%I #15 Jan 25 2016 14:22:42
%S 1,1,1,1,1,1,1,2,2,1,1,1,2,1,1,1,4,4,4,4,1,1,2,8,4,8,2,1,1,6,12,24,24,
%T 12,6,1,1,1,6,6,24,6,6,1,1,1,2,2,6,12,12,6,2,2,1,1,4,8,4,24,12,24,4,8,
%U 4,1,1,10,40,40,40,60,60,40,40,40,10,1,1,2
%N Triangle read by rows: T(n,k) = A239682(n)/(A239682(k)* A239682(n-k)).
%C A239682(0) = 1 since it is the empty product.
%C These are the generalized binomial coefficients associated with the sequence A173557.
%H Tom Edgar, <a href="http://www.emis.de/journals/INTEGERS/papers/o62/o62.Abstract.html">Totienomial Coefficients</a>, INTEGERS, 14 (2014), #A62.
%H Tom Edgar and Michael Z. Spivey, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Edgar/edgar3.html">Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
%H Donald E. Knuth and Herbert S. Wilf, <a href="http://www.math.upenn.edu/~wilf/website/dm36.pdf">The power of a prime that divides a generalized binomial coefficient</a>, J. Reine Angew. Math., 396:212-219, 1989.
%F T(n,k) = A239682(n)/(A239682(k)* A239682(n-k)).
%F T(n,k) = prod_{i=1..n} A173557(i)/(prod_{i=1..k} A173557(i)*prod_{i=1..n-k} A173557(i)).
%F T(n,k) = A173557(n)/n*(k/A173557(k)*T(n-1,k-1)+(n-k)/A173557(n-k)*T(n-1,k)).
%e The first six terms A173557 are 1,1,2,1,4,2 and so T(4,2) = 1*2*1*1/((1*1)*(1*1))=2 and T(6,3) = 2*4*1*2*1*1/((2*1*1)*(2*1*1))=4.
%e The triangle begins
%e 1
%e 1 1
%e 1 1 1
%e 1 2 2 1
%e 1 1 2 1 1
%e 1 4 4 4 4 1
%e 1 2 8 4 8 2 1
%e 1 6 12 24 24 12 6 1
%o (Sage)
%o q=100 #change q for more rows
%o P=[0]+[prod([(x-1) for x in prime_divisors(n)]) for n in [1..q]]
%o [[prod(P[1:n+1])/(prod(P[1:k+1])*prod(P[1:(n-k)+1])) for k in [0..n]] for n in [0..len(P)-1]] # generates the triangle up to q rows.
%Y Cf. A239682, A173557, A048804, A238453.
%K nonn,tabl
%O 0,8
%A _Tom Edgar_, Mar 24 2014