A047874
Triangle of numbers T(n,k) = number of permutations of (1,2,...,n) with longest increasing subsequence of length k (1<=k<=n).
Original entry on oeis.org
1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 41, 61, 16, 1, 1, 131, 381, 181, 25, 1, 1, 428, 2332, 1821, 421, 36, 1, 1, 1429, 14337, 17557, 6105, 841, 49, 1, 1, 4861, 89497, 167449, 83029, 16465, 1513, 64, 1, 1, 16795, 569794, 1604098, 1100902, 296326, 38281, 2521, 81, 1
Offset: 1
Eric Rains (rains(AT)caltech.edu)
T(3,2) = 4 because 132, 213, 231, 312 have longest increasing subsequences of length 2.
Triangle T(n,k) begins:
1;
1, 1;
1, 4, 1;
1, 13, 9, 1;
1, 41, 61, 16, 1;
1, 131, 381, 181, 25, 1;
1, 428, 2332, 1821, 421, 36, 1;
...
- Leonid Petrov, Rows n = 1..120, flattened (first 60 rows from Alois P. Heinz)
- P. Diaconis, Group Representations in Probability and Statistics, IMS, 1988; see p. 112.
- FindStat - Combinatorial Statistic Finder, The length of the longest increasing subsequence of the permutation.
- Ira M. Gessel, Symmetric functions and P-recursiveness, J. Combin. Theory Ser. A 53 (1990), no. 2, 257-285.
- J. M. Hammersley, A few seedings of research, in Proc. Sixth Berkeley Sympos. Math. Stat. and Prob., ed. L. M. le Cam et al., Univ. Calif. Press, 1972, Vol. I, pp. 345-394.
- Guo-Niu Han, A promenade in the garden of hook length formulas, Slides, 61st SLC Curia, Portugal - September 22, 2008. [From _Wouter Meeussen_, Sep 16 2010]
- J. Hunt and T. Szymanski, A fast algorithm for computing longest common subsequences, Commun. ACM, 20 (1977), 350-353.
- E. Irurozki, B. Calvo, and J. A. Lozano, Sampling and learning the Mallows model under the Ulam distance, Technical Report, 2014.
- S. Pilpel, Descending subsequences of random permutations, J. Combin. Theory, A 53 (1990), 96-116.
- A. Regev, Asymptotic values for degrees associated with strips of Young diagrams, Adv. in Math. 41 (1981), 115-136.
- C. Schensted, Longest increasing and decreasing subsequences, Canadian J. Math. 13 (1961), 179-191.
- Richard P. Stanley, Increasing and Decreasing Subsequences of Permutations and Their Variants, arXiv:math/0512035 [math.CO], 2005.
- Wikipedia, Longest increasing subsequence problem
Cf.
A224652 (Table II "Distribution of F_n" on p. 99 of the Pilpel reference).
-
h:= proc(l) local n; n:= nops(l); add(i, i=l)! /mul(mul(1+l[i]-j
+add(`if`(l[k]>=j, 1, 0), k=i+1..n), j=1..l[i]), i=1..n) end:
g:= (n, i, l)-> `if`(n=0 or i=1, h([l[], 1$n])^2, `if`(i<1, 0,
add(g(n-i*j, i-1, [l[], i$j]), j=0..n/i))):
T:= n-> seq(g(n-k, min(n-k, k), [k]), k=1..n):
seq(T(n), n=1..12); # Alois P. Heinz, Jul 05 2012
-
Table[Total[NumberOfTableaux[#]^2&/@ IntegerPartitions[n,{k}]],{n,7},{k,n}] (* Wouter Meeussen, Sep 16 2010, revised Nov 19 2013 *)
h[l_List] := Module[{n = Length[l]}, Total[l]!/Product[Product[1+l[[i]]-j+Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]]; g[n_, i_, l_List] := If[n == 0 || i == 1, h[Join[l, Array[1&, n]]]^2, If[i<1, 0, Sum[g[n-i*j, i-1, Join[l, Array[i&, j]]], {j, 0, n/i}]]]; T[n_] := Table[g[n-k, Min[n-k, k], {k}], {k, 1, n}]; Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Mar 06 2014, after Alois P. Heinz *)
A214015
Number of permutations A(n,k) in S_n with longest increasing subsequence of length <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 5, 1, 0, 1, 1, 2, 6, 14, 1, 0, 1, 1, 2, 6, 23, 42, 1, 0, 1, 1, 2, 6, 24, 103, 132, 1, 0, 1, 1, 2, 6, 24, 119, 513, 429, 1, 0, 1, 1, 2, 6, 24, 120, 694, 2761, 1430, 1, 0, 1, 1, 2, 6, 24, 120, 719, 4582, 15767, 4862, 1, 0
Offset: 0
A(4,2) = 14 because 14 permutations of {1,2,3,4} do not contain an increasing subsequence of length > 2: 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312, 4321. Permutation 1423 is not counted because it contains the noncontiguous increasing subsequence 123.
A(4,2) = 14 = 2^2 + 3^2 + 1^2 because the partitions of 4 with <= 2 parts are [2,2], [3,1], [4] with 2, 3, 1 standard Young tableaux, respectively:
+------+ +------+ +---------+ +---------+ +---------+ +------------+
| 1 3 | | 1 2 | | 1 3 4 | | 1 2 4 | | 1 2 3 | | 1 2 3 4 |
| 2 4 | | 3 4 | | 2 .-----+ | 3 .-----+ | 4 .-----+ +------------+
+------+ +------+ +---+ +---+ +---+
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, 1, ...
0, 1, 1, 1, 1, 1, 1, 1, ...
0, 1, 2, 2, 2, 2, 2, 2, ...
0, 1, 5, 6, 6, 6, 6, 6, ...
0, 1, 14, 23, 24, 24, 24, 24, ...
0, 1, 42, 103, 119, 120, 120, 120, ...
0, 1, 132, 513, 694, 719, 720, 720, ...
0, 1, 429, 2761, 4582, 5003, 5039, 5040, ...
Columns k=0-10 give:
A000007,
A000012,
A000108,
A005802,
A047889,
A047890,
A052399,
A072131,
A072132,
A072133,
A072167.
A(2n,n-1) gives
A269042(n) for n>0.
-
h:= proc(l) local n; n:=nops(l); add(i, i=l)! /mul(mul(1+l[i]-j
+add(`if`(l[k]>=j, 1, 0), k=i+1..n), j=1..l[i]), i=1..n)
end:
g:= (n, i, l)-> `if`(n=0 or i=1, h([l[], 1$n])^2, `if`(i<1, 0,
add(g(n-i*j, i-1, [l[], i$j]), j=0..n/i))):
A:= (n, k)-> `if`(k>=n, n!, g(n, k, [])):
seq(seq(A(n, d-n), n=0..d), d=0..14);
-
h[l_] := With[{n = Length[l]}, Sum[i, {i, l}]! / Product[Product[1+l[[i]]-j + Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]];
g[n_, i_, l_] := If[n == 0 || i == 1, h[Join[l, Array[1&, n]]]^2, If[i < 1, 0, Sum[g[n - i*j, i-1, Join[l, Array[i&, j]]], {j, 0, n/i}]]];
A[n_, k_] := If[k >= n, n!, g[n, k, {}]];
Table [Table [A[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Dec 09 2013, translated from Maple *)
A047888
Rectangular array of numbers a(n,k) = number of permutations of n things with longest increasing subsequence of length <= k (1 <= k <= oo), read by antidiagonals.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 5, 2, 1, 1, 14, 6, 2, 1, 1, 42, 23, 6, 2, 1, 1, 132, 103, 24, 6, 2, 1, 1, 429, 513, 119, 24, 6, 2, 1, 1, 1430, 2761, 694, 120, 24, 6, 2, 1, 1, 4862, 15767, 4582, 719, 120, 24, 6, 2, 1, 1, 16796, 94359, 33324, 5003, 720, 120, 24, 6, 2, 1, 1, 58786, 586590
Offset: 1
Square array a(n,k) begins:
1, 1, 1, 1, 1, 1, ...
1, 2, 2, 2, 2, 2, ...
1, 5, 6, 6, 6, 6, ...
1, 14, 23, 24, 24, 24, ...
1, 42, 103, 119, 120, 120, ...
1, 132, 513, 694, 719, 720, ...
-
rows = 12; h[l_List] := Module[{n = Length[l]}, Total[l]!/Product[Product[1+l[[i]]-j+Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]] ; g[n_, i_, l_List] := If[n == 0 || i == 1, h[Join[l, Array[1&, n]]]^2, If[i<1, 0, Sum[g[n-i*j, i-1, Join[l, Array[i&, j]]], {j, 0, n/i}]]]; T[n_] := Table[g[n-k, Min[n-k, k], {k}], {k, 1, rows}] // Accumulate; A047888 = Table[T[n], {n, 1, rows}]; Table[A047888[[n-k+1, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 06 2014, after Alois P. Heinz *)
-
b(n, k) = {
my(x = 'x + O('x^(2*n)));
sum(i = 0, n, x^(2*i+k)/(i!*(i+k)!));
};
u(n, k) = {
my(v = Vec(matdet(matrix(k, k, i, j, b(n, abs(i-j))))));
return(vector((#v-1)\2, i, v[2*i+1] * i!^2));
};
A(n, k) = {
my(m = [;]);
for (i = 1, k, m = concat(m, u(n, i)~));
return(m);
};
A(6, 6) \\ Gheorghe Coserea, Feb 02 2016
A331883
The number of permutations in the symmetric group S_n in which it is possible to find two disjoint increasing subsequences each with length equal to the length of the longest increasing subsequence of the permutation.
Original entry on oeis.org
0, 1, 1, 5, 26, 132, 834, 6477, 56242
Offset: 1
a(3) = 1 because the only permutation whose longest increasing subsequence is 1 is [3,2,1] and this contains two disjoint increasing subsequences of length 1.
The a(4) = 5 permutations are:
[2,1,4,3],
[2,4,1,3],
[3,1,4,2],
[3,4,1,2],
[4,3,2,1].
Showing 1-4 of 4 results.
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