + 10n^4 + 35n^3 + 50n^2 + 29n + 10)/5. - _Charles R Greathouse IV_, Nov 02 2009
%F A166942 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 24 for n > 4; a(0)=2, a(1)=27, a(2)=148, a(3)=509, a(4)=1350. - _Klaus Brockhaus_, Nov 14 2009
%F A166942 G.f.: (2+15*x+16*x^2-14*x^3+6*x^4-x^5)/(1-x)^6. - _Klaus Brockhaus_, Nov 14 2009
%e A166942 a(0) = (0*1*2*3*4 + 0 + 1 + 2 + 3 + 4)/5 = (0 + 10)/5 = 2.
%e A166942 a(1) = (1*2*3*4*5 + 1 + 2 + 3 + 4 + 5)/5 = (120 + 15)/5 = 27.
%t A166942 Table[((n+4)*(n+3)*(n+2)*(n+1)*n+(n+4)+(n+3)+(n+2)+(n+1)+n)/5, {n,0,100}]
%t A166942 (Total[#]+Times@@#)/5&/@Partition[Range[0,100],5,1]  (* _Harvey P. Dale_, Mar 05 2011 *)
%o A166942 (Magma) [ (&*s + &+s)/5 where s is [n..n+4]: n in [0..29] ]; // _Klaus Brockhaus_, Nov 14 2009
%Y A166942 Cf. A001477 (nonnegative integers), A062938 (squares of the form n(n+1)(n+2)(n+3)+1), A028387 (n+(n+1)^2), A167875, A166941, A166943.
%K A166942 nonn,easy
%O A166942 0,1
%A A166942 _Vladimir Joseph Stephan Orlovsky_, Oct 24 2009
%E A166942 Edited and offset corrected by _Klaus Brockhaus_, Nov 14 2009