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%I #24 Aug 10 2019 11:32:39
%S 0,16,63,141,250,390,561,763,996,1260,1555,1881,2238,2626,3045,3495,
%T 3976,4488,5031,5605,6210,6846,7513,8211,8940,9700,10491,11313,12166,
%U 13050,13965,14911,15888,16896,17935
%N a(n) = n*(31*n + 1)/2.
%H G. C. Greubel, <a href="/A022289/b022289.txt">Table of n, a(n) for n = 0..5000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 31*n + a(n-1) - 15, for n>0, a(0)=0. - _Vincenzo Librandi_, Aug 04 2010
%F G.f.: x*(16 + 15*x)/(1 - x)^3 . - _R. J. Mathar_, Sep 02 2016
%F a(n) = A000217(16*n) - A000217(15*n). In general, n*((2*k+1)*n + 1)/2 = A000217((k+1)*n) - A000217(k*n). - _Bruno Berselli_, Oct 13 2016
%F E.g.f.: (x/2)*(31*x + 32)*exp(x). - _G. C. Greubel_, Aug 23 2017
%t Table[n (31 n + 1)/2, {n, 0, 40}] (* _Bruno Berselli_, Oct 13 2016 *)
%t LinearRecurrence[{3,-3,1},{0,16,63},40] (* _Harvey P. Dale_, Aug 10 2019 *)
%o (PARI) a(n)=n*(31*n+1)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. similar sequences of the form n*((2*k+1)*n + 1)/2: A000217 (k=0), A005449 (k=1), A005475 (k=2), A022265 (k=3), A022267 (k=4), A022269 (k=5), A022271 (k=6), A022273 (k=7), A022275 (k=8), A022277 (k=9), A022279 (k=10), A022281 (k=11), A022283 (k=12), A022285 (k=13), A022287 (k=14), this sequence (k=15).
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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