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%I #17 Jan 25 2016 14:21:42
%S 1,1,1,1,31,1,1,242,242,1,1,992,7744,992,1,1,3124,99968,99968,3124,1,
%T 1,7502,756008,3099008,756008,7502,1,1,16806,4067052,52501944,
%U 52501944,4067052,16806,1,1,31744,17209344,533489664,1680062208,533489664,17209344,31744
%N Triangle read by rows: T(n,k) = A059384(n)/(A059384(k)*A059384(n-k)).
%C We assume that A059384(0)=1 since it would be the empty product.
%C These are the generalized binomial coefficients associated with the Jordan totient function J_5 given in A059378.
%C Another name might be the 5-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) = A059384(n)/(A059384(k)* A059384(n-k)).
%F T(n,k) = prod_{i=1..n} A059378(i)/(prod_{i=1..k} A059378(i)*prod_{i=1..n-k} A059378(i)).
%F T(n,k) = A059378(n)/n*(k/A059378(k)*T(n-1,k-1)+(n-k)/A059378(n-k)*T(n-1,k)).
%e The first five terms in the fifth Jordan totient function are 1,31,242,992,3124 and so T(4,2) = 992*242*31*1/((31*1)*(31*1))=7744 and T(5,3) = 3124*992*242*31*1/((242*31*1)*(31*1))=99968.
%e The triangle begins
%e 1
%e 1 1
%e 1 31 1
%e 1 242 242 1
%e 1 992 7744 992 1
%e 1 3124 99968 99968 3124 1
%o (Sage)
%o q=100 #change q for more rows
%o P=[0]+[i^5*prod([1-1/p^5 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. A059378, A059384, A238453, A238688, A238743, A238754.
%K nonn,tabl
%O 0,5
%A _Tom Edgar_, Mar 22 2014