A176646 a(n) is the number of convex pentagons in an n-triangular net.
0, 0, 3, 21, 78, 216, 498, 1014, 1884, 3264, 5349, 8379, 12642, 18480, 26292, 36540, 49752, 66528, 87543, 113553, 145398, 184008, 230406, 285714, 351156, 428064, 517881, 622167, 742602, 880992, 1039272, 1219512, 1423920, 1654848, 1914795, 2206413, 2532510, 2896056
Offset: 1
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Jun-Ming Zhu, The number of convex pentagons and hexagons in an n-triangular net, arXiv:1012.4058 [math.CO], 2010. See P(n), formula (1), page 2.
- Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1).
Crossrefs
Cf. A166189 (for the hexagons).
Programs
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Magma
[(1/320)*(12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n): n in [1..40]]; // G. C. Greubel, Jul 02 2021
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Maple
A176646:= n-> (12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n)/320; seq(A176646(n), n=1..40); # R. J. Mathar, Dec 21 2010
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Mathematica
LinearRecurrence[{5,-9,5,5,-9,5,-1}, {0,0,3,21,78,216,498}, 40] (* Harvey P. Dale, Jan 14 2015 *) -
PARI
f(k) = (12*k^5 + 25*k^4 + 5*k^3 - 10*k^2 - 2*k)/10; g(k) = (12*k^5 - 5*k^4 - 15*k^3 + 5*k^2 + 3*k)/10; a(n) = if (n%2, f((n-1)/2), g(n/2)); \\ Michel Marcus, Jul 04 2021
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Sage
[(1/320)*(12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n) for n in (1..40)] # G. C. Greubel, Jul 02 2021
Formula
From G. C. Greubel, Jul 03 2021: (Start)
a(n) = (1/320)*(12*n^5 - 10*n^4 - 60*n^3 + 40*n^2 + 48*n - 15 + 15*(-1)^n).
a(2*n+1) = n*(n+1)*(12*n^3 + 13*n^2 - 8*n - 2)/10.
a(2*n) = n*(4*n-3)*(3*n+1)*(n-1)*(n+1)/10.
G.f.: 3*x^3*(1 + 2*x)/((1 + x)*(1 - x)^6).
E.g.f.: (1/320)*(15*exp(-x) - (15 -30*x +30*x^2 -180*x^3 -110*x^4 -12*x^5)*exp(x)). (End)
Extensions
Definition corrected and edited by Michel Marcus and G. C. Greubel, Jul 03 2021
Comments