A173010 a(n) = round((2^n - n - 1)/4).

0, 0, 1, 3, 7, 14, 30, 62, 126, 253, 509, 1021, 2045, 4092, 8188, 16380, 32764, 65531, 131067, 262139, 524283, 1048570, 2097146, 4194298, 8388602, 16777209, 33554425, 67108857, 134217721, 268435448, 536870904, 1073741816, 2147483640

Offset

1,4

Comments

The variance v(n) = Sum_{k=0..2^n-n-1} (k - m(n))^2*p(n,k) of the distribution function p(n,k) = binomial(2^n -n-1, k)/2^(2^n -n-1) with m(n) its mean value is 0., 0.25, 1., 2.75, 6.5, 14.25, 30., 61.75, 125.5, 253.25, 509., 1020.75, 2044.5, 4092.25, 8188... We set a(n) = round(v(n)).

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (3,-2,0,1,-3,2).

Formula

a(n) = round((2^n -n -1)/4).

G.f.: x^3*(1 -x^3 +x^4)/(1 -3*x +2*x^2 -x^4 +3*x^5 -2*x^6). [sign corrected by Georg Fischer, Apr 17 2020]

v(n) = (1/8)*2^n -1/4 + v(-1+n) with v(1) = 0 and a(n) = round(v(n)).

a(n) = round(A000295(n)/4). - G. C. Greubel, Feb 20 2021

Maple

A173010:= round((2^n -n-1)/4); seq(A173010(n), n=1..40); # G. C. Greubel, Feb 20 2021

Mathematica

nn:=33; Rest[CoefficientList[Series[x^3*(1-x^3+x^4)/(1-3*x+2*x^2-2*x^6-x^4+3*x^5), {x, 0, nn}], x]] (* Georg Fischer, Apr 17 2020 *)
LinearRecurrence[{3, -2, 0, 1, -3, 2}, {0, 0, 1, 3, 7, 14, 30}, 40] (* Harvey P. Dale, Feb 06 2023 *)

Prog

(Sage) [round((2^n -n -1)/4) for n in (1..40)] # G. C. Greubel, Feb 20 2021
(Magma) [Round((2^n -n-1)/4): n in [1..40]]; // G. C. Greubel, Feb 20 2021

Crossrefs

Cf. A000295, A173009.

Sequence in context: A066225 A305777 A139817 * A036892 A123707 A011947

Adjacent sequences: A173007 A173008 A173009 * A173011 A173012 A173013

Keyword

nonn

Author

Thomas Wieder, Feb 07 2010

Extensions

Edited by Georg Fischer and Joerg Arndt, Apr 17 2020

Status

approved