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a(n) = 9*binomial(n,4) = 3n*(n-1)*(n-2)*(n-3)/8.
5

%I #27 Jul 19 2022 05:46:55

%S 0,0,0,0,9,45,135,315,630,1134,1890,2970,4455,6435,9009,12285,16380,

%T 21420,27540,34884,43605,53865,65835,79695,95634,113850,134550,157950,

%U 184275,213759,246645,283185,323640,368280,417384,471240,530145,594405

%N a(n) = 9*binomial(n,4) = 3n*(n-1)*(n-2)*(n-3)/8.

%C Number of permutations of n letters where exactly four change position.

%H Harry J. Smith, <a href="/A060008/b060008.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F Equals 3*A050534. - _Zerinvary Lajos_, Feb 12 2007

%F G.f.: 9*x^4/(1-x)^5. - _Colin Barker_, Jul 02 2012

%F From _Amiram Eldar_, Jul 19 2022: (Start)

%F Sum_{n>=4} 1/a(n) = 4/27.

%F Sum_{n>=4} (-1)^n/a(n) = 32*log(2)/9 - 64/27. (End)

%e 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.

%t 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 *)

%o (PARI) { for (n=0, 1000, write("b060008.txt", n, " ", 3*n*(n - 1)*(n - 2)*(n - 3)/8); ) } \\ _Harry J. Smith_, Jul 01 2009

%Y 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.

%Y A diagonal of A008291.

%K easy,nonn

%O 0,5

%A _Henry Bottomley_, Mar 16 2001