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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.)