Edward Omey - cp's OEIS frontend

A191010 a(n) = (n4^(n+1) + (64^(n+1)+(-1)^n)/5)/5.

1, 7, 41, 215, 1065, 5079, 23593, 107479, 482345, 2139095, 9395241, 40936407, 177167401, 762356695, 3264175145, 13915694039, 59098749993, 250138895319, 1055531162665, 4442026976215, 18647717207081, 78109306037207, 326510972984361, 1362338887279575

Offset: 0

Edward Omey, Jun 16 2011

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)).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-8,-16).

Cf. A191008.

seq((n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5, n=0..50); # Robert Israel, May 03 2017

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 *)

a(n)= (n*4^(n+1)+(6*4^(n+1)+(-1)^n)/5)/5; \\ Michel Marcus, Oct 16 2014

Vec(1 / ((1 + x)*(1 - 4*x)^2) + O(x^30)) \\ Colin Barker, May 03 2017

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

Formula corrected and more terms from Michel Marcus, Oct 16 2014

Edited by M. F. Hasler, Oct 16 2014

A191007 a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.

Original entry on oeis.org

3, 9, 27, 69, 171, 405, 939, 2133, 4779, 10581, 23211, 50517, 109227, 234837, 502443, 1070421, 2271915, 4805973, 10136235, 21321045, 44739243, 93672789, 195734187, 408245589, 850045611, 1767200085, 3668617899, 7605671253, 15748213419, 32570168661, 67287820971

Offset: 0

Edward Omey, Jun 16 2011

Another renewal type of sequence: Let X, X(1),X(2),... denote independent random variables with pdf P(X=1) = P(X=2) = P(X=4) = 1/3. Let N(x) denote the first value of k such that X(1)*X(2)...*X(k) > x, and let H(x) = E(N(x)). The sequence a(n) is given by a(n) = 2^(n+1)*H(2^n).

Index entries for linear recurrences with constant coefficients, signature (3,0,-4).

3 times A045883.

[n*2^(n+1)+(2^(n+3)+(-1)^n)/3: n in [0..30]]; // Vincenzo Librandi, Oct 16 2014

Table[n 2^(n + 1) + (2^(n + 3) + (-1)^n)/3, {n, 0, 70}] (* Vincenzo Librandi, Oct 16 2014 *)
LinearRecurrence[{3,0,-4},{3,9,27},40] (* Harvey P. Dale, Feb 11 2024 *)

a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3; \\ Michel Marcus, Oct 16 2014

a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.
a(n) = 3 * A045883(n+1).
G.f.: 3/((1 + x)*(1 - 2*x)^2). [Bruno Berselli, Oct 16 2014]

Formula corrected and more terms from Michel Marcus, Oct 16 2014

A191008 a(n) = (n3^(n+1)+((53^(n+1)+(-1)^(n))/4))/4.

Original entry on oeis.org

1, 5, 22, 86, 319, 1139, 3964, 13532, 45517, 151313, 498226, 1627538, 5281195, 17039327, 54705208, 174877304, 556916953, 1767605981, 5593383310, 17651846030, 55570626391, 174557144075, 547207226932, 1712229064916, 5348509347109, 16680994498409, 51949382866474

Offset: 0

Edward Omey, Jun 16 2011

Another renewal type of sequence. Let X, X(1), X(2),... denote random variables with pdf P(X = 1) = P(X = 4 ) = 1/4 and P(X = 2) = 1/2. Let N(x) denote the first value of k such that X(1)*X(2)*...*X(k) > x and let H(x)= E(N(x)). The sequence is given by a(n) = 3^(n+1)*H(2^n)/4.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-3,-9).

A191008:=n->(n*3^(n+1)+((5*3^(n+1)+(-1)^(n))/4))/4: seq(A191008(n), n=0..40); # Wesley Ivan Hurt, May 03 2017

LinearRecurrence[{5, -3, -9}, {1, 5, 22}, 27] (* or *)
CoefficientList[Series[1/((1 + x) (1 - 3 x)^2), {x, 0, 26}], x] (* Michael De Vlieger, May 03 2017 *)

a(n)=(n*3^(n+1)+((5*3^(n+1)+(-1)^(n))/4))/4; \\ Michel Marcus, Oct 16 2014

Vec(1 / ((1 + x)*(1 - 3*x)^2) + O(x^30)) \\ Colin Barker, May 03 2017

a(n) = (n*3^(n+1)+((5*3^(n+1)+(-1)^(n))/4))/4.
From Colin Barker, May 03 2017: (Start)
G.f.: 1 / ((1 + x)*(1 - 3*x)^2).
a(n) = 5*a(n-1) - 3*a(n-2) - 9*a(n-3) for n>2.
(End)

More terms from Michel Marcus, Oct 16 2014

A191682 Twice A113473.

Original entry on oeis.org

2, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 1

Edward Omey, Jun 11 2011

Arises in a renewal problem. Suppose X(1), X(2), ... are independent and identically distributed random variables with P(X = 1) = P(X = 2) = 0.5, and let N(n) denote the first value of k such that X(1)*X(2)*...*X(k) > x. Then a(n) gives the expected value of N(n), n = 1, 2, 3, ...

Cf. A113473.

cp's OEIS Frontend

User: Edward Omey

A191010 a(n) = (n*4^(n+1) + (6*4^(n+1)+(-1)^n)/5)/5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

Extensions

A191007 a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A191008 a(n) = (n*3^(n+1)+((5*3^(n+1)+(-1)^(n))/4))/4.

Views

Author

Keywords

Comments

Links

Programs

Maple

Mathematica

PARI

PARI

Formula

Extensions

A191682 Twice A113473.

Views

Author

Keywords

Comments

Crossrefs

A191010 a(n) = (n4^(n+1) + (64^(n+1)+(-1)^n)/5)/5.

A191008 a(n) = (n3^(n+1)+((53^(n+1)+(-1)^(n))/4))/4.