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%I #30 Dec 14 2023 09:04:05
%S 1,2,3,5,11,35,155,875,5915,46235,409115,4037915,43954715,522956315,
%T 6749977115,93928268315,1401602636315,22324392524315,378011820620315,
%U 6780385526348315,128425485935180315,2561327494111820315
%N a(1) = 1, a(2) = 2, a(n) = (n-1)*a(n-1) - (n-2)*a(n-2).
%C Successive differences are factorials, or (n+1)st successive difference divided by n-th successive difference = n. I.e., {a(n+2)-a(n+1)}/{a(n+1)-a(n)} = n. - _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 14 2003
%C Equals the row sums of A165680. - _Johannes W. Meijer_, Oct 16 2009
%H Reinhard Zumkeller, <a href="/A067078/b067078.txt">Table of n, a(n) for n = 1..100</a>
%F a(n) = 1 + Sum_{i=0..n-2} i! = 2*A014288(n-1)+1 = A007489(n-2)+2 (n>1). - _Henry Bottomley_, Oct 23 2002; corrected by _M. F. Hasler_, Dec 16 2007
%F a(n) = 1+!(n-1) = 1+A003422(n-1); a(n+1)=a(n)+(n-1)!. - _M. F. Hasler_, Dec 16 2007
%F E.g.f.: A(x)=x*B(x) satisfies the differential equation B'(x)=B(x)+log(1/(1-x))+1. - _Vladimir Kruchinin_, Jan 19 2011
%e a(6) = 35, a(5)= 11 hence a(7) = 6*35 - 5*11 = 155.
%t a[1] = 1; a[2] = 2; a[n_] := a[n] = (n - 1)*a[n - 1] - (n - 2)*a[n - 2]; Table[ a[n], {n, 1, 25} ]
%t a=FoldList[Plus,2,(Range@40)! ];PrependTo[a,1] (* _Vladimir Joseph Stephan Orlovsky_, May 21 2010 *)
%o (PARI) A067078(n)=sum(k=0, n-2, k!, 1) \\ _M. F. Hasler_, Dec 16 2007
%o (Haskell)
%o a067078 n = a067078_list !! (n-1)
%o a067078_list = scanl (+) 1 a000142_list
%o -- _Reinhard Zumkeller_, Dec 27 2011
%Y Cf. A003422, A014288, A007489, A165680.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Jan 05 2002
%E More terms from _Robert G. Wilson v_, Jan 07 2002
%E Edited by _M. F. Hasler_, Dec 16 2007
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