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Search: id:A143947
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| A143947 |
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Triangle read by rows: T(n,k) is the number of permutations of [n] for which the sum of the positions of the right-to-left minima is k (n<=k<=n(n+1)/2). |
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3
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| 1, 0, 1, 1, 0, 0, 2, 1, 2, 1, 0, 0, 0, 6, 2, 3, 7, 2, 3, 1, 0, 0, 0, 0, 24, 6, 8, 14, 27, 10, 9, 14, 3, 4, 1, 0, 0, 0, 0, 0, 120, 24, 30, 46, 68, 142, 41, 53, 50, 73, 23, 17, 23, 4, 5, 1, 0, 0, 0, 0, 0, 0, 720, 120, 144, 204, 270, 436, 834, 260, 256, 351, 310, 463, 148, 145, 118, 148, 40
(list; graph; listen)
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OFFSET
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1,7
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COMMENT
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Row n contains n(n+1)/2 entries, first n-1 of which are 0. Sum of entries in row n = n!=A000142(n).
Sum of entries in column n = A143948(n).
T(n,n)=(n-1)!
Sum(k*T(n,k),k=n..n(n+1)/2)=A001705(n).
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FORMULA
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Generating polynomial of row n is (n-1+t)(n-2+t^2)(n-3+t^3)...(1+t^(n-1))t^n.
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EXAMPLE
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T(4,6)=3 because we have 4132, 3142 and 2143 with right-to-left minima at positions 2 and 4.
Triangle starts:
1;
0,1,1;
0,0,2,1,2,1;
0,0,0,6,2,3,7,2,3,1;
0,0,0,0,24,6,8,14,27,10,9,14,3,4,1;
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MAPLE
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P:=proc(n) options operator, arrow: sort(expand(product(t^(n-j)+j, j=0..n-1))) end proc: for n to 7 do seq(coeff(P(n), t, i), i=1..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
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CROSSREFS
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A000142, A001705, A143948
Sequence in context: A055135 A121310 A024356 this_sequence A073781 A048622 A105661
Adjacent sequences: A143944 A143945 A143946 this_sequence A143948 A143949 A143950
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 22 2008
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