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A008577 Crystal ball sequence for planar net 4.8.8. 4

%I #58 Sep 30 2021 01:30:14

%S 1,4,9,17,28,41,57,76,97,121,148,177,209,244,281,321,364,409,457,508,

%T 561,617,676,737,801,868,937,1009,1084,1161,1241,1324,1409,1497,1588,

%U 1681,1777,1876,1977,2081,2188,2297,2409,2524,2641,2761,2884,3009,3137,3268

%N Crystal ball sequence for planar net 4.8.8.

%H T. D. Noe, <a href="/A008577/b008577.txt">Table of n, a(n) for n = 0..1000</a>

%H Brian Galebach, <a href="http://probabilitysports.com/tilings.html?u=0&amp;n=1&amp;t=2">Uniform Tiling 2 of 11</a>

%H W. M. Meier and H. J. Moeck, <a href="http://dx.doi.org/10.1016/0022-4596(79)90177-4">Topology of 3-D 4-connected nets ...</a>, J. Solid State Chem 27 1979 349-355, esp. p. 351.

%H <a href="/index/Cor#crystal_ball">Index entries for crystal ball sequences</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).

%F G.f.: ((1+x)^2*(1+x^2)) / ((1-x)^3*(1+x+x^2)). - _Ralf Stephan_, Apr 24 2004

%F a(n) = 4*(n/3)*(n+1)+10/9+A099837(n+2)/9. - _R. J. Mathar_, Nov 20 2010

%F The above g.f. and formula were originally stated as conjectures, but I now have a proof. This also justifies the b-file. Details will be added later. - _N. J. A. Sloane_, Dec 29 2015

%F From _Michael Somos_, May 02 2020: (Start)

%F Euler transform of length 3 sequence [4, -1, 1, -1].

%F a(n) = a(-1-n) = floor((n^2+n+1)*4/3) for all n in Z.

%F a(n) - 2*a(n+1) + a(n+2) = A164359(n) unless n=0.

%F (End)

%e G.f. = 1 + 4*x + 9*x^2 + 17*x^3 + 28*x^4 + 41*x^5 + 67*x^6 + ... - _Michael Somos_, May 02 2020

%t A099837[0] = 1; A099837[n_] := Mod[n+2, 3] - Mod[n, 3]; a[n_] := 4*n/3*(n+1) + 10/9 + A099837[n+2]/9; Table[a[n], {n, 0, 39}] (* _Jean-François Alcover_, Feb 15 2012, after _R. J. Mathar_ *)

%t CoefficientList[Series[((1 + x)^2 (1 + x^2))/((1 - x)^3 (1 + x + x^2)), {x, 0, 50}], x] (* _Vincenzo Librandi_, Dec 31 2015 *)

%t LinearRecurrence[{2,-1,1,-2,1},{1,4,9,17,28},40] (* _Harvey P. Dale_, Dec 17 2017 *)

%t a[ n_] := Quotient[(n^2 + n + 1)*4, 3]; (* _Michael Somos_, May 02 2020 *)

%o (PARI) {a(n) = (n^2 + n + 1)*4\3)}; /* _Michael Somos_, May 02 2020 */

%Y Partial sums of A008576.

%Y Cf. A099837, A164359.

%K nonn,easy,nice

%O 0,2

%A _N. J. A. Sloane_

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