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%I #60 Jan 02 2023 12:30:54
%S 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,
%T 0,576,0,31680,0,31680,0,2880,0,288000,0,201600,0,14400,2505600,0,
%U 2764800,0,1987200,0,86400,30067200,0,28512000,0,14515200,0,518400
%N Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.
%C 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
%H Peter J. C. Moses, <a href="/A293783/b293783.txt">Table of the first 100 rows</a>
%H Vladimir Shevelev, <a href="http://list.seqfan.eu/oldermail/seqfan/2017-October/018053.html">Basis in subsets of permutations described in A293783</a>, SeqFan post Oct 27 2017.
%F Row sums of triangle give A293857.
%F 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
%e Triangle begins
%e 0, 1;
%e 0, 1;
%e 2, 0, 2;
%e 8, 0, 4;
%e 0, 24, 0, 12;
%e 0, 108, 0, 36;
%e 576, 0, 720, 0, 144;
%e 4608, 0, 4032, 0, 576;
%e 0, 31680, 0, 31680, 0, 2880;
%e 0, 288000, 0, 201600, 0, 14400;
%e 2505600, 0, 2764800, 0, 1987200, 0, 86400;
%e 30067200, 0, 28512000, 0, 14515200, 0, 518400;
%e 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))!.
%e 0, 1;
%e 0, 1;
%e 1, 0, 1;
%e 2, 0, 1;
%e 0, 2, 0, 1;
%e 0, 3, 0, 1;
%e 4, 0, 5, 0, 1;
%e 8, 0, 7, 0, 1;
%e 0, 11, 0, 11, 0, 1;
%e 0, 20, 0, 14, 0, 1;
%e 29, 0, 32, 0, 23, 0, 1;
%e 58, 0, 55, 0, 28, 0, 1;
%e 0, 88, 0, 94, 0, 46, 0, 1;
%e 0, 169, 0, 146, 0, 53, 0, 1;
%e 263, 0, 282, 0, 283, 0, 86, 0, 1;
%e 526, 0, 515, 0, 383, 0, 97, 0, 1;
%e 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
%t 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 *)
%Y Cf. A008794, A010551, A293857, A293984.
%K nonn,tabf
%O 1,5
%A _Peter J. C. Moses_ and _Vladimir Shevelev_, Oct 18 2017
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