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%I #122 Jan 11 2016 21:44:40
%S -1,-1,0,-1,-1,-1,-1,-1,-1,0,1,2,2,1,-1,-3,-6,-10,-14,-17,-17,-14,-9,
%T -4,-1,-1,-3,-5,-6,-5,-2,3,9,14,16,14,9,4,1,0,0,-1,-4,-9,-15,-20,-22,
%U -19,-10,5,24,43,58,67,70,67,58,44,28,14,5,1,0,-1,-5,-14,-29,-49,-71,-90,-100,-95,-70,-23,44,126,216,305,382,436,459,449,411
%N Recursive 2-parameter sequence allowing calculation of the Möbius function.
%C The a(n,m) forms a table where each row has (n-2)*(n-3)/2+1 = A000124(n-3) elements.
%C The index of the first row is n=3 and the index of the first column is m=0.
%C The right diagonal a(n, A000217(n-3)) = A008683(n), Möbius numbers, for n>=3.
%H Gevorg Hmayakyan, <a href="https://ghmath.files.wordpress.com/2010/06/mobius.pdf">Recursive Formula Related To The Mobius Function</a>
%F a(n,m) = a(n, m-1)+a(n-1, m)-a(n-1, m-n+1)-a(n-1, nu(n-1))*(T(n-2, m-1)-T(n-2, m-2)), where T(n,m) are coefficients of A008302, nu(n)=(n-2)*(n-3)/2, a(n,0)=-1, a(n,m)=0 if m<0 and m>nu(n).
%F Möbius(n) = a(n,nu(n)).
%F Easy to prove:
%F a(n+1,1)-a(n,1) = -1 - Möbius(n), n>=3 and accordingly
%F a(n,1) = 3 - n - A002321(n-1).
%F a(n+1,2)-a(n,2) = 2 - n - A002321(n) - Möbius(n)*(n-3), n>3 and accordingly
%F a(n,2) = -1 - (n-1)*(n-4)/2 - (n-3)*A002321(n-1).
%F a(n,-1+(n-2)*(n-3)/2) = Möbius(n+1) + Möbius(n)*(n-3).
%F a(n,-2+(n-2)*(n-3)/2) = Möbius(n+2) + Möbius(n+1)*(n-3) + (1/2)*Möbius(n)*(n-1)*(n-4) and in general:
%F a(n,-k+(n-2)*(n-3)/2)=f(n,k), for n>k.
%F a(n,-k-n+(n-2)*(n-3)/2)=f(n,n+k) + f(n,k) + h(n, k), for n>k,
%F where:
%F f(n,k)=Sum_{m=0..k}{Möbius(n+k-m)*(T(n-1,m)-T(n-1,m-1))}
%F h(n,s)=Sum_{k=1..n}{Sum_{i=0..k+1)}{(-1)^(i+1)*binomial(k+1, i)*f(n+k, s-2*k+1-i)}}
%F Easy to see:
%F a(n,2)-(n-3)*a(n,1)=(n-3)*(n-4)/2
%F Conjecture:
%F Sum_{m=0..(n-2)*(n-3)/2}{a(n,m)} = -A068337(n-1).
%F Sum_{m=0..(n-2)*(n-3)/2}{a(n,m)*m^k*(-1)^m} = 0, n>2*k+4.
%e The first few rows are
%e -1;
%e -1, 0;
%e -1, -1, -1, -1;
%e -1, -1, 0, 1, 2, 2, 1;
%e -1, -3, -6, -10, -14, -17, -17, -14, -9, -4, -1;
%p T := proc(n, k) option remember; if k=0 then 1 else if (n=1 and k=1) then 0 else if (k<0 or k>binomial(n, 2)) then 0 else T(n-1, k)+T(n, k-1)-T(n-1, k-n) end if end if end if end proc:
%p nu := n->((n-2)*(n-3))/(2):
%p a := proc (n, m) option remember; if m = 0 then -1 elif m < 0 or nu(n) < m then 0 else procname(n, m-1)+procname(n-1, m)-procname(n-1, m-n+1)-procname(n-1, nu(n-1))*(T(n-2, m-1)-T(n-2, m-2)) end if end proc:
%p L := seq(seq(a(N, r), r = 0 .. nu(N)), N = 3 .. 20);
%Y Cf. A002321, A008302, A008683, A068337.
%K sign,tabf
%O 3,12
%A _Gevorg Hmayakyan_, Dec 28 2015
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