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A238754
Triangle read by rows: T(n,k) = A059383(n)/(A059383(k)*A059383(n-k)).
2
1, 1, 1, 1, 15, 1, 1, 80, 80, 1, 1, 240, 1280, 240, 1, 1, 624, 9984, 9984, 624, 1, 1, 1200, 49920, 149760, 49920, 1200, 1, 1, 2400, 192000, 1497600, 1497600, 192000, 2400, 1, 1, 3840, 614400, 9216000, 23961600, 9216000, 614400, 3840, 1, 1, 6480, 1658880
OFFSET
0,5
COMMENTS
We assume that A059383(0)=1 since it would be the empty product.
These are the generalized binomial coefficients associated with the Jordan totient function J_4 given in A059377.
Another name might be the 4-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) = A059383(n)/(A059383(k)* A059383(n-k)).
T(n,k) = prod_{i=1..n} A059377(i)/(prod_{i=1..k} A059377(i)*prod_{i=1..n-k} A059377(i)).
T(n,k) = A059377(n)/n*(k/A059377(k)*T(n-1,k-1)+(n-k)/A059377(n-k)*T(n-1,k)).
EXAMPLE
The first five terms in the fourth Jordan totient function are 1,15,80,240,624 and so T(4,2) = 240*80*15*1/((15*1)*(15*1))=1280 and T(5,3) = 624*240*80*15*1/((80*15*1)*(15*1))=9984.
The triangle begins
1
1 1
1 15 1
1 80 80 1
1 240 1280 240 1
1 624 9984 9984 624 1
PROG
(Sage)
q=100 #change q for more rows
P=[0]+[i^4*prod([1-1/p^4 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 04 2014
STATUS
approved