A253196 Irregular array read by rows. T(n,k) is the number of divisors d of n such that k^2 is the greatest square that divides d, n>=1, 1<=k<=A000188(n).
1, 2, 2, 2, 1, 2, 4, 2, 2, 2, 2, 0, 1, 4, 2, 4, 2, 2, 4, 4, 2, 2, 0, 1, 2, 4, 0, 2, 2, 4, 2, 4, 4, 2, 4, 4, 2, 0, 0, 0, 1, 4, 2, 0, 2, 4, 2, 2, 8, 2, 2, 2, 0, 2, 4, 4, 4, 4, 2, 2, 0, 0, 1, 2, 4, 4, 4, 4, 2, 8, 2, 4, 2, 4, 0, 2, 4, 2, 4, 4, 0, 2, 2, 0, 0, 0, 0, 0, 1, 4, 0, 0, 0, 2, 4, 4, 2, 2, 4, 0, 4, 4, 4, 4, 4, 4, 2, 8, 4
Offset: 1
Examples
1 2 2 2,1 2 4 2 2,2 2,0,1 4 2 4,2 2 4 4 2,2,0,1 2 4,0,2 For n=18, The divisors are: 1,2,3,6,9,18. T(18,1)=4 because 1 is the largest square that divides 1,2,3,6. T(18,3) = 2 because 9 is the largest square that divides 9,18.
Links
- Alois P. Heinz, Rows n = 1..6000, flattened
Programs
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Maple
with(numtheory): T:= n-> (p-> seq(coeff(p, x, j), j=1..degree(p)))(add( x^mul(i[1]^iquo(i[2], 2), i=ifactors(d)[2]), d=divisors(n))): seq(T(n), n=1..70); # Alois P. Heinz, Mar 25 2015
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Mathematica
nn = 60;g[list_] := list /. {j___, 0 ...} -> {j}; f[list_, i_] := list[[i]];Map[g, Transpose[Table[a = Table[If[n == k^2, 1, 0], {n, 1, nn}]; b = Table[2^PrimeNu[n], {n, 1, nn}];Table[DirichletConvolve[f[a, n], f[b, n], n, m], {m, 1, nn}], {k,1, nn}]]] // Grid
Formula
Dirichlet g.f. for column k: 1/k^(2*s) * zeta(s)^2/zeta(2*s).
Comments