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A259703
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Triangle read by rows: T(n,k) = number of permutations without overlaps in which the first increasing run has length k.
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1
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1, 1, 1, 2, 1, 1, 5, 2, 2, 1, 12, 5, 4, 2, 1, 33, 13, 12, 4, 3, 1, 87, 35, 30, 12, 6, 3, 1, 252, 98, 90, 32, 21, 6, 4, 1, 703, 278, 243, 94, 54, 21, 8, 4, 1, 2105, 812, 745, 270, 175, 57, 32, 8, 5, 1, 6099, 2385, 2108, 808, 485, 181, 84, 32, 10, 5, 1
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OFFSET
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2,4
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COMMENTS
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The 12th row of the triangle (as given in the reference) is definitely wrong, since the first column of this triangle is known (it is A000560). The row sums are also known - see A000682.
To determine the first increasing run of the permutation 176852943 start on the left and move to the right counting the consecutive integers.
(1)7685(2)94(3). This permutation a has a first run of (3-1)=2. The permutation 123465 has a first run of (5-1)=4. (1)(2)(3)(4)6(5). (End)
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REFERENCES
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A. Sade, Sur les Chevauchements des Permutations, published by the author, Marseille, 1949
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LINKS
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EXAMPLE
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Triangle begins:
1;
1, 1;
2, 1, 1;
5, 2, 2, 1;
12, 5, 4, 2, 1;
33, 13, 12, 4, 3, 1;
87, 35, 30, 12, 6, 3, 1;
252, 98, 90, 32, 21, 6, 4, 1;
703, 278, 243, 94, 54, 21, 8, 4, 1;
2105, 812, 745, 270, 175, 57, 32, 8, 5, 1;
6099, 2385, 2108, 808, 485, 181, 84, 32, 10, 5, 1;
...
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PROG
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(PARI)
Overlapfree(v)={for(i=1, #v, for(j=i+1, v[i]-1, if(v[j]>v[i], return(0)))); 1}
Chords(u)={my(n=2*#u, v=vector(n), s=u[#u]); if(s%2==0, s=n+1-s); for(i=1, #u, my(t=n+1-s); s=u[i]; if(s%2==0, s=n+1-s); v[s]=t; v[t]=s); v}
FirstRunLen(v)={my(e=1); for(i=1, #v, if(v[i]==e, e++)); e-2}
row(n)={my(r=vector(n-1)); if(n>=2, forperm(n, v, if(v[1]<>1, break); if(Overlapfree(Chords(v)), r[FirstRunLen(v)]++))); r}
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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