%I
%S 0,3,7,13,19,27,39,53,63,71,83,99,119,147,183,217,235,243,255,271,291,
%T 319,355,391,419,447,487,539,607,699,803,885,919,927,939,955,975,1003,
%U 1039,1075,1103,1131,1171,1223,1291,1383,1487,1571,1615
%N a(n) = number of grid points that are covered after nth stage of A139250, assuming the toothpicks have length 2.
%C a(n) is also the number of "ON" cells at nth stage in simple 2dimensional cellular automaton whose virtual skeleton is a polyedge as the toothpick structure of A139250. [From _Omar E. Pol_, May 18 2009]
%C It appears that the number of grid points that are covered after nth stage of A139250, assuming the toothpicks have length 2*k, is equal to (2*k2) * A139250(n) + a(n), k>0. See formulas in A160420 and A160422. [From _Omar E. Pol_, Nov 15 2010]
%C More generally, it appears that a(n) is also the number of grid points that are covered by the "special points" of the toothpicks of A139250, after nth stage, assuming the toothpicks have length 2*k, k>0 and that each toothpick has three special points: the midpoint and two endpoints.
%C Note that if k>1 then also there are 2*k2 grid points that are covered by each toothpick, but these points are not considered for this sequence. [From _Omar E. Pol_, Nov 15 2010]
%C Contribution from _Omar E. Pol_, Sep 16 2012 (Start):
%C It appears that a(n)/A139250(n) converge to 4/3.
%C It appears that a(n)/A160124(n) converge to 2.
%C It appears that a(n)/A139252(n) converge to 4.
%C (End)
%H David Applegate, <a href="/A147614/b147614.txt">Table of n, a(n) for n = 0..10135</a>
%H David Applegate, Omar E. Pol and N. J. A. Sloane, <a href="http://neilsloane.com/doc/tooth.pdf">The Toothpick Sequence and Other Sequences from Cellular Automata</a>, Congressus Numerantium, Vol. 206 (2010), 157191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n1)1) for n >= 2.]
%H N. J. A. Sloane, <a href="/wiki/Catalog_of_Toothpick_and_CA_Sequences_in_OEIS">Catalog of Toothpick and Cellular Automata Sequences in the OEIS</a>
%F Since A160124(n) = 1+2*A139250(n)A147614(n), n>0 (see A160124), and we have recurrences for A160125 (hence A160124) and A139250, we have a recurrence for this sequence.  _N. J. A. Sloane_, Feb 02 2010
%F a(n) = A187220(n+1)A160124(n), n>0.  _Omar E. Pol_, Feb 15 2013
%Y Cf. A139250, A139251, A139252, A160124, A160420, A160422, A170884.
%K nonn,look
%O 0,2
%A _David Applegate_, Apr 29 2009
