%I #7 Nov 04 2020 05:23:39
%S 0,0,0,0,1,0,0,1,1,0,0,2,2,2,0,0,1,2,2,1,0,0,2,2,3,2,2,0,0,1,2,2,2,2,
%T 1,0,0,3,3,4,3,4,3,3,0,0,2,4,4,4,4,4,4,2,0,0,2,3,5,4,5,4,5,3,2,0,0,1,
%U 2,3,4,4,4,4,3,2,1,0,0,3,3,4,4,6,5,6,4,4,3,3,0
%N Triangle read by rows: T(n,k)=number of prime divisors of C(n,k), counted with multiplicity (0<=k<=n).
%H T. D. Noe, <a href="/A132896/b132896.txt">Rows n=0..100 of triangle, flattened</a>
%F T(n, k) = A001222(A007318(n, k)). - _Michel Marcus_, Nov 04 2020
%e T(8,3)=4 because C(8,3)=56=2*2*2*7.
%e Triangle begins:
%e 0;
%e 0,0;
%e 0,1,0;
%e 0,1,1,0;
%e 0,2,2,2,0;
%e 0,1,2,2,1,0;
%p with(numtheory): T:=proc(n,k) if k <= n then bigomega(binomial(n,k)) else x end if end proc: for n from 0 to 12 do seq(T(n,k),k=0..n) end do; # yields sequence in triangular form
%Y Cf. A048571, which counts only distinct factors.
%Y Cf. A001222, A007318.
%K nonn,tabl
%O 0,12
%A _Emeric Deutsch_, Oct 16 2007