A051286 Whitney number of level n of the lattice of the ideals of the fence of order 2n.
1, 1, 2, 5, 11, 26, 63, 153, 376, 931, 2317, 5794, 14545, 36631, 92512, 234205, 594169, 1510192, 3844787, 9802895, 25027296, 63972861, 163701327, 419316330, 1075049011, 2758543201, 7083830648, 18204064403, 46812088751, 120452857976
Offset: 0
Examples
a(3) = 5 because the ideals of size 3 of the fence F(6) = { x1 < x2 > x3 < x4 > x5 < x6 } are x1*x3*x5, x1*x2*x3, x3*x4*x5, x1*x5*x6, x3*x5*x6.
Links
- Vincenzo Librandi and Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Andrei Asinowski, Axel Bacher, Cyril Banderier and Bernhard Gittenberger, Analytic Combinatorics of Lattice Paths with Forbidden Patterns: Enumerative Aspects, in International Conference on Language and Automata Theory and Applications, S. Klein, C. Martín-Vide, D. Shapira (eds), Springer, Cham, pp 195-206, 2018.
- Andrei Asinowski, Axel Bacher, Cyril Banderier and Bernhard Gittenberger, Analytic combinatorics of lattice paths with forbidden patterns, the vectorial kernel method, and generating functions for pushdown automata, Laboratoire d'Informatique de Paris Nord (LIPN 2019).
- Cyril Banderier and Paweł Hitczenko, Enumeration and asymptotics of restricted compositions having the same number of parts, Disc. Appl. Math. 160 (18) (2012) 2542-2554. See Puzzle 3.1.
- Jean-Luc Baril, Nathanaël Hassler, Sergey Kirgizov, and José L. Ramírez, Grand zigzag knight's paths, arXiv:2402.04851 [math.CO], 2024.
- Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Grand Dyck paths with air pockets, arXiv:2211.04914 [math.CO], 2022.
- Jean-Luc Baril and José L. Ramírez, Fibonacci and Catalan paths in a wall, 2023.
- Paul Barry, On a Generalization of the Narayana Triangle, J. Int. Seq. 14 (2011) # 11.4.5.
- Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
- Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
- Miklós Bóna and Arnold Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
- Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
- Steffen Eger, The Combinatorics of Weighted Vector Compositions, arXiv:1704.04964 [math.CO], 2017.
- Sergi Elizalde, The degree of symmetry of lattice paths, arXiv:2002.12874 [math.CO], 2021.
- Edyta Hetmaniok, Barbara Smoleń and Roman Wituła, The Stirling triangles, Proceedings of the Symposium for Young Scientists in Technology, Engineering and Mathematics (SYSTEM 2017), Kaunas, Lithuania, April 28, 2017, p. 35-41.
- Ivo L. Hofacker, Christian M. Reidys, and Peter F. Stadler, Symmetric circular matchings and RNA folding. Discr. Math., 312:100-112, 2012. See Prop. 5, C_l^{1}(z).
- Emanuele Munarini and Norma Zagaglia Salvi, On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns, Discrete Mathematics 259 (2002), 163-177.
- Jesús Sistos Barrón and Hua Wang, Balanced n-color compositions, Integers (2024) Vol. 24A, Art. No. A2. See pp. 4, 5, 8, 14.
Crossrefs
Programs
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Maple
seq( sum('binomial(i-k,k)*binomial(i-k,k)', 'k'=0..floor(i/2)), i=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001 # second Maple program: a:= proc(n) option remember; `if`(n<4, [1$2, 2, 5][n+1], ((2*n-1)*a(n-1)+(n-1)*a(n-2)+(2*n-3)*a(n-3)-(n-2)*a(n-4))/n) end: seq(a(n), n=0..35); # Alois P. Heinz, Aug 11 2016
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Mathematica
Table[Sum[Binomial[n-k,k]^2,{k,0,Floor[n/2]}],{n,0,40}] (* Emanuele Munarini, Mar 01 2011; corrected by Harvey P. Dale, Sep 12 2012 *) CoefficientList[Series[1/Sqrt[1-2*x-x^2-2*x^3+x^4], {x, 0, 20}], x] (* Vaclav Kotesovec, Jan 05 2013 *) a[n_] := HypergeometricPFQ[ {(1-n)/2, (1-n)/2, -n/2, -n/2}, {1, -n, -n}, 16]; Table[a[n], {n, 0, 29}] (* Jean-François Alcover, Feb 26 2013 *)
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Maxima
makelist(sum(binomial(n-k,k)^2,k,0,floor(n/2)),n,0,40); /* Emanuele Munarini, Mar 01 2011 */
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PARI
a(n)=polcoeff(1/sqrt((1+x+x^2)*(1-3*x+x^2)+x*O(x^n)),n)
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PARI
a(n)=sum(k=0,n,binomial(n-k,k)^2) /* Paul D. Hanna */
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PARI
{a(n)=polcoeff( exp(sum(m=1,n, sum(k=0,m, binomial(2*m,2*k)*x^k) *x^m/m) +x*O(x^n)), n)} /* Paul D. Hanna, Mar 18 2011 */
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PARI
{a(n)=local(A=1); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2*x^k) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
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PARI
{a(n)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, n, binomial(m+k, k)^2*x^k) * (1-x)^(2*m+1) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
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PARI
{a(n)=local(A=1+x); A=sum(m=0, n\2, x^(2*m) * sum(k=0, n, binomial(m+k, k)^2*x^k) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
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PARI
{a(n)=local(A=1+x); A=sum(m=0, n\2, x^(2*m) * sum(k=0, m, binomial(m, k)^2*x^k) / (1-x +x*O(x^n))^(2*m+1) ); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
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Python
from sympy import binomial def a(n): return sum(binomial(n - k, k)**2 for k in range(n//2 + 1)) print([a(n) for n in range(31)]) # Indranil Ghosh, Apr 18 2017
Formula
G.f.: 1/sqrt(1 - 2*x - x^2 - 2*x^3 + x^4).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*A002426(n-2k). - Paul Barry, Jan 31 2005
From Paul D. Hanna, May 22 2005: (Start)
a(n) = Sum_{k=0..n} C(n-k, k)^2.
Limit_{n->oo} a(n+1)/a(n) = (sqrt(5)+3)/2.
G.f.: 1/sqrt((1+x+x^2)*(1-3*x+x^2)). (End)
a(n) = Sum_{k=0..n} A049310(n, k)^2. - Philippe Deléham, Nov 21 2005
a(n) = Sum_{k=0..n} (C(k,k/2)*(1+(-1)^k)/2) * Sum_{j=0..n} (-1)^((n-j)/2)*C((n+j)/2,j)*((1+(-1)^(n-j))/2)*C(j,k). - Paul Barry, Mar 31 2010
G.f.: exp( Sum_{n>=1} (x^n/n)*Sum_{k=0..n} C(2n,2k)*x^k ). - Paul D. Hanna, Mar 18 2011
Logarithmic derivative equals A185828. - Paul D. Hanna, Mar 18 2011
D-finite with recurrence: n*a(n) - (2*n-1)*a(n-1) - (n-1)*a(n-2) - (2*n-3)*a(n-3) + (n-2)*a(n-4) = 0. - R. J. Mathar, Dec 17 2011
The g.f. A(x) satisfies the differential equation (1-2*x-x^2-2*x^3+x^4)*A'(x) = (1+x+3*x^2-2*x^3)*A(x), from which the recurrence conjectured by Mathar follows. - Emanuele Munarini, Dec 18 2017
a(n) ~ phi^(2*n + 2) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jan 05 2013, simplified Dec 18 2017
From Paul D. Hanna, Sep 05 2014: (Start)
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^k.
G.f.: Sum_{n>=0} x^n *[Sum_{k>=0} C(n+k,k)^2 * x^k] * (1-x)^(2*n+1).
G.f.: Sum_{n>=0} x^(2*n) * [Sum_{k>=0} C(n+k,k)^2 * x^k].
G.f.: Sum_{n>=0} x^(2*n) * [Sum_{k=0..n} C(n,k)^2 * x^k] /(1-x)^(2n+1).
(End)
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