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A096000
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Cupolar numbers: a(n) = (n+1)*(5*n^2+7*n+3)/3.
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7
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1, 10, 37, 92, 185, 326, 525, 792, 1137, 1570, 2101, 2740, 3497, 4382, 5405, 6576, 7905, 9402, 11077, 12940, 15001, 17270, 19757, 22472, 25425, 28626, 32085, 35812, 39817, 44110, 48701, 53600, 58817, 64362, 70245, 76476, 83065, 90022, 97357, 105080, 113201, 121730
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OFFSET
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0,2
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COMMENTS
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Number of equal balls that will fill a triangular cupola, formed by splitting a cuboctahedron along one of its four "equilateral" hexagons.
Also as a(n)=(1/6)*(10*n^3-6*n^2+10*n), n>0: structured pentagonal anti-prism numbers (Cf. A100185 = structured anti-prisms); and structured tetragonal anti-diamond numbers (vertex structure 7) (Cf. A000447 = alternate vertex; A100188 = structured anti-diamonds). Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
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REFERENCES
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H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
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LINKS
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FORMULA
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a(n) = (1/2)*(Q(n) + 3n^2 + 3n + 1), where Q(n) are the cuboctahedral numbers, A005902.
G.f.: (1+6*x+3*x^2)/(1-x)^4. - Paul Barry, Oct 28 2006
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MAPLE
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MATHEMATICA
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CoefficientList[Series[(1 + 6 x + 3 x^2)/(1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, May 23 2015 *)
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PROG
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(PARI) a(n) = (1/3)*(n+1)*(5*n^2+7*n+3) \\ Michel Marcus, Jul 11 2013
(Magma) [(n+1)*(5*n^2+7*n+3)/3 : n in [0..50]]; // Wesley Ivan Hurt, May 23 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane, in memory of Harold Scott MacDonald Coxeter [Feb 09 1907 - Mar 31 2003], May 08 2004
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STATUS
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approved
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