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1, 1, 1, 2, 3, 3, 4, 10, 16, 23, 35, 85, 142, 229, 369, 895, 1522, 2614, 4348, 10467, 18038, 32160, 54488, 130148, 226594, 414130, 710880, 1685496, 2958666, 5503780, 9544629, 22476690, 39724867, 74884360, 130949625, 306457174, 544777361, 1037587152, 1827129712
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OFFSET
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0,4
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COMMENTS
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Or row sums of the compressed triangle in A293783.
Conjecture: all terms are positive integers.
Conjecture is true. Proof.
1) Let C={c_1..c_n} be a permutation of {1..n}, d(C) be alternating sum c_1 - c_2 + ... +(-1)^(n-1)*c_n. Then max_{C in S_n}d(C) = A008794(n+1). Indeed, if n = 2*m, then evidently the maximum is reached on a C={2*m,1,2*m-1,2,...,m+1,m}; if n=2*m - 1, then the maximum is reached on a C={2*m-1,1,2*m-2,2,...,m-1,m}. In both cases max_{C in S_n}d(C) = m^2 = A008794(n+1). The number of distinct reaches of the maximum is, evidently, floor(n/2)!*floor((n+1)/2)! which is also Avi Peretz's representation (2001) of A010551(n). So, A293857(n) >= A010551(n) and a(n)>=1.
2) Consider two cases: a) there are no C in S_n for which d(C) = k^2 < A008794(n+1). Then A293857(n) = A010551(n) and a(n) = 1; b) there is C for which d(C) = k^2 < A008794(n+1). Then, as in 1) to reach k^2 in case n=2*m consider all (n/2)! permutations of {c_1,c_3,...,c_n} and all (n/2)! permutations of {c_2, c_4, ... , c_(n+1)), or in case n = 2*m-1, all ((n+1)/2)! permutations of {c_1,c_3,...,c_(2*m-1)} and ((n-1)/2)! permutations of {c_2,c_4,...,c_(2m-2)}. So we again have A010551(n) distinct reaches. If the same k^2 could be reached by another permutation C_1 (other than above permutations of C), then we again obtain A010551 distinct reaches, etc. So, A293857(n) is always divisible by A010551(n). (End)
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LINKS
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MAPLE
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b:= proc(p, m, s) option remember; (n-> `if`(n=0, `if`(issqr(s), 1, 0),
`if`(p>0, b(p-1, m, s+n), 0)+`if`(m>0, b(p, m-1, s-n), 0)))(p+m)
end:
a:= n-> (t-> b(n-t, t, 0))(iquo(n, 2)):
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MATHEMATICA
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a293984=Table[
possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2, Floor[(n+1)/2]^2];
filteredSums=Select[possibleSums, IntegerQ[Sqrt[#]]&];
positions=Map[Flatten[{#, Position[possibleSums, #, 1]-1}]&, filteredSums];
Total[Map[SeriesCoefficient[QBinomial[n, Floor[(n+1)/2], q], {q, 0, #[[2]]/2}]&, positions]], {n, 20}] (* Peter J. C. Moses, Nov 05 2017 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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