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%I #33 Apr 23 2017 01:01:53
%S 1,1,1,1,1,1,2,1,1,1,2,1,1,1,1,3,1,2,1,2,1,1,1,1,3,2,1,1,1,2,1,1,2,1,
%T 1,4,1,1,1,3,2,1,1,1,3,2,1,2,1,1,2,1,1,1,1,4,3,1,2,1,2,1,1,3,1,3,2,1,
%U 1,1,3,3,1,1,1,1,5,1,2,1,1,2,1,1,2,1,1,4,4,1,1,1,2,1,1,2,1,1,4,3,1,1,1,3,3
%N Triangle T(n,m) read by rows (n >= 1, 0 <= m <= A001221(n)), giving the number of divisors of n with m distinct prime factors.
%C The formula used in obtaining the n-th row (see below) also gives the number of divisors of the k-th power of n.
%C Two numbers have identical rows in the table if and only if they have the same prime signature.
%C T(n,0)=1.
%H G. C. Greubel, <a href="/A146289/b146289.txt">Table of n, a(n) for the first 500 rows, flattened</a>
%H Anonymous?, <a href="http://xrjunque.nom.es/precis/polycalc.aspx">Polynomial calculator</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DistinctPrimeFactors.html">Distinct Prime Factors</a>
%H G. Xiao, WIMS server, <a href="http://wims.unice.fr/~wims/en_tool~algebra~factor.en.html">Factoris</a> (both expands and factors polynomials)
%F If the canonical factorization of n into prime powers is Product p^e(p), then T(n, m) is the coefficient of k^m in the polynomial expansion of Product_p (1 + e(p) k).
%e Rows begin:
%e 1;
%e 1,1;
%e 1,1;
%e 1,2;
%e 1,1;
%e 1,2,1;
%e 1,1;
%e 1,3;
%e 1,2;
%e 1,2,1;
%e ...
%e 12 has 1 divisor with 0 distinct prime factors (1); 3 with 1 (2, 3 and 4); and 2 with 2 (6 and 12), for a total of 6. The 12th row of the table therefore reads (1, 3, 2). These are the positive coefficients of the polynomial equation 1 + 3k + 2k^2 = (1 + 2k)(1 + k), derived from the prime factorization of 12 (namely, 2^2*3^1).
%p f:= proc(n)
%p local F,G,f,t,k;
%p F:= ifactors(n)[2];
%p G:= mul(1+f[2]*t, f= F);
%p seq(coeff(G,t,k),k=0..nops(F));
%p end proc:
%p seq(f(n),n=1..100); # _Robert Israel_, Feb 10 2015
%t Join[{{1}}, Table[nn = DivisorSigma[0, n];CoefficientList[Series[Product[1 + i x, {i, FactorInteger[n][[All, 2]]}], {x, 0,nn}], x], {n, 2, 100}]] // Grid (* _Geoffrey Critzer_, Feb 09 2015 *)
%o (PARI) tabf(nn) = {for (n=1, nn, vd = divisors(n); vo = vector(#vd, k, omega(vd[k])); for (k=0, vecmax(vo), print1(#select(x->x==k, vo), ", ");); print(););} \\ _Michel Marcus_, Apr 22 2017
%Y Row sums equal A000005(n).
%Y T(n, 1) = A001222(n) for n>1. T(n, A001221(n)) = A005361(n).
%Y Row n of A007318 is identical to row A002110(n) of this table and also identical to the row for any squarefree number with n prime factors.
%Y Cf. A146290. Also cf. A146291, A146292.
%K nonn,tabf
%O 1,7
%A _Matthew Vandermast_, Nov 11 2008
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