%I M1433 N0567 #39 Sep 14 2021 04:43:50
%S 1,1,-1,2,-5,13,-33,80,-184,402,-840,1699,-3382,6750,-13716,28550,
%T -60587,129579,-275915,579828,-1197649,2431775,-4870105,9672634,
%U -19173013,38151533,-76521331,154941608,-316399235,649807589,-1337598675,2751021907,-5640238583,11513062785,-23389948481,47310801199,-95345789479,191616365385
%N Expansion of bracket function.
%C Inverse binomial transform of A006218.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H G. C. Greubel, <a href="/A001659/b001659.txt">Table of n, a(n) for n = 1..1000</a>
%H H. W. Gould, <a href="http://www.fq.math.ca/Scanned/2-4/gould.pdf">Binomial coefficients, the bracket function and compositions with relatively prime summands</a>, Fib. Quart. 2, issue 4, (1964), 241-260.
%F a(n) = Sum_{j=0..n} ((-1)^(n-j)*binomial(n,j)*Sum_{k=1..j} floor(j/k)).
%F G.f.: Sum_{k>0} x^k/((1+x)^k-x^k).
%F G.f.: Sum_{k>0} tau(k)*x^k/(1+x)^k. - _Vladeta Jovovic_, Jun 24 2003
%F G.f.: Sum_{n>=1} z^n/(1-z^n) (Lambert series) where z=x/(1+x). - _Joerg Arndt_, Jan 30 2011
%F a(n) = Sum_{k=1..n} (-1)^(n-k)*binomial(n-1,k-1)*tau(k). - _Ridouane Oudra_, Aug 21 2021
%t Table[Sum[(-1)^(n - k)*Binomial[n, k]*Sum[Floor[k/j], {j, 1, k}], {k, 0, n}], {n, 1, 50}] (* _G. C. Greubel_, Jul 02 2017 *)
%o (PARI) a(n)=sum(j=0,n,(-1)^(n-j)*binomial(n,j)*sum(k=1,j,j\k))
%o (PARI) a(n)=polcoeff(sum(k=1,n,x^k/((1+x)^k-x^k),x*O(x^n)),n)
%Y Cf. A000005, A000748, A000749, A000750, A006090, A006218.
%Y Equals A038200(n-1) + A038200(n), n>1.
%K sign
%O 1,4
%A _N. J. A. Sloane_
%E Edited by _Michael Somos_, Jun 14 2003