A047874 Triangle of numbers T(n,k) = number of permutations of (1,2,...,n) with longest increasing subsequence of length k (1<=k<=n).
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
Examples
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; ...
Links
- 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
Crossrefs
Columns k=1-10 give: A000012, A001453, A001454, A001455, A001456, A001457, A001458, A239432, A245665, A245666.
Row sums give A000142.
Cf. A047884. - Wouter Meeussen, Sep 16 2010
Cf. A224652 (Table II "Distribution of F_n" on p. 99 of the Pilpel reference).
Cf. A245667.
T(2n,n) gives A267433.
Cf. A003316.
Programs
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Maple
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 -
Mathematica
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 *)
Formula
Sum_{k=1..n} k * T(n,k) = A003316(n). - Alois P. Heinz, Nov 04 2018
Comments