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A238743 Triangle read by rows: T(n,k) = A059382(n)/(A059382(k)*A059382(n-k)). 3
1, 1, 1, 1, 7, 1, 1, 26, 26, 1, 1, 56, 208, 56, 1, 1, 124, 992, 992, 124, 1, 1, 182, 3224, 6944, 3224, 182, 1, 1, 342, 8892, 42408, 42408, 8892, 342, 1, 1, 448, 21888, 153216, 339264, 153216, 21888, 448, 1, 1, 702, 44928, 590976, 1920672, 1920672, 590976, 44928 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,5
COMMENTS
We assume that A059382(0)=1 since it would be the empty product.
These are the generalized binomial coefficients associated with the Jordan totient function J_3 given in A059376.
Another name might be the 3-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) = A059382(n)/(A059382(k)* A059382(n-k)).
T(n,k) = prod_{i=1..n} A059376(i)/(prod_{i=1..k} A059376(i)*prod_{i=1..n-k} A059376(i)).
T(n,k) = A059376(n)/n*(k/A059376(k)*T(n-1,k-1)+(n-k)/A059376(n-k)*T(n-1,k)).
EXAMPLE
The first five terms in the third Jordan totient function are 1,7,26,56,124 and so T(4,2) = 56*26*7*1/((7*1)*(7*1))=208 and T(5,3) = 124*56*26*7*1/((26*7*1)*(7*1))=992.
The triangle begins
1
1 1
1 7 1
1 26 26 1
1 56 208 56 1
1 124 992 992 124 1
1 182 3224 6944 3224 182 1
PROG
(Sage)
q=100 #change q for more rows
P=[0]+[i^3*prod([1-1/p^3 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
Sequence in context: A168290 A218695 A179837 * A168517 A142465 A248829
KEYWORD
nonn,tabl
AUTHOR
Tom Edgar, Mar 04 2014
STATUS
approved

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Last modified April 23 05:37 EDT 2024. Contains 371906 sequences. (Running on oeis4.)