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A239633
Triangle read by rows: T(n,k) = A059384(n)/(A059384(k)*A059384(n-k)).
1
1, 1, 1, 1, 31, 1, 1, 242, 242, 1, 1, 992, 7744, 992, 1, 1, 3124, 99968, 99968, 3124, 1, 1, 7502, 756008, 3099008, 756008, 7502, 1, 1, 16806, 4067052, 52501944, 52501944, 4067052, 16806, 1, 1, 31744, 17209344, 533489664, 1680062208, 533489664, 17209344, 31744
OFFSET
0,5
COMMENTS
We assume that A059384(0)=1 since it would be the empty product.
These are the generalized binomial coefficients associated with the Jordan totient function J_5 given in A059378.
Another name might be the 5-totienomial coefficients.
LINKS
Tom Edgar, Totienomial Coefficients, INTEGERS, 14 (2014), #A62.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
Donald E. Knuth and Herbert S. Wilf, The power of a prime that divides a generalized binomial coefficient, J. Reine Angew. Math., 396:212-219, 1989.
FORMULA
T(n,k) = A059384(n)/(A059384(k)* A059384(n-k)).
T(n,k) = prod_{i=1..n} A059378(i)/(prod_{i=1..k} A059378(i)*prod_{i=1..n-k} A059378(i)).
T(n,k) = A059378(n)/n*(k/A059378(k)*T(n-1,k-1)+(n-k)/A059378(n-k)*T(n-1,k)).
EXAMPLE
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.
The triangle begins
1
1 1
1 31 1
1 242 242 1
1 992 7744 992 1
1 3124 99968 99968 3124 1
PROG
(Sage)
q=100 #change q for more rows
P=[0]+[i^5*prod([1-1/p^5 for p in prime_divisors(i)]) for i in [1..q]]
[[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.
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Tom Edgar, Mar 22 2014
STATUS
approved