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A191662
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a(n) = n! / A000034(n-1).
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4
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1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000
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OFFSET
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1,3
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COMMENTS
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The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1 = 1, 3, 5, 7, 9, ... (k=1)
A000290 = (n^2)/1 = 1, 4, 9, 16, 25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6 = 1, 5, 14, 30, 55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12 = 1, 6, 20, 50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27, 77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360 = 1, 8, 35, 112, 294, ... (k=6)
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011
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LINKS
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FORMULA
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a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
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MAPLE
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MATHEMATICA
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CROSSREFS
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Cf. A000290, A000330, A002415, A005408, A005585, A029651, A040977, A050486, A053347, A054333, A054334, A057788.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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