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 A191010 a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5. 1
 1, 7, 41, 215, 1065, 5079, 23593, 107479, 482345, 2139095, 9395241, 40936407, 177167401, 762356695, 3264175145, 13915694039, 59098749993, 250138895319, 1055531162665, 4442026976215, 18647717207081, 78109306037207, 326510972984361, 1362338887279575 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) = 4^(n+1)*H(2^n)/5 with H(2^n) = n+(6+(-1)^n/4^(n+1))/5 = E(N(2^n)), where X, X(1), X(2),... denote random variables with pdf P(X = 1) = P(X = 4) = 1/5 and P(X = 2) = 3/5, N(x) is the first value of k such that X(1)*X(2)*...*X(k) > x and H(x)= E(N(x)). LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (7,-8,-16). FORMULA a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5. From Colin Barker, May 03 2017: (Start) G.f.: 1 / ((1 + x)*(1 - 4*x)^2). a(n) = 7*a(n-1) - 8*a(n-2) - 16*a(n-3) for n>2. (End) E.g.f.: (80*x*exp(4*x)+24*exp(4*x)+exp(-x))/25. - Robert Israel, May 03 2017 MAPLE seq((n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5, n=0..50); # Robert Israel, May 03 2017 MATHEMATICA CoefficientList[Series[1/((1 + x) (1 - 4 x)^2), {x, 0, 23}], x] (* or *) LinearRecurrence[{7, -8, -16}, {1, 7, 41}, 24] (* Michael De Vlieger, May 03 2017 *) PROG (PARI) a(n)= (n*4^(n+1)+(6*4^(n+1)+(-1)^n)/5)/5; \\ Michel Marcus, Oct 16 2014 (PARI) Vec(1 / ((1 + x)*(1 - 4*x)^2) + O(x^30)) \\ Colin Barker, May 03 2017 CROSSREFS Cf. A191008. Sequence in context: A266887 A237664 A168584 * A239041 A081625 A144635 Adjacent sequences: A191007 A191008 A191009 * A191011 A191012 A191013 KEYWORD nonn,easy AUTHOR Edward Omey, Jun 16 2011 EXTENSIONS Formula corrected and more terms from Michel Marcus, Oct 16 2014 Edited by M. F. Hasler, Oct 16 2014 STATUS approved

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Last modified January 30 07:19 EST 2023. Contains 359939 sequences. (Running on oeis4.)