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%I #90 Jan 21 2023 03:24:37
%S 0,5,28,81,176,325,540,833,1216,1701,2300,3025,3888,4901,6076,7425,
%T 8960,10693,12636,14801,17200,19845,22748,25921,29376,33125,37180,
%U 41553,46256,51301,56700,62465,68608,75141,82076,89425,97200,105413,114076,123201,132800,142885
%N a(n) = 2*n^3 + 3*n^2.
%C Apart from the initial zero this sequence gives the 2nd pentagonal number, the 4th hexagonal number, the 6th heptagonal number, the 8th octagonal number, the 10th nonagonal number, etc. as well as the 5th nonnegative number, the 7th triangular number, the 9th square, the 11th pentagonal number, the 13th hexagonal number, etc. This is a reliable pattern that does not seem to appear on any other pairs of polygonal numbers (see link).
%C a(n) is the maximal determinant of a 3 X 3 matrix with integer elements from {1, ..., n+1}, so (for example) the maximum determinant of a 3 X 3 matrix with integer elements from {1, ..., 5} = det(1, 5, 5; 5, 1, 5; 5, 5, 1) = a(4) = 176. - _Matthew Scroggs_, Dec 31 2022
%H Carauleanu Marc and Colin Barker, <a href="/A275709/b275709.txt">Table of n, a(n) for n = 0..3030</a> (first 1000 terms from Colin Barker)
%H Joshua Giambalvo, <a href="http://imgur.com/gallery/2BeBR">Illustration of initial terms in a square array</a>, Imgur, (2016).
%H Petro Kolosov, <a href="https://arxiv.org/abs/1603.02468">Another power identity involving binomial theorem and Faulhaber's formula</a>, arXiv:1603.02468 [math.NT], 2018. (In particular p. 3)
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F From _Colin Barker_, Aug 06 2016: (Start)
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4.
%F G.f.: x*(5+8*x-x^2) / (1-x)^4.
%F (End)
%F a(n) = A033431(n) + A033428(n). - _Omar E. Pol_, Aug 09 2016
%F a(n) = A000290(n) * A005408(n+1). - _Robert Israel_, Aug 09 2016
%F a(n) = A320047(1, n, 0). - _Kolosov Petro_, Oct 04 2018
%F E.g.f.: x*(5 + 9*x + 2*x^2)*exp(x). - _G. C. Greubel_, Oct 19 2018
%F From _Amiram Eldar_, Jan 21 2023: (Start)
%F Sum_{n>=1} 1/a(n) = Pi^2/18 + 4*log(2)/9 - 16/27.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/36 + Pi/9 -2*log(2)/9 - 8/27. (End)
%p seq(2*n^3+3*n^2, n=0..30); # _Robert Israel_, Aug 09 2016
%t Table[2 n^3 + 3 n^2, {n, 0, 41}] (* or *)
%t CoefficientList[Series[x (5 + 8 x - x^2)/(1 - x)^4, {x, 0, 41}], x] (* _Michael De Vlieger_, Aug 11 2016 *)
%o (PARI) concat(0, Vec(x*(5+8*x-x^2)/(1-x)^4 + O(x^50))) \\ _Colin Barker_, Aug 28 2016
%o (PARI) a(n)=n^2*(2*n+3) \\ _Charles R Greathouse IV_, Aug 28 2016
%o (Magma) [n^2*(2*n + 3): n in [0..30]]; // _G. C. Greubel_, Oct 19 2018
%o (Python) for n in range(0,50): print(n**2*(2*n+3), end=' ') # _Stefano Spezia_, Oct 19 2018
%Y Cf. A000290, A005408, A005563, A033428, A033431, A057145, A139600.
%K nonn,easy
%O 0,2
%A _Joshua Giambalvo_, Aug 06 2016
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