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%I #28 Jan 18 2017 23:08:02
%S 1,0,0,-1,0,0,0,0,-1,1,0,0,0,0,0,0,0,-1,2,-1,0,0,0,0,0,0,0,0,-1,2,-1,
%T 0,1,-2,1,0,0,0,0,0,0,0,0,0,0,1,-2,0,1,1,-1,1,-1,-2,3,-1,0,0,0,0,0,0,
%U 0,0,0,0,0,0,-1,1,1,1,-2,-1,0,-1,3,-1,1,-1,0,1,-2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-2,0,0,2,2,-3,-1,0,0,1,1,-2,2,-1,1,-1
%N Recursive 1-parameter sequence a(n) allowing calculation of the Möbius function.
%C This sequence is generated from A266378 by excluding the second recursion parameter.
%F l(n) = floor((1/3)*(81+81*n+3*sqrt(729*n^2+1458*n+1104))^(1/3)-5/(81+81*n+3*sqrt(729*n^2+1458*n+1104))^(1/3))
%F c(n) = n*(n^2+3*n+8)/6 = A003600(n)
%F K(n) = n - 1 - c(l(n) - 1)
%F T(n,m) are coefficients of A008302
%F p(n) = c(l(n)-2)
%F u(n) = a(p(n)+K(n)+1)
%F v(n) = a(p(n)+K(n)-l(n)+2)
%F x(n) = a(p(n)+l(n)-1)*T(l(n)-1,l(n)*(l(n)-1)/2-K(n))
%F a(1) = 1
%F a(2) = 0
%F if (l(n)-2 >= K(n) or (1/2)*l(n)*(l(n)-1) < K(n)) then a(n) = 0 else a(n) = u(n)-v(n)-x(n)
%F Möbius(n) = a(c(n-1)+n)
%F A100198(n-2) = a(c(n-1)-n), for n>3.
%e Möbius(2) = a(c(1)+2) and because the c(1)=2 => a(c(1)+2)= a(4). l(4)=2, K(4)=1 so l(4)-2<K(4) and l(4)*(l(4)-1)/2>=K(4) and a(4)=u(4)-v(4)-x(4)
%e p(4)=c(l(4)-2)=c(0)=0
%e u(4)=a(p(4)+K(4)+1)=a(2)=0
%e v(4)=a(p(4)+K(4)-l(4)+2)=a(1)=1
%e x(4)=a(p(4)+l(4)-1)*T(l(4)-1,l(4)*(l(4)-1)/2-K(4))=a(1)*T(1,0)=0, as T(1,0)=0.
%e a(4)=u(4)-v(4)-x(4)=0-1-0=-1.
%p l := n->floor((1/3)*(81+81*n+3*sqrt(1104+1458*n+729*n^2))^(1/3)-5/(81+81*n+3*sqrt(1104+1458*n+729*n^2))^(1/3)):
%p c := n->(1/6)*n*(n^2+3*n+8):
%p K := n->n-1-c(l(n)-1):
%p A := (n, z)->z*(product(z^i-1, i = 1 .. n-1)):
%p T := (n, k)->coeff(eval(A(n, z)), z, k):
%p p := n->c(l(n)-2):
%p u := n->a(p(n)+K(n)+1):
%p v := n->a(p(n)+K(n)-l(n)+2):
%p x := n->a(p(n)+l(n)-1)*T(l(n)-1, (1/2)*l(n)*(l(n)-1)-K(n)):
%p a := proc (n) option remember; if K(n) <= l(n)-2 or (1/2)*l(n)*(l(n)-1) < K(n) then 0 else u(n)-v(n)-x(n) end if end proc:
%p a(2) := 0:
%p a(1) := 1:
%Y Cf. A002321, A003600, A008302, A266378.
%K sign
%O 1,19
%A _Gevorg Hmayakyan_, Jan 06 2017
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