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%I #25 Feb 24 2021 02:48:18
%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 n-th stage of A139250, assuming the toothpicks have length 2.
%C a(n) is also the number of "ON" cells at n-th stage in simple 2-dimensional 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 n-th stage of A139250, assuming the toothpicks have length 2*k, is equal to (2*k-2) * 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 n-th 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*k-2 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="/A000695/a000695_1.pdf">The Toothpick Sequence and Other Sequences from Cellular Automata</a>, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-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
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