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 A113861 a(n) = (1/9)*((6*n - 7)*2^(n-1) - (-1)^n). 4
 0, 1, 5, 15, 41, 103, 249, 583, 1337, 3015, 6713, 14791, 32313, 70087, 151097, 324039, 691769, 1470919, 3116601, 6582727, 13864505, 29127111, 61050425, 127693255, 266571321, 555512263, 1155763769, 2401006023, 4980969017, 10319851975, 21355531833, 44142719431 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS This sequence is connected with the Collatz problem (see the sequences A045883 and A001045). - Michel Lagneau, Jan 13 2012 LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 T. Etzion, On the stopping redundancy of Reed-Muller codes, IEEE Trans. Information Theory 52 (11) (2006) 4867-4879; arXiv:cs.IT/0511056. Index entries for linear recurrences with constant coefficients, signature (3,0,-4). FORMULA a(n+1) - 2*a(n) = A001045(n+2), Jacobsthal numbers. - Paul Curtz, Jul 05 2008 3*a(n) - a(n+1) = -1, -2, 4*a(n). - Paul Curtz, Jul 05 2008 From R. J. Mathar, Nov 11 2008: (Start) G.f.: x^2*(1+2*x)/((1+x)*(1-2*x)^2). a(n) + a(n+1) = A014480(n-1). (End) a(n) = 4*a(n-1) - 4*a(n-2) + (-1)^(n+1), n>2. - Gary Detlefs, Dec 19 2010 a(n) = 3*a(n-1) - 4*a(n-3), n>3. - Gary Detlefs, Dec 19 2010 a(n) = n*2^n - A045883(n). - Michel Lagneau, Jan 13 2012 Starting with "1" = triangle A059260 * A016813 as a vector, where A016813 = (4n + 1): [ 1, 5, 9, 13,...]. - Gary W. Adamson, Mar 06 2012 MATHEMATICA Join[{0}, Numerator[CoefficientList[Series[(x+1)/((x-1)*(x^2+x-2)), {x, 0, 40}], x]]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *) PROG (PARI) a(n)=((6*n-7)<<(n-1)-(-1)^n)/9 \\ Charles R Greathouse IV, Jan 13 2012 CROSSREFS Cf. A102301. Cf. A059260, A016813. Sequence in context: A230955 A038066 A306159 * A080870 A288414 A102620 Adjacent sequences:  A113858 A113859 A113860 * A113862 A113863 A113864 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Jan 25 2006 STATUS approved

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Last modified January 23 04:16 EST 2020. Contains 331168 sequences. (Running on oeis4.)