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A368945
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a(n) is the length of the central extent of width 0 for the symmetric representation of sigma, SRS(n), when that has an even number of parts otherwise -1.
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3
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-1, -1, 0, -1, 2, -1, 4, -1, -1, 0, 8, -1, 10, 2, -1, -1, 14, -1, 16, -1, 0, 6, 20, -1, -1, 8, 2, -1, 26, -1, 28, -1, 4, 12, -1, -1, 34, 14, 6, -1, 38, -1, 40, 2, -1, 18, 44, -1, -1, -1, 10, 4, 50, -1, 0, -1, 12, 24, 56, -1, 58, 26, -1, -1, 2, -1, 64, 8, 16, -1, 68, -1, 70, 32, 4, 10, -1, 0, 76, -1
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OFFSET
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1,5
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COMMENTS
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The central extent of width 0 for SRS(n) is that uninterrupted section where both boundary Dyck paths coincide, includes the point on the diagonal and thus has even length.
The maximum possible extent of width 0 in SRS(n) for odd numbers n is 2n - (n+1) - 2 = n - 3. This is achieved only by odd prime numbers p so that the values a(p) = p - 3 form the subsequence of records in this sequence; in particular, p-3 is the largest instantiated width 0 extent in the interval 1..p.
Conjecture: Every nonnegative even number occurs in this sequence.
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LINKS
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FORMULA
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a(n) = 2*ceiling((n+1)/(s(n)+1) - (s(n)+1)/2) - 2, where s(n) = position of the rightmost 1 in row n of the triangle in sequence A249223 when the last entry in that row is 0, and a(n) = -1 otherwise.
a(p) = p-3, p >= 3 prime, since s(p) = 1.
a(2*p) = p-5, p >= 5 prime, since s(p) = 3.
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EXAMPLE
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a(3) = 3 - 3 = 0 and a(2*5) = 5 - 5 = 0 by the formulas (see the illustrations in A237270 and A237593).
a(27) = 2 since SRS(a(27)) consists of the 4 parts of width 1 that have lengths 14, 6, 6, 14 and leave a section of length 2 across the diagonal of width 0 (see the illustration in A237593).
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MATHEMATICA
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t249223[n_] := FoldList[#1+(-1)^(#2+1)KroneckerDelta[Mod[n-#2 (#2+1)/2, #2]]&, 1, Range[2, Floor[(Sqrt[8n+1]-1)/2]]] (* row n in triangle of A249223 *)
extent0[n_] := Module[{t=t249223[n], s}, s=Position[t, 1][[-1, -1]]; If[Last[t]==0, 2Ceiling[(n+1)/(s+1)-(s+1)/2]-2, -1]]a368945[n_] := Map[extent0, Range[n]]a368945[80]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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