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, 4, 11, 41, 199, 1184, 8273, 66163, 595439, 5954354, 65497849, 785974133, 10217663663, 143047291204, 2145709367969, 34331349887399, 583632948085663, 10505393065541798, 199602468245294009, 3992049364905880009, 83833036663023479999

Offset

1,1

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..200

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

Adjacent sequences: A067350 A067351 A067352 * A067354 A067355 A067356

Keyword

easy,nonn

Author

Floor van Lamoen, Jan 17 2002

Extensions

More terms from Vincenzo Librandi, May 06 2013

Status

approved