OFFSET
1,3
COMMENTS
The mean value m(n) = Sum_{k=0..(2^n -n-1)} k*p(n,k) of the distribution function p(n,k) = binomial(2^n-n-1, k)/2^(2^n-n-1) is 0., 0.5, 2., 5.5, 13., 28.5, 60., 123.5, 251., 506.5, 1018., 2041.5, 4089., 8184.5... We set a(n) = round(m(n)).
The half-integer sequence h(n) = 0, 1/2, 2, 11/2, 13, 57/2, 60, 247/2, 251, 1013/2, 1018, 4083/2, 4089, 16369/2, 16376, 65519/2, 65527, ... is the binomial transform of 0, 1/2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).
FORMULA
G.f.: x*(1 - x + x^2)/(1 -3*x +x^2 +3*x^3 -2*x^4).
m(n) = (1/4)*2^n - 1/2 + m*(n-1) with m(1)=0 and a(n) = round(m(n)).
a(1)=0, a(2)=1, a(3)=2, a(4)=6, a(n) = 3*a(n-1) -a(n-2) -3*a(n-3) +2*a(n-4). - Harvey P. Dale, Nov 16 2011
a(n) = round(A000295(n)/2). - G. C. Greubel, Feb 20 2021
MAPLE
MATHEMATICA
Table[Ceiling[(2^n-n-1)/2], {n, 30}] (* or *) LinearRecurrence[{3, -1, -3, 2}, {0, 1, 2, 6}, 30] (* Harvey P. Dale, Nov 16 2011 *)
PROG
(PARI) a(n)=([0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1; 2, -3, -1, 3]^(n-1)*[0; 1; 2; 6])[1, 1] \\ Charles R Greathouse IV, Apr 18 2020
(Sage) [round((2^n -n-1)/2) for n in (1..40)] # G. C. Greubel, Feb 20 2021
(Magma) [Round((2^n -n-1)/2): n in [1..40]]; // G. C. Greubel, Feb 20 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Thomas Wieder, Feb 07 2010
STATUS
approved