%I #10 Mar 02 2015 16:08:48

%S 2,4,11,41,199,1184,8273,66163,595439,5954354,65497849,785974133,

%T 10217663663,143047291204,2145709367969,34331349887399,

%U 583632948085663,10505393065541798,199602468245294009,3992049364905880009,83833036663023479999

%N Divide the natural numbers in sets of consecutive numbers starting with {1,2} as the first set. The number of elements of the n-th set is equal to the sum of the n-1 final numbers in the (n-1)st set. The final number of the n-th set gives a(n).

%H Vincenzo Librandi, <a href="/A067353/b067353.txt">Table of n, a(n) for n = 1..200</a>

%F a(n)=n*a(n-1)-(n-1)(n-2)/2 with a(1)=2. a(n)=b(1)+b(2)+...+b(n) with b(n) as in A067352.

%F E.g.f. (with a(0)=2): (exp(x)*(x^2-2*x+2)-6)/(2*(x-1)). - _Vaclav Kotesovec_, Oct 21 2012

%e The sets begin {1,2},{3,4},{5,6,...,9,10,11},{12,13,...,38,39,40,41},...

%t Rest[CoefficientList[Series[(E^x*(x^2-2*x+2)-6)/(2*(x-1)),{x,0,20}],x]*Range[0,20]!] (* _Vaclav Kotesovec_, Oct 21 2012 *)

%o (PARI) x='x+O('x^66); Vec(serlaplace((exp(x)*(x^2-2*x+2)-6)/(2*(x-1))) -2 ) \\ _Joerg Arndt_, May 06 2013

%K easy,nonn

%O 1,1

%A _Floor van Lamoen_, Jan 17 2002

%E More terms from _Vincenzo Librandi_, May 06 2013