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a(n) = n*(a(n-1)-4), a(0) = 1.
1

%I #19 May 23 2016 02:47:41

%S 1,-3,-14,-54,-232,-1180,-7104,-49756,-398080,-3582756,-35827600,

%T -394103644,-4729243776,-61480169140,-860722368016,-12910835520300,

%U -206573368324864,-3511747261522756,-63211450707409680

%N a(n) = n*(a(n-1)-4), a(0) = 1.

%H Vincenzo Librandi, <a href="/A166680/b166680.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = n! - 4*floor(e*n!) + 4, n>0. - _Gary Detlefs_, Jun 19 2010

%t RecurrenceTable[{a[0]==1,a[n]==n(a[n-1]-4)},a,{n,20}] (* _Harvey P. Dale_, Apr 18 2014 *)

%t FoldList[#1*#2 - 4 #2 &, 1, Range[50]] (* _G. C. Greubel_, May 23 2016 *)

%Y Cf. A000142, A166554, A165793, A165814.

%K easy,sign

%O 0,2

%A _Philippe Deléham_, Oct 18 2009