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A080924
Jacobsthal gap sequence.
6
0, 1, 3, 1, 15, 1, 63, 1, 255, 1, 1023, 1, 4095, 1, 16383, 1, 65535, 1, 262143, 1, 1048575, 1, 4194303, 1, 16777215, 1, 67108863, 1, 268435455, 1, 1073741823, 1, 4294967295, 1, 17179869183, 1, 68719476735, 1, 274877906943, 1, 1099511627775, 1
OFFSET
0,3
COMMENTS
Inverse binomial transform of A080925
From Peter Bala, Dec 26 2012: (Start)
Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(1/4) = 0.79761 68651 30459 16010 ... = 1/(1 + 1/(3 + 1/(1 + 1/(15 + 1/(1 + 1/(63 + 1/(1 + 1/(255 + ...)))))))). See A111317. (End)
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 3", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Apr 19 2017
REFERENCES
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.
FORMULA
a(2n) = 3*A001045(2n) = 3*A002450(n) = 4^n-1, a(2n+1)=1.
a(n) = (2^n-2*(-1)^n+(-2)^n)/2.
G.f.: x*(1+4*x)/((1+x)*(1+2*x)*(1-2*x)).
E.g.f.: (exp(2*x)-2*exp(-x)+exp(-2*x))/2.
MATHEMATICA
CoefficientList[Series[x (1 + 4 x) / ((1 + x) (1 + 2 x) (1 - 2 x)), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 05 2013 *)
LinearRecurrence[{-1, 4, 4}, {0, 1, 3}, 42] (* Jean-François Alcover, Sep 21 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Feb 26 2003
STATUS
approved