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A004068
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Number of atoms in a decahedron with n shells.
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29
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0, 1, 7, 23, 54, 105, 181, 287, 428, 609, 835, 1111, 1442, 1833, 2289, 2815, 3416, 4097, 4863, 5719, 6670, 7721, 8877, 10143, 11524, 13025, 14651, 16407, 18298, 20329, 22505, 24831, 27312, 29953, 32759, 35735, 38886, 42217, 45733, 49439
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OFFSET
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0,3
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COMMENTS
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Also as a(n)=(n/6)*(5*n^2+1), n>0: structured pentagonal diamond numbers (vertex structure 6) (cf. A081436 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of atoms in decahedron with n shells, number = 5/6*(n^3) + 1/6*(n) (T. P. Martin, Shells of atoms, eq.(3)). - Brigitte Stepanov, Jul 02 2011
a(n+1) is the number of triples (w,x,y) having all terms in {0,...,n} and x+y >= w. - Clark Kimberling, Jun 14 2012
a(n) is also a figurate number representing a cube of side n with a vertex cut off by a tetrahedron of side n-1. As such, a(n) = A000578(n) - A000292(n-1), n > 0. - Jean M. Morales, Aug 11 2013
The sequence starting with 1 is the third partial sum of (1, 4, 5, 5, 5, ...) and the binomial transform of (1, 6, 10, 5, 0, 0, 0, ...). - Gary W. Adamson, Sep 27 2015
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LINKS
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T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (3).
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FORMULA
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a(n) = 5*binomial(n + 1, 3) + binomial(n, 1).
a(n) = 5*n^3/6 + n/6.
a(n) = Sum_{i=0..n-1} A005891(i). - Xavier Acloque, Oct 08 2003
E.g.f.: (x/6)*(5x^2 + 15x + 6)*exp(x). - G. C. Greubel, Sep 27 2015
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i/sqrt(5)) + polygamma(0, 1+i/sqrt(5)) = 1.233988011257952852492845364799197179252... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023
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MATHEMATICA
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PROG
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CROSSREFS
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(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
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KEYWORD
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nonn,easy
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AUTHOR
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Albert D. Rich (Albert_Rich(AT)msn.com)
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EXTENSIONS
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STATUS
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approved
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