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A245182
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Irregular triangle read by rows: T(n,k) (n>=2, 1<=k<=n) gives number of arrangements of the elements from the multiset M(n, 2) into exactly k disjoint cycles.
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2
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1, 1, 1, 2, 1, 3, 6, 4, 1, 12, 26, 20, 7, 1, 60, 140, 121, 51, 11, 1, 360, 894, 849, 410, 110, 16, 1, 2520, 6594, 6763, 3634, 1135, 211, 22, 1, 20160, 55152, 60304, 35336, 12404, 2723, 371, 29, 1, 181440, 515808, 595404, 374940, 144613, 36001, 5866, 610, 37, 1
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OFFSET
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2,4
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COMMENTS
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The multiset M(n, 2) consists of 2 1's and one copy of each of 2..n-1. For example, M(5,2) is the multiset {1,1,2,3,4}. See Griffiths reference for formula. - Andrew Howroyd, Feb 24 2020
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LINKS
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FORMULA
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T(n,k) = Sum_{m=0..k} |Stirling1(n-2, m)| * Sum_{j=0, 2-k+m} binomial(n+j-3, j) * A008284(2-j, k-m) for n >= 2. - Andrew Howroyd, Feb 24 2020
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EXAMPLE
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Triangle begins:
1 1
1 2 1
3 6 4 1
12 26 20 7 1
60 140 121 51 11 1
360 894 849 410 110 16 1
2520 6594 6763 3634 1135 211 22 1
...
For n = 4, arrangements of the multiset {1,1,2,3} into cycles are:
T(4,1) = 3: (1123), (1132), (1213);
T(4,2) = 6: (112)(3), (113)(2), (123)(1), (132)(1), (11)(23) (12)(13);
T(4,3) = 4: (11)(2)(3), (12)(1)(3), (13)(1)(2), (23)(1)(1);
T(4,4) = 1: (1)(1)(2)(3).
(End)
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PROG
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(PARI)
T(n)={my(P=matrix(3, 3, n, k, #partitions(n-k, k-1))); matrix(n, n, n, k, if(n<2, 0, sum(m=0, k, abs(stirling(n-2, m, 1)) * sum(j=0, 2-k+m, binomial(n+j-3, j)*P[1+2-j, 1+k-m]))))}
{my(A=T(10)); for(n=2, #A, print(A[n, 1..n]))} \\ Andrew Howroyd, Feb 24 2020
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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