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%I #20 Jan 25 2016 14:22:15
%S 1,1,1,1,3,1,1,8,8,1,1,12,32,12,1,1,24,96,96,24,1,1,24,192,288,192,24,
%T 1,1,48,384,1152,1152,384,48,1,1,48,768,2304,4608,2304,768,48,1,1,72,
%U 1152,6912,13824,13824,6912,1152,72,1,1,72,1728,10368,41472,41472
%N Triangle read by rows: T(n,k) = A059381(n)/(A059381(k)*A059381(n-k)).
%C We assume that A059381(0)=1 since it would be the empty product.
%C These are the generalized binomial coefficients associated with the Jordan totient function J_2 given in A007434.
%C Another name might be the 2-totienomial coefficients.
%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) = A059381(n)/(A059381(k)* A059381(n-k)).
%F T(n,k) = prod_{i=1..n} A007434(i)/(prod_{i=1..k} A007434(i)*prod_{i=1..n-k} A007434(i)).
%F T(n,k) = A007434(n)/n*(k/A007434(k)*T(n-1,k-1)+(n-k)/A007434(n-k)*T(n-1,k)).
%e The first five terms in the second Jordan totient function are 1,3,8,12,24 and so T(4,2) = 12*8*3*1/((3*1)*(3*1))=32 and T(5,3) = 24*12*8*3*1/((8*3*1)*(3*1))=96.
%e The triangle begins
%e 1
%e 1 1
%e 1 3 1
%e 1 8 8 1
%e 1 12 32 12 1
%e 1 24 96 96 24 1
%e 1 24 192 288 192 24 1
%o (Sage)
%o q=100 #change q for more rows
%o P=[0]+[i^2*prod([1-1/p^2 for p in prime_divisors(i)]) for i 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. A007434, A059381, A238453.
%K nonn,tabl
%O 0,5
%A _Tom Edgar_, Mar 02 2014
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