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A092580
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Triangle read by rows: T(n,k) is the number of permutations p of [n] in which exactly the first k terms satisfy the up-down property, i.e., p(1)<p(2), p(2)>p(3), p(3)<p(4), ...
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4
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1, 1, 1, 3, 1, 2, 12, 4, 3, 5, 60, 20, 15, 9, 16, 360, 120, 90, 54, 35, 61, 2520, 840, 630, 378, 245, 155, 272, 20160, 6720, 5040, 3024, 1960, 1240, 791, 1385, 181440, 60480, 45360, 27216, 17640, 11160, 7119, 4529, 7936, 1814400, 604800, 453600, 272160, 176400, 111600, 71190, 45290, 28839, 50521
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OFFSET
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1,4
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COMMENTS
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LINKS
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FORMULA
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T(n, k) = n!*[(k+1)*E(k)-E(k+1)]/(k+1)! for k<n and T(n, n) = E(n), where tan(x)+sec(x) = Sum_{n>=0} [E(n)x^n/n!] (i.e., E(n) = A000111(n)).
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EXAMPLE
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T(4,3)=3 because 1432, 2431, 3421 are the only permutations of [4] in which exactly the first 3 entries satisfy the up-down property.
Triangle starts:
1;
1, 1;
3, 1, 2;
12, 4, 3, 5;
60, 20, 15, 9, 16;
360, 120, 90, 54, 35, 61;
...
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MAPLE
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b:= proc(u, o) option remember;
`if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
end:
E:= n-> b(n, 0):
T:= (n, k)-> `if`(n=k, E(n), n!*((k+1)*E(k)-E(k+1))/(k+1)!):
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MATHEMATICA
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b[u_, o_] := b[u, o] = If[u + o == 0, 1, Sum[b[o - 1 + j, u - j], {j, 1, u}]]; e[n_] := b[n, 0]; T[n_, k_] := If[n == k, e[n], n!*((k + 1)*e[k] - e[k + 1])/(k + 1)!]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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AUTHOR
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Emeric Deutsch and Warren P. Johnson (wjohnson(AT)bates.edu), Apr 10 2004
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STATUS
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approved
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