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%I #30 Jul 31 2021 03:31:00
%S 1,7,41,215,1065,5079,23593,107479,482345,2139095,9395241,40936407,
%T 177167401,762356695,3264175145,13915694039,59098749993,250138895319,
%U 1055531162665,4442026976215,18647717207081,78109306037207,326510972984361,1362338887279575
%N a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5.
%C 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)).
%H Colin Barker, <a href="/A191010/b191010.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-8,-16).
%F a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5.
%F From _Colin Barker_, May 03 2017: (Start)
%F G.f.: 1 / ((1 + x)*(1 - 4*x)^2).
%F a(n) = 7*a(n-1) - 8*a(n-2) - 16*a(n-3) for n>2.
%F (End)
%F E.g.f.: (80*x*exp(4*x)+24*exp(4*x)+exp(-x))/25. - _Robert Israel_, May 03 2017
%p seq((n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5, n=0..50); # _Robert Israel_, May 03 2017
%t CoefficientList[Series[1/((1 + x) (1 - 4 x)^2), {x, 0, 23}], x] (* or *)
%t LinearRecurrence[{7, -8, -16}, {1, 7, 41}, 24] (* _Michael De Vlieger_, May 03 2017 *)
%o (PARI) a(n)= (n*4^(n+1)+(6*4^(n+1)+(-1)^n)/5)/5; \\ _Michel Marcus_, Oct 16 2014
%o (PARI) Vec(1 / ((1 + x)*(1 - 4*x)^2) + O(x^30)) \\ _Colin Barker_, May 03 2017
%Y Cf. A191008.
%K nonn,easy
%O 0,2
%A _Edward Omey_, Jun 16 2011
%E Formula corrected and more terms from _Michel Marcus_, Oct 16 2014
%E Edited by _M. F. Hasler_, Oct 16 2014
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