A294811 Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are triangular numbers (A000217). Then a(n) = b(n)/A010551(n).

1, 1, 1, 1, 2, 4, 6, 11, 16, 30, 48, 97, 157, 322, 524, 1077, 1777, 3684, 6157, 12876, 21684, 45520, 77212, 162533, 277608, 585993, 1006784, 2129433, 3677453, 7788711, 13514487, 28654668, 49933938, 105964856, 185377690, 393631445, 691101516, 1468137470

Offset: 0

Author

Vladimir Shevelev and Peter J. C. Moses, Nov 09 2017

Keywords

• nonn

Comments

All terms are positive integers (for a proof, cf. comment in A293984). Note that a(1), a(2), a(3), a(4) remain the same if in the definition the triangular numbers are replaced by k-gonal numbers for k >= 5.

Examples

			Let n=3. For a permutation C={c_1,c_2,c_3}, set s = s(C) = c_1 - c_2 + c_3. We have the permutations:
1,2,3; s=2
1,3,2; s=0
2,1,3; s=4
2,3,1; s=0
3,1,2; s=4
3,2,1; s=2
Here there are 2 permutations for which {s} are triangular numbers (when s = 0). Further, since A010551(3) = 2, then a(3) = 1.
Let n=4. For a permutation C={c_1,c_2,c_3,c_4}, set s = s(C) = c_1 - c_2 + c_3 - c_4. We have the permutations:
1,2,3,4; s=-2
1,3,2,4; s=-4
2,1,3,4; s=0
2,3,1,4; s=-4
3,1,2,4; s=0
3,2,1,4; s=-2
1,2,4,3; s=0
1,3,4,2; s=0
2,1,4,3; s=2
2,3,4,1; s=2
3,1,4,2; s=4
3,2,4,1; s=4
1,4,2,3; s=-4
1,4,3,2; s=-2
2,4,1,3; s=-4
2,4,3,1; s=0
3,4,1,2; s=-2
3,4,2,1; s=0
4,1,2,3; s=2
4,1,3,2; s=4
4,2,1,3; s=0
4,2,3,1; s=4
4,3,1,2; s=0
4,3,2,1; s=2
Here there are 8 permutations for which {s} are triangular numbers (when s = 0). Further, since A010551(4) = 4, then a(4) = 8/4 = 2.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..250 (terms n=1..200 from Peter J. C. Moses)

Crossrefs

Cf. A010551, A293857, A293984.

Programs

Maple

b:= proc(p, m, s) option remember; (n-> `if`(n=0, `if`(issqr(8*s+1), 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

polyQ[order_,n_]:=If[n==0,True,IntegerQ[(#-4+Sqrt[(#-4)^2+8 n (#-2)])/(2 (#-2))]&[order]];(*is a number polygonal?*)
Map[Total,Table[
possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2,Floor[(n+1)/2]^2];
filteredSums=Select[possibleSums,polyQ[3,#]&&#>-1&];
positions=Map[Flatten[{#,Position[possibleSums,#,1]-1}]&,filteredSums];
Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#[[2]]/2}]&,positions],{n,25}]] (* Peter J. C. Moses, Jan 02 2018 *)

Extensions

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