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A191010
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a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5.
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1
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1, 7, 41, 215, 1065, 5079, 23593, 107479, 482345, 2139095, 9395241, 40936407, 177167401, 762356695, 3264175145, 13915694039, 59098749993, 250138895319, 1055531162665, 4442026976215, 18647717207081, 78109306037207, 326510972984361, 1362338887279575
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OFFSET
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0,2
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COMMENTS
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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)).
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LINKS
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Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-8,-16).
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FORMULA
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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
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MAPLE
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seq((n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5, n=0..50); # Robert Israel, May 03 2017
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A191008.
Sequence in context: A266887 A237664 A168584 * A239041 A081625 A144635
Adjacent sequences: A191007 A191008 A191009 * A191011 A191012 A191013
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KEYWORD
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nonn,easy
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AUTHOR
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Edward Omey, Jun 16 2011
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EXTENSIONS
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Formula corrected and more terms from Michel Marcus, Oct 16 2014
Edited by M. F. Hasler, Oct 16 2014
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STATUS
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approved
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