cp's OEIS Frontend

A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.

%I A293783 #61 Dec 23 2024 14:53:45
%S A293783 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 A293783 0,576,0,31680,0,31680,0,2880,0,288000,0,201600,0,14400,2505600,0,
%U A293783 2764800,0,1987200,0,86400,30067200,0,28512000,0,14515200,0,518400
%N A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.
%C A293783 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 A293783 Peter J. C. Moses, <a href="/A293783/b293783.txt">Table of the first 100 rows</a>
%H A293783 Vladimir Shevelev, <a href="https://web.archive.org/web/*/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 A293783 Row sums of triangle give A293857.
%F A293783 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 A293783 Triangle begins
%e A293783          0,      1;
%e A293783          0,      1;
%e A293783          2,      0,        2;
%e A293783          8,      0,        4;
%e A293783          0,     24,        0,     12;
%e A293783          0,    108,        0,     36;
%e A293783        576,      0,      720,      0,      144;
%e A293783       4608,      0,     4032,      0,      576;
%e A293783          0,  31680,        0,  31680,        0,  2880;
%e A293783          0, 288000,        0, 201600,        0, 14400;
%e A293783    2505600,      0,  2764800,      0,  1987200,     0,  86400;
%e A293783   30067200,      0, 28512000,      0, 14515200,     0, 518400;
%e A293783 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 A293783     0,   1;
%e A293783     0,   1;
%e A293783     1,   0,   1;
%e A293783     2,   0,   1;
%e A293783     0,   2,   0,   1;
%e A293783     0,   3,   0,   1;
%e A293783     4,   0,   5,   0,   1;
%e A293783     8,   0,   7,   0,   1;
%e A293783     0,  11,   0,  11,   0,   1;
%e A293783     0,  20,   0,  14,   0,   1;
%e A293783    29,   0,  32,   0,  23,   0,   1;
%e A293783    58,   0,  55,   0,  28,   0,   1;
%e A293783     0,  88,   0,  94,   0,  46,   0,   1;
%e A293783     0, 169,   0, 146,   0,  53,   0,   1;
%e A293783   263,   0, 282,   0, 283,   0,  86,   0,   1;
%e A293783   526,   0, 515,   0, 383,   0,  97,   0,   1;
%e A293783 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 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 A293783 Cf. A008794, A010551, A293857, A293984.
%K A293783 nonn,tabf
%O A293783 1,5
%A A293783 _Peter J. C. Moses_ and _Vladimir Shevelev_, Oct 18 2017