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A091156
Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n, having k long ascents (i.e., ascents of length at least 2). Rows are of length 1,1,2,2,3,3,... .
5
1, 1, 1, 1, 1, 4, 1, 11, 2, 1, 26, 15, 1, 57, 69, 5, 1, 120, 252, 56, 1, 247, 804, 364, 14, 1, 502, 2349, 1800, 210, 1, 1013, 6455, 7515, 1770, 42, 1, 2036, 16962, 27940, 11055, 792, 1, 4083, 43086, 95458, 57035, 8217, 132, 1, 8178, 106587, 305812, 257257
OFFSET
0,6
COMMENTS
Also number of ordered trees with n edges, having k branch nodes (i.e., vertices of outdegree at least 2).
Also number of Łukasiewicz paths of length n having k fall steps (1,-1) that start at an odd level. A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(4,2)=2 because we have U(D)U(D) and U(3)(D)D(D), where U=(1,1), D=(1,-1), U(3)=(1,3) and the fall steps that start at an odd level are shown between parentheses. Row n has 1+floor(n/2) terms. Row sums are the Catalan numbers (A000108). T(2n,n)=A000108(n). T(2n+1,n)=A001791(n+1)=binomial(2n+2,n). - Emeric Deutsch, Jan 06 2005
Also number of Dyck paths of semilength n with k UUD's - I. Tasoulas (jtas(AT)unipi.gr), Feb 19 2006
T(n,k) = number of Dyck n-paths whose decomposition into 2-step subpaths contains k UUs. For example, T(4,2)=2 counts UU|UU|DD|DD, UU|DD|UU|DD (vertical bars indicate path decomposition). - David Callan, Jun 07 2006
T(n,k) = number of binary trees on n-1 edges containing k right edges whose child vertex has no right child. Under Knuth's "natural" correspondence, such a vertex in binary (n-1)-tree ~ a vertex of outdegree >=2 in ordered n-tree. - David Callan, Sep 25 2006
T(n,k) = number of binary trees on n-1 edges containing k left edges whose child vertex has no left child. Under "natural" correspondence, such a vertex in binary (n-1)-tree ~ a leaf edge with no left neighbor edge and not incident to the root in ordered n-tree ~ a UUD in Dyck n-path. - David Callan, Sep 25 2006
T(n,k) = number of permutations of length n avoiding 321 (classically) with k descents. - Andrew Baxter, May 17 2011.
REFERENCES
R. P. Stanley, Enumerative Combinatorics, Vol. 1, 1986; See Exercise 3.71(f).
LINKS
M. Barnabei et al., The descent statistic over 123-avoiding permutations, arXiv:0910.0963 [math.CO], 2009.
A. M. Baxter, Algorithms for Permutation Statistics, Ph. D. Dissertation, Rutgers University, May 2011. See p. 88.
Michael Bukata, Ryan Kulwicki, Nicholas Lewandowski, Lara Pudwell, Jacob Roth, Teresa Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv:1812.07112 [math.CO], 2018.
Colin Defant, Stack-Sorting Preimages of Permutation Classes, arXiv:1809.03123 [math.CO], 2018.
Katie R. Gedeon, Kazhdan-Lusztig polynomials of thagomizer matroids, arXiv:1610.05349 [math.CO], 2016.
Y. Park, S. Park, Avoiding permutations and the Narayana numbers, J. Korean Math. Soc. 50 (2013), No. 3, pp. 529-541.
Lara Pudwell, On the distribution of peaks (and other statistics), 16th International Conference on Permutation Patterns, Dartmouth College, 2018.
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
FORMULA
T(n,k) = (1/(n+1)) * binomial(n+1, k) * Sum_{j=0..n-2k} binomial(k+j-1, k-1)*binomial(n+1-k, n-2k-j).
G.f. G(t, z) satisfies z*(1-z+t*z)*G^2 - G + 1 = 0.
T(n,k) = n!*(1+k)/((n-2*k)!*(1+k)!^2)*hypergeom([k,2*k-n], [k+2], -1). - Peter Luschny, Oct 16 2015
T(n,k) = A055151(n,k)*hypergeom([k,2*k-n],[k+2],-1). - Peter Luschny, Oct 16 2015
EXAMPLE
T(4,1) = 11 because among the 14 Dyck paths of semilength 4, the paths that do not have exactly one long ascent are UDUDUDUD (no long ascent), UUDDUUDD (two long ascents) and UUDUUDDD (two long ascents). Here U=(1,1) and D=(1,-1).
Triangle begins:
1;
1;
1, 1;
1, 4;
1, 11, 2;
1, 26, 15;
1, 57, 69, 5;
1, 120, 252, 56;
1, 247, 804, 364, 14;
1, 502, 2349, 1800, 210;
1, 1013, 6455, 7515, 1770, 42;
...
MAPLE
a := (n, k)->binomial(n+1, k)* add(binomial(k+j-1, k-1)*binomial(n+1-k, n-2*k-j), j=0..n-2*k)/(n+1); seq(seq(a(n, k), k=0..floor(n/2)), n=0..15);
seq(seq(simplify(n!*(1+k)/((n-2*k)!*(1+k)!^2)*hypergeom([k, 2*k-n], [k+2], -1)), k=0.. floor(n/2)), n=0..15); # Peter Luschny, Oct 16 2015
# alternative Maple program:
b:= proc(x, y) option remember; `if`(y>x or y<0, 0,
`if`(x=0, 1, expand(b(x-1, y)*`if`(y=0, 1, 2)*z+
b(x-1, y+1) +b(x-1, y-1))))
end:
T:= n-> (p-> seq(coeff(p, z, n-2*i), i=0..n/2))(b(n, 0)):
seq(T(n), n=0..15); # Alois P. Heinz, Aug 07 2018
MATHEMATICA
T[n_, k_] := Binomial[n+1, k]*Sum[Binomial[k+j-1, k-1]*Binomial[n+1-k, n- 2*k-j], {j, 0, n-2*k}]/(n+1); Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/2 ]}] // Flatten (* Jean-François Alcover, Jan 31 2016 *)
PROG
(PARI)
tabf(nn) = {for(n=-1, nn, for(k=0, floor(n/2), if(binomial(n+1, k) * sum(j=0, n-2*k, binomial(k+j-1, k-1) * binomial(n+1-k, n-2*k-j))/(n+1)==0, print1("1, "), print1(binomial(n+1, k) * sum(j=0, n-2*k, binomial(k+j-1, k-1) * binomial(n+1-k, n-2*k-j))/(n+1), ", ")); ); print(); ); };
tabf(16); \\ Indranil Ghosh, Mar 05 2017
CROSSREFS
T(n,k) are rational multiples of A055151.
Sequence in context: A115022 A230534 A177822 * A092288 A111964 A242351
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Feb 22 2004
EXTENSIONS
Edited by Andrew Baxter, May 17 2011
STATUS
approved