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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, 1, 1, 48, 384, 1152, 1152, 384, 48, 1, 1, 48, 768, 2304, 4608, 2304, 768, 48, 1, 1, 72, 1152, 6912, 13824, 13824, 6912, 1152, 72, 1, 1, 72, 1728, 10368, 41472, 41472
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OFFSET
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0,5
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COMMENTS
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We assume that A059381(0)=1 since it would be the empty product.
These are the generalized binomial coefficients associated with the Jordan totient function J_2 given in A007434.
Another name might be the 2-totienomial coefficients.
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LINKS
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Table of n, a(n) for n=0..60.
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.
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FORMULA
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T(n,k) = A059381(n)/(A059381(k)* A059381(n-k)).
T(n,k) = prod_{i=1..n} A007434(i)/(prod_{i=1..k} A007434(i)*prod_{i=1..n-k} A007434(i)).
T(n,k) = A007434(n)/n*(k/A007434(k)*T(n-1,k-1)+(n-k)/A007434(n-k)*T(n-1,k)).
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EXAMPLE
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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.
The triangle begins
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 1
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PROG
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(Sage)
q=100 #change q for more rows
P=[0]+[i^2*prod([1-1/p^2 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.
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CROSSREFS
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Cf. A007434, A059381, A238453.
Sequence in context: A124469 A094816 A097712 * A174117 A157210 A034801
Adjacent sequences: A238685 A238686 A238687 * A238689 A238690 A238691
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KEYWORD
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nonn,tabl
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AUTHOR
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Tom Edgar, Mar 02 2014
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STATUS
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approved
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