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%I #20 Mar 25 2015 18:43:09
%S 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,
%T 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,
%U 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
%N 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).
%C Row sums are A000005.
%C Column 1 is A034444.
%H Alois P. Heinz, <a href="/A253196/b253196.txt">Rows n = 1..6000, flattened</a>
%F Dirichlet g.f. for column k: 1/k^(2*s) * zeta(s)^2/zeta(2*s).
%e 1
%e 2
%e 2
%e 2,1
%e 2
%e 4
%e 2
%e 2,2
%e 2,0,1
%e 4
%e 2
%e 4,2
%e 2
%e 4
%e 4
%e 2,2,0,1
%e 2
%e 4,0,2
%e 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.
%p with(numtheory):
%p T:= n-> (p-> seq(coeff(p, x, j), j=1..degree(p)))(add(
%p x^mul(i[1]^iquo(i[2], 2), i=ifactors(d)[2]), d=divisors(n))):
%p seq(T(n), n=1..70); # _Alois P. Heinz_, Mar 25 2015
%t 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
%Y Cf. A000005, A000188, A034444.
%K nonn,tabf
%O 1,2
%A _Geoffrey Critzer_, Mar 24 2015
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