OFFSET
1,5
COMMENTS
From Shevelev's comment in A008794 it follows that the last entries of rows, corresponding to the maximal possible i^2, form sequence A010551, n >= 1. Also note that the last entry of each row is the gcd of all its entries (for a proof see comment in A293857). - Vladimir Shevelev, Oct 26 2017
LINKS
Peter J. C. Moses, Table of the first 100 rows
Vladimir Shevelev, Basis in subsets of permutations described in A293783, SeqFan post Oct 27 2017.
FORMULA
Row sums of triangle give A293857.
If C = {c_1..c_n} is a permutation of {1..n}, then c_1 - c_2 + ... has the same parity as 1 + 2 + ... + n = n*(n+1)/2. So adjacent rows in the triangle for odd and even n have the same positions of 0's. These positions follow through one, beginning from the first position for n == 1,2 (mod 4) and from the second position for n == 3,0 (mod 4). - David A. Corneth and Vladimir Shevelev, Oct 19 2017
EXAMPLE
Triangle begins
0, 1;
0, 1;
2, 0, 2;
8, 0, 4;
0, 24, 0, 12;
0, 108, 0, 36;
576, 0, 720, 0, 144;
4608, 0, 4032, 0, 576;
0, 31680, 0, 31680, 0, 2880;
0, 288000, 0, 201600, 0, 14400;
2505600, 0, 2764800, 0, 1987200, 0, 86400;
30067200, 0, 28512000, 0, 14515200, 0, 518400;
The compressed triangle resulting from the division of each entry by the last entry of its row begins as follows. If i is the index of the row, starting with i = 1 then this last entry is floor(i/2)! * (i - floor(i/2))!.
0, 1;
0, 1;
1, 0, 1;
2, 0, 1;
0, 2, 0, 1;
0, 3, 0, 1;
4, 0, 5, 0, 1;
8, 0, 7, 0, 1;
0, 11, 0, 11, 0, 1;
0, 20, 0, 14, 0, 1;
29, 0, 32, 0, 23, 0, 1;
58, 0, 55, 0, 28, 0, 1;
0, 88, 0, 94, 0, 46, 0, 1;
0, 169, 0, 146, 0, 53, 0, 1;
263, 0, 282, 0, 283, 0, 86, 0, 1;
526, 0, 515, 0, 383, 0, 97, 0, 1;
For the sense of the entries of this triangle see the [Shevelev] link (with a continuation there). Let B(n,i) be the set of permutations C of 1..n for which c_1 - c_2 + ... + (-1)^(n-1)*c_n = i^2, i >= 0. Then |B(n,i)| is the entry in the n-th row and i-th column of the first triangle. Let us call two permutations C_1 and C_2 equivalent if one of them is obtained from another by a permutation of its elements with odd indices and/or separately with even indices. Let b(n,i) be the entry in the n-th row and i-th column of the second triangle. Then b(n,i) is the maximal possible number of pairwise non-equivalent permutations which could be chosen in B(n,i). On the other hand, it is the smallest number of non-equivalent permutations in B(n,i) such that every other permutation in B(n,i) is equivalent to one of them. So in some sense b(n,i) is the dimension of B(n,i). In particular, b(n,i) = 0 corresponds to empty B(n,i). - Vladimir Shevelev, Nov 13 2017
MATHEMATICA
a293783=Flatten[Table[PadLeft[Riffle[#, Table[0, {Floor[(n-1)/4]}]/.{}->0], 1+Floor[(1+n)/2]](Floor[n/2]!*(n-Floor[n/2])!)&[Reverse[Map[SeriesCoefficient[QBinomial[n, Floor[(n+1)/2], q], {q, 0, #}]&, Map[2#(Floor[(n+1)/2] - #)&, Range[0, Floor[(n+1)/4]]]]]], {n, 20}]] (* Peter J. C. Moses, Nov 01 2017 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Peter J. C. Moses and Vladimir Shevelev, Oct 18 2017
STATUS
approved