

A053121


Catalan triangle (with 0's) read by rows.


108



1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 5, 0, 4, 0, 1, 5, 0, 9, 0, 5, 0, 1, 0, 14, 0, 14, 0, 6, 0, 1, 14, 0, 28, 0, 20, 0, 7, 0, 1, 0, 42, 0, 48, 0, 27, 0, 8, 0, 1, 42, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 0, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 132, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0
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OFFSET

0,8


COMMENTS

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of nstep walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,1). (This is basically a different formulation of the previous  walks with a wall  property.)  Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bellsubgroup of the Riordangroup. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bellmatrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(1)}(x))/x, with f(x) := x*G(x) and f^{(1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231.  Emeric Deutsch, Oct 12 2006
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n1,0)+T(n1,1), T(n,k)=T(n1,k1)+y*T(n1,k)+T(n1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) > A053121; (0,1) > A089942; (0,2) > A126093; (0,3) > A126970; (1,0) > A061554; (1,1) > A064189; (1,2) > A039599; (1,3) > A110877; (1,4) > A124576; (2,0) > A126075; (2,1) > A038622; (2,2) > A039598; (2,3) > A124733; (2,4) > A124575; (3,0) > A126953; (3,1) > A126954; (3,2) > A111418; (3,3) > A091965; (3,4) > A124574; (4,3) > A126791; (4,4) > A052179; (4,5) > A126331; (5,5) > A125906.  Philippe Deléham, Sep 25 2007
By columns without the zeros, nth row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then nth row = coefficients of A^(n+1).  Gary W. Adamson, May 13 2009
As an upper right triangle, rows represent powers of 5sqrt(24):
5  sqrt(24)^1 = 0.101020514...
5  sqrt(24)^2 = 0.010205144...
5  sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,1), H=1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH.  Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the Nth Chebyshev Spolynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x).  L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular Ngon:
rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310.  Wolfdieter Lang, Sep 21 2013
The socalled Asequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n1, m1) + a(n1, m+1) for n>=1 and m>=1, with inputs. The Zsequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n1,1), n>=1, a(0,0) = 1. For A and Zsequences for Riordan triangles see the W. Lang link under A006232.  Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the mth ((m+1)dimensional) irreducible representation in the nth tensor power of the standard one.  Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the BoasBuck class (see a comment and references in A046521), hence they satisfy the BoasBuck identity: (E_x  n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n1} (1/2)*(1  (1)^j)*binomial(j+1, (j+1)/2)*p(n1j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section.  Wolfdieter Lang, Aug 11 2017
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the xaxis with the following constraints: The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches. If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3. c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
top UuUdUddd c=3 top UdUuUddd c=1 top UdUdUudd c=1
/\ /\
//\\ / \
// \\ / /\ \ /\
// \\ / / \ \ / \
///\ /\\\ /\ / / /\ \ \ /\ /\ / /\ \
\\\ \/ /// \ \ \ \/ / / / \ \ \ \/ / / /
\\\ /// \ \ \ / / / \ \ \ / / /
\\\/// \ \ \/ / / \ \ \/ / /
\\// \ \ / / \ \ / /
\/ \ \/ / \ \/ /
\ / \ /
\/ \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)


REFERENCES

J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137147.


LINKS

W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318328.
L. W. Shapiro, S. Getu, WenJin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229239.


FORMULA

a(n, m) := 0 if n<m or nm odd, else a(n, m) = (m+1)*binomial(n+1, (nm)/2)/(n+1);
a(n, m) = (4*(n1)*a(n2, m) + 2*(m+1)*a(n1, m1))/(n+m+2), a(n, m)=0 if n<m, a(n, 1) := 0, a(0, 0)=1=a(1, 1), a(1, 0)=0.
G.f. for mth column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.
G.f.: G(t,z) = c(z^2)/(1  t*z*c(z^2)), where c(z) = (1  sqrt(14*z))/(2*z) is the g.f. for the Catalan numbers (A000108).  Emeric Deutsch, Jun 16 2011
a(n, m) = a(n1, m1) + a(n1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0.  Henry Bottomley, Jan 25 2001
Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even.  Philippe Deléham, May 26 2005
T(n,k)=sum{i=0..n, (1)^(ni)*C(n,i)*sum{j=0..i, C(i,j)*(C(ij,j+k)C(ij,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)BesselI(k+2,2x).  Paul Barry, Feb 16 2006
Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively.  Philippe Deléham, Nov 22 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = 1, 0, 1, 2, 3 respectively.  Philippe Deléham, Nov 28 2009
Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n1k, x), n >= 0, with C(1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108.  Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015
The BoasBuck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(nm))*Sum_{j=0..n1m} (1/2)*(1  (1)^j)*binomial(j+1, (j+1)/2)* a(n1j, k), for n > m >= 0 and input a(m, m) = 1.  Wolfdieter Lang, Aug 11 2017


EXAMPLE

Triangle a(n,m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 0 1
2: 1 0 1
3: 0 2 0 1
4: 2 0 3 0 1
5: 0 5 0 4 0 1
6: 5 0 9 0 5 0 1
7: 0 14 0 14 0 6 0 1
8: 14 0 28 0 20 0 7 0 1
9: 0 42 0 48 0 27 0 8 0 1
10: 42 0 90 0 75 0 35 0 9 0 1
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
Production matrix is
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
BoasBuck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9.  Wolfdieter Lang, Aug 11 2017


MAPLE

T:=proc(n, k): if n+k mod 2 = 0 then (k+1)*binomial(n+1, (nk)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
F:=proc(l, p) if ((lp) mod 2) = 1 then 0 else (p+1)*l!/( ( (lp)/2 )! * ( (l+p)/2 +1)! ); fi; end;
r:=n>[seq( F(n, p), p=0..n)]; [seq(r(n), n=0..15)]; # N. J. A. Sloane, Jan 29 2011
A053121 := proc(n, k) option remember; `if`(k>n or k<0, 0, `if`(n=k, 1,
procname(n1, k1)+procname(n1, k+1))) end proc:


MATHEMATICA

a[n_, m_] /; n < m  OddQ[nm] = 0; a[n_, m_] = (m+1) Binomial[n+1, (nm)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* JeanFrançois Alcover, May 18 2011 *)


PROG

(Haskell)
a053121 n k = a053121_tabl !! n !! k
a053121_row n = a053121_tabl !! n
a053121_tabl = iterate
(\row > zipWith (+) ([0] ++ row) (tail row ++ [0, 0])) [1]
(Sage)
M = matrix(ZZ, dim, dim)
for n in (0..dim1): M[n, n] = 1
for n in (1..dim1):
for k in (0..n1):
M[n, k] = M[n1, k1] + M[n1, k+1]
return M
(PARI) T(n, m)=if(n<m(nm)%2, return(0)); (m+1)*binomial(n+1, (nm)/2)/(n+1)


CROSSREFS

Variant without zerodiagonals: A033184 and with rows reversed: A009766.


KEYWORD



AUTHOR



EXTENSIONS



STATUS

approved



