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A191662
a(n) = n! / A000034(n-1).
4
1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000
OFFSET
1,3
COMMENTS
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)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
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
FORMULA
a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
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
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023
MAPLE
A191662:= proc(n): n!/A000034(n-1) end: A000034 := proc(n) op((n mod 2)+1, [1, 2]) ; end proc: seq(A191662(n), n=1..17); # Johannes W. Meijer, Jun 22 2011
MATHEMATICA
Array[If[EvenQ[#], #!/2, #!]&, 20] (* Harvey P. Dale, Mar 14 2014 *)
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Jun 10 2011
EXTENSIONS
More terms from Harvey P. Dale, Mar 14 2014
STATUS
approved