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A060008
a(n) = 9*binomial(n,4) = 3*n*(n-1)*(n-2)*(n-3)/8.
5
0, 0, 0, 0, 9, 45, 135, 315, 630, 1134, 1890, 2970, 4455, 6435, 9009, 12285, 16380, 21420, 27540, 34884, 43605, 53865, 65835, 79695, 95634, 113850, 134550, 157950, 184275, 213759, 246645, 283185, 323640, 368280, 417384, 471240, 530145, 594405, 664335, 740259
OFFSET
0,5
COMMENTS
Number of permutations of n letters where exactly four change position.
FORMULA
Equals 3*A050534. - Zerinvary Lajos, Feb 12 2007
G.f.: 9*x^4/(1-x)^5. - Colin Barker, Jul 02 2012
From Amiram Eldar, Jul 19 2022: (Start)
Sum_{n>=4} 1/a(n) = 4/27.
Sum_{n>=4} (-1)^n/a(n) = 32*log(2)/9 - 64/27. (End)
From Elmo R. Oliveira, Nov 28 2025: (Start)
E.g.f.: 3*x^4*exp(x)/8.
a(n) = 9*A000332(n).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
EXAMPLE
a(6) = 135 since there are 15 ways to choose the four points that move and 9 ways to move them and 15*9 = 135.
MATHEMATICA
9*Binomial[Range[0, 40], 4] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 0, 0, 9}, 40] (* Harvey P. Dale, Jun 09 2014 *)
PROG
(PARI) a(n) = { 3*n*(n - 1)*(n - 2)*(n - 3)/8 } \\ Harry J. Smith, Jul 01 2009
CROSSREFS
For changing 0, 1, 2, 3, 4, 5, n-4, n elements see A000012, A000004, A000217 (offset), A007290, A060008, A060836, A000475, A000166. Also see A000332, A008290.
A diagonal of A008291.
Cf. A050534.
Sequence in context: A188351 A220443 A289721 * A212089 A212142 A095166
KEYWORD
easy,nonn
AUTHOR
Henry Bottomley, Mar 16 2001
STATUS
approved