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%I #11 Mar 27 2021 23:28:28
%S 1,2,3,4,8,13,8,20,44,75,16,48,132,308,541,32,112,368,1076,2612,4683,
%T 64,256,976,3408,10404,25988,47293,128,576,2496,10096,36848,116180,
%U 296564,545835,256,1280,6208,28480,120400,454608,1469892,3816548
%N Triangular array read by rows: a(n, k) = number of ordered factorizations of a "hook-type" number with n total prime factors and k distinct prime factors. "Hook-type" means that only one prime can have multiplicity > 1.
%C The first three columns are A000079, A001792 and A098385.
%C The first two diagonals are A000670 and A005649.
%C A070175 gives the smallest representative of each hook-type prime signature, so this sequence is a rearrangement of A074206(A070175).
%F a(n, k) = 1 + (Sum_{i=1..k-1} binomial(k-1, i)*a(i, i)) + (Sum_{j=1..k} Sum_{i=j..j+n-k-1} binomial(k-1, j-1)*a(i, j)) + (Sum_{j=1..k-1} binomial(k-1,j-1)*a(j+n-k, j)). - _David Wasserman_, Feb 21 2008
%F a(n, k) = A074206(2^(n+1-k)*A070826(k)). - _David Wasserman_, Feb 21 2008
%F The following conjectural formula for the triangle entries agrees with the values listed above: T(n,k) = Sum_{j = 0..n-k} 2^(n-k-j)*binomial(n-k,j)*a(k,j), where a(k,j) = 2^j*Sum_{i = j+1..k+1} binomial(i,j+1)*(i-1)!*Stirling2(k+1,i). See A098384 for related conjectures. - _Peter Bala_, Apr 20 2012
%e a(4, 2) = 20 because 24=2*2*2*3 has 20 ordered factorizations and so does any other number with the same prime signature.
%Y Cf. A050324, A070175, A070826, A074206, A095705. A098349 gives the row sums. A098384.
%K nonn,tabl,easy
%O 1,2
%A _Alford Arnold_, Sep 04 2004
%E Edited and extended by _David Wasserman_, Feb 21 2008
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