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A067353
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).
2
2, 4, 11, 41, 199, 1184, 8273, 66163, 595439, 5954354, 65497849, 785974133, 10217663663, 143047291204, 2145709367969, 34331349887399, 583632948085663, 10505393065541798, 199602468245294009, 3992049364905880009, 83833036663023479999
OFFSET
1,1
LINKS
FORMULA
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.
E.g.f. (with a(0)=2): (exp(x)*(x^2-2*x+2)-6)/(2*(x-1)). - Vaclav Kotesovec, Oct 21 2012
EXAMPLE
The sets begin {1,2},{3,4},{5,6,...,9,10,11},{12,13,...,38,39,40,41},...
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Sequence in context: A094762 A295772 A099934 * A105996 A213937 A328437
KEYWORD
easy,nonn
AUTHOR
Floor van Lamoen, Jan 17 2002
EXTENSIONS
More terms from Vincenzo Librandi, May 06 2013
STATUS
approved