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%I #25 Sep 08 2022 08:45:48
%S 1,7,29,90,232,524,1072,2030,3613,6111,9905,15484,23464,34608,49848,
%T 70308,97329,132495,177661,234982,306944,396396,506584,641186,804349,
%U 1000727,1235521,1514520,1844144,2231488,2684368,3211368
%N Number of 3 X 3 X 3 triangular nonnegative integer arrays with all sums of an element and its neighbors <= n.
%C a(n) gives the number of hexagons that have vertices at the lattice points and sides on lattice lines of a triangular lattice with sides n+3. Note that the hexagons can be non-regular. This problem appeared as ConvexHexagons in Single Round Match 455 in TopCoder. - _Dmitry Kamenetsky_, Dec 17 2009
%H G. C. Greubel, <a href="/A166189/b166189.txt">Table of n, a(n) for n = 0..1000</a>
%H Jun-Ming Zhu, <a href="http://arxiv.org/abs/1012.4058">The number of convex pentagons and hexagons in an n-triangular net</a>, arXiv:1012.4058 [math.CO], 2010; See H(n), formula 3, on page 4.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,14,0,-14,14,-6,1).
%F From _G. C. Greubel_, Jul 02 2021: (Start)
%F a(n) = (1/1920)*(4*n^6 +72*n^5 +530*n^4 +2040*n^3 +4296*n^2 +4608*n +1905 +15*(-1)^n).
%F a(2*n+1) = (1/15)*binomial(n+2, 2)*binomial(n+3, 2)*(8*n^2 + 32*n + 35).
%F a(2*n) = (1/30)*binomial(n+2, 2)*(8*n^4 + 48*n^3 + 105*n^2 + 99*n + 30).
%F G.f.: (1 - x^3)/((1-x^2)*(1-x)^7).
%F E.g.f.: (1/1920)*((1905 +2*x*(5775 +7665*x +3690*x^2 +755*x^3 +66*x^4 +2*x^5))*exp(x) + 15*exp(-x)). (End)
%F a(n) = A001779(n)-A001779(n-3). - _R. J. Mathar_, Jul 04 2021
%t LinearRecurrence[{6,-14,14,0,-14,14,-6,1}, {1,7,29,90,232,524,1072,2030}, 51] (* _G. C. Greubel_, Jul 02 2021 *)
%o (Magma)
%o R<x>:=PowerSeriesRing(Integers(), 50);
%o Coefficients(R!( (1-x^3)/((1-x^2)*(1-x)^7) )); // _G. C. Greubel_, Jul 02 2021
%o (Sage)
%o def a(n): return (n+2)*(n+4)*(2*n^4 +24*n^3 +105*n^2 +198*n +120)/960 if (n%2==0) else (n+1)*(n+3)^2*(n+5)*(2*n*(n+6) +21)/960
%o [a(n) for n in (0..50)] # _G. C. Greubel_, Jul 02 2021
%o (PARI) \\ using Zhu expressions
%o f(k) = (8*k^6 + 24*k^5 + 25*k^4 + 10*k^3 - 3*k^2 -4*k)/60;
%o g(k) = (8*k^6 - 5*k^4 - 3*k^2)/60;
%o a(n) = n+=3; if (n%2, f((n-1)/2), g(n/2)); \\ _Michel Marcus_, Jul 04 2021
%Y Cf. A176646, A011888 (first differences).
%K nonn,easy
%O 0,2
%A _R. H. Hardin_, Oct 09 2009
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