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a(n)=12*sum(1<=i<=j<=k<=n,i*j/k).
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%I #16 Jun 17 2017 03:06:05

%S 0,12,54,154,349,685,1217,2009,3134,4674,6720,9372,12739,16939,22099,

%T 28355,35852,44744,55194,67374,81465,97657,116149,137149,160874,

%U 187550,217412,250704,287679,328599,373735,423367,477784,537284,602174,672770

%N a(n)=12*sum(1<=i<=j<=k<=n,i*j/k).

%C Always an integer.

%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)=(n/24)*(9*n^3 + 58*n^2 + 123*n + 98).

%F G.f.: x*(x^3-4*x^2+6*x-12) / (x-1)^5. - _Colin Barker_, Jun 15 2013

%t LinearRecurrence[{5,-10,10,-5,1},{0,12,54,154,349},40] (* _Harvey P. Dale_, May 01 2014 *)

%o (PARI) a(n)=if(n<0,0,a(n)=n/24*(9*n^3 + 58*n^2 + 123*n + 98))

%Y Cf. A088942, A088943.

%Y A051798(n) - 1.

%K nonn,easy

%O 0,2

%A _Benoit Cloitre_, Oct 25 2003