OFFSET
0,3
COMMENTS
From Omar E. Pol, Oct 01 2011: (Start)
On the infinite square grid, consider only the first three quadrants and count only the toothpicks of length 2.
At stage 0, we start from a vertical half toothpick at [(0,0),(0,1)]. This half toothpick represents one of the two components of the first toothpick placed in the toothpick structure of A139250, so a(0) = 0.
At stage 1, we place an orthogonal toothpick of length 2 centered at the end, so a(1) = 1. Also we place half toothpick at [(0,-1),(1,-1)]. This last half toothpick represents one of the two components of the third toothpick placed in the toothpick structure of A139250.
At stage 2, we place three toothpicks, so a(2) = 1+3 = 4.
In each subsequent stage, for every exposed toothpick end, place an orthogonal toothpick centered at that end.
The sequence gives the number of toothpicks after n stages. A153004 (the first differences) gives the number of toothpicks added to the structure at n-th stage.
LINKS
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, 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.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
FORMULA
MATHEMATICA
A139250[n_] := A139250[n] = Module[{m, k}, If[n == 0, Return[0]]; m = 2^(Length[IntegerDigits[n, 2]] - 1); k = (2 m^2 + 1)/3; If[n == m, k, k + 2 A139250[n - m] + A139250[n - m + 1] - 1]];
a[n_] := If[n == 0, 0, (3/4)(A139250[n + 1] - 3) + 1];
a /@ Range[0, 49] (* Jean-François Alcover, Apr 06 2020 *)
PROG
(Python)
def msb(n):
t=0
while n>>t>0: t+=1
return 2**(t - 1)
def a139250(n):
k=(2*msb(n)**2 + 1)/3
return 0 if n==0 else k if n==msb(n) else k + 2*a139250(n - msb(n)) + a139250(n - msb(n) + 1) - 1
def a(n): return 0 if n==0 else (a139250(n + 1) - 3)*3/4 + 1
[a(n) for n in range(51)] # Indranil Ghosh, Jul 01 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Omar E. Pol, Jan 02 2009
STATUS
approved