OFFSET
1
COMMENTS
Obtained by replacing the elements on the diagonal of A051731 by zero.
Apart from the initial T(1,1), the same as A176890.
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...
FORMULA
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.
Lim_{s -> Infinity} T(n,k) where Ts(n,k) is the table defined by the recurrence:
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
EXAMPLE
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,
1,1,0,1,0,0,0,0,
1,0,1,0,0,0,0,0,0,
1,1,0,0,1,0,0,0,0,0,
MATHEMATICA
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 *)
Flatten[Table[Which[n==k, 0, Divisible[n, k], 1, True, 0], {n, 20}, {k, n}]] (* Harvey P. Dale, Oct 03 2014 *)
(* recurrence *)
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 *)
CROSSREFS
KEYWORD
AUTHOR
Mats Granvik, Dec 06 2010
STATUS
approved