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a(n) = Sum_{i=1..n} (3i)^2.
1

%I #22 Jun 15 2020 14:38:07

%S 0,9,45,126,270,495,819,1260,1836,2565,3465,4554,5850,7371,9135,11160,

%T 13464,16065,18981,22230,25830,29799,34155,38916,44100,49725,55809,

%U 62370,69426,76995,85095,93744,102960,112761,123165,134190,145854,158175,171171,184860,199260

%N a(n) = Sum_{i=1..n} (3i)^2.

%H T. A. Gulliver, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/37-40-2012/gulliverIJCMS37-40-2012.pdf">Sums of Powers of Integers Divisible by Three</a>, Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 38, pp. 1895-1901

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = (3*n*(1+3*n+2*n^2))/2. G.f.: 9*x*(x+1) / (x-1)^4. - _Colin Barker_, Jul 22 2013

%F a(n) = A045943(n) * A005408(n). - _Bruce J. Nicholson_, Jun 12 2020

%o (PARI) a(n) = sum(k=1, n, (3*k)^2); \\ _Michel Marcus_, Aug 29 2015

%Y Equals 9*A000330.

%Y Cf. A045943, A005408.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Dec 22 2012