%I #27 Apr 10 2016 12:23:04
%S 0,1,0,1,0,0,1,1,0,0,1,0,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,1,1,0,1,0,0,
%T 0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,
%U 1,1,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N Triangle T(n,k) read by rows. If n=k then 0, else if k divides n then 1 else 0.
%C Obtained by replacing the elements on the diagonal of A051731 by zero.
%C Apart from the initial T(1,1), the same as A176890.
%C The inverse of A051731 can be calculated with matrix powers of this matrix T(n,k) as follows: T(n,k)^0-T(n,k)^1+T(n,k)^2-T(n,k)^3+T(n,k)^4 ... Compare this to the binomial series for (1+x)^-1=1-x+x^2-x^3+x^4...
%F T(n,n) = 0. T(n,k) = 1 if k|n and k<n. T(n,k) = 0 if k does not divide n and k<n.
%F Lim_{s -> Infinity} T(n,k) where Ts(n,k) is the table defined by the recurrence:
%F Ts(n,k)=if k=1 then zeta[s]*(1-1/n^(s-1))-Sum_{i=1..n-1} Ts(n,k+i)/(i+1)^(s-1) else if k divides n then Ts(n/k,1) else Ts(n,k)=0 else Ts(n,k)=0. - _Mats Granvik_, Apr 09 2016
%e 0,
%e 1,0,
%e 1,0,0,
%e 1,1,0,0,
%e 1,0,0,0,0,
%e 1,1,1,0,0,0,
%e 1,0,0,0,0,0,0,
%e 1,1,0,1,0,0,0,0,
%e 1,0,1,0,0,0,0,0,0,
%e 1,1,0,0,1,0,0,0,0,0,
%t Flatten[Table[Table[If[Mod[n, k] == 0, If[n == k, 0, 1], 0], {k, 1, n}], {n, 1, 19}]] (* _Mats Granvik_, May 25 2013 *)
%t Flatten[Table[Which[n==k,0,Divisible[n,k],1,True,0],{n,20},{k,n}]] (* _Harvey P. Dale_, Oct 03 2014 *)
%t (* recurrence *)
%t Clear[t, s, n, k, z, nn]; nn = 19; t[n_, k_] := t[n, k] = If[k == 1, Zeta[s]*(1 - 1/n^(s - 1)) -Sum[t[n, k + i]/(i + 1)^(s - 1), {i, 1, n - 1}], If[Mod[n, k] == 0, t[n/k, 1], 0], 0]; Flatten[Table[Table[Limit[t[n, k], s -> Infinity], {k, 1, n}], {n, 1, nn}]] (* _Mats Granvik_, Apr 09 2016 *)
%Y Cf. A051731, A176890, A008683.
%K nonn,tabl,easy
%O 1
%A _Mats Granvik_, Dec 06 2010