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T(n)^3 - n^3 where T(n) is a triangular number.
1

%I #22 Sep 08 2022 08:45:58

%S 0,19,189,936,3250,9045,21609,46144,90396,165375,286165,472824,751374,

%T 1154881,1724625,2511360,3576664,4994379,6852141,9253000,12317130,

%U 16183629,21012409,26986176,34312500,43225975,53990469,66901464,82288486,100517625,121994145

%N T(n)^3 - n^3 where T(n) is a triangular number.

%H Vincenzo Librandi, <a href="/A193575/b193575.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F a(n) = (n^3*(n^3+3*n^2+3*n-7)/8) = (1/8)*(n-1)*(n^2+4*n+7)*n^3.

%F From _Wesley Ivan Hurt_, Aug 23 2014: (Start)

%F G.f.: x^2*(19+56*x+12*x^2+2*x^3+x^4)/(1-x)^7.

%F a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7).

%F a(n) = sum_{i=1..n} sum_{j=1..n} sum_{k=1..n} (i*j*k-1).

%F a(n) = A000217(n)^3 - A000578(n), n > 0.

%F (End)

%p A193575:=n->n^3*(n^3+3*n^2+3*n-7)/8: seq(A193575(n), n=1..40); # _Wesley Ivan Hurt_, Aug 23 2014

%t Table[n^3*(n^3 + 3*n^2 + 3*n - 7)/8, {n, 40}] (* _Wesley Ivan Hurt_, Aug 23 2014 *)

%t CoefficientList[Series[x*(19 + 56 x + 12 x^2 + 2 x^3 + x^4)/(1 - x)^7, {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Aug 23 2014 *)

%t LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,19,189,936,3250,9045,21609},40] (* _Harvey P. Dale_, Oct 24 2020 *)

%o (Magma) [(n^3*(n^3+3*n^2+3*n-7)/8): n in [1..40]]

%Y Cf. A000217, A000578.

%K nonn,easy

%O 1,2

%A _Vincenzo Librandi_, Sep 08 2011