A293984 a(n) = A293857(n)/A010551(n).

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

Offset: 0

Author

Vladimir Shevelev, Oct 21 2017

Keywords

• nonn

Comments

Or row sums of the compressed triangle in A293783.

Conjecture: all terms are positive integers.

From David A. Corneth (with participation of Vladimir Shevelev), Oct 24 2017: (Start)

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_(2*m-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)

Links

• Peter J. C. Moses, Table of n, a(n) for n = 0..250

Crossrefs

Cf. A008794, A010551, A293783, A293857.

Programs

Maple

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)):
seq(a(n), n=0..40);  # Alois P. Heinz, Sep 17 2020

Mathematica

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 *)

Extensions

a(13)-a(30) from David A. Corneth, Oct 21 2017; a(31)-a(38) from Peter J. C. Moses, Nov 02 2017

a(0)=1 prepended by Alois P. Heinz, Sep 17 2020