cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A134264 Coefficients T(j, k) of a partition transform for Lagrange compositional inversion of a function or generating series in terms of the coefficients of the power series for its reciprocal. Enumeration of noncrossing partitions and primitive parking functions. T(n,k) for n >= 1 and 1 <= k <= A000041(n-1), an irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1, 1, 7, 7, 7, 21, 42, 21, 21, 35, 105, 35, 35, 70, 21, 1, 1, 8, 8, 8, 4, 28, 56, 56, 28, 28, 56, 168, 84, 168, 14, 70, 280, 140, 56, 140, 28, 1, 1, 9, 9, 9, 9, 36, 72
Offset: 1

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Author

Tom Copeland, Jan 14 2008

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal to 0, form h(t) = t / f(t) and let h_n denote the coefficient of t^n in h(t).
Lagrange inversion gives the compositional inverse about t=0 as g(t) = Sum_{j>=1} ( t^j * (1/j) * Sum_{permutations s with s(1) + s(2) + ... + s(j) = j - 1} h_s(1) * h_s(2) * ... * h_s(j) ) = t * T(1,1) * h_0 + Sum_{j>=2} ( t^j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) ), where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.
Denoting h_n by (n') for brevity, to 8th order in t,
g(t) = t * (0')
+ t^2 * [ (0') (1') ]
+ t^3 * [ (0')^2 (2') + (0') (1')^2 ]
+ t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ]
+ t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ]
+ t^6 * [ (0')^5 (5') + 5 (0')^4 (1') (4') + 5 (0')^4 (2') (3') + 10 (0')^3 (1')^2 (3') + 10 (0')^3 (1') (2')^2 + 10 (0')^2 (1')^3 (2') + (0') (1')^5 ]
+ t^7 * [ (0')^6 (6') + 6 (0')^5 (1') (5') + 6 (0')^5 (2') (4') + 3 (0')^5 (3')^2 + 15 (0')^4 (1')^2 (4') + 30 (0')^4 (1') (2') (3') + 5 (0')^4 (2')^3 + 20 (0')^3 (1')^3 (3') + 30 (0')^3 (1')^2 (2')^2 + 15 (0')^2 (1')^4 (2') + (0') (1')^6]
+ t^8 * [ (0')^7 (7') + 7 (0')^6 (1') (6') + 7 (0')^6 (2') (5') + 7 (0')^6 (3') (4') + 21 (0')^5 (1')^2* (5') + 42 (0')^5 (1') (2') (4') + 21 (0')^5 (1') (3')^2 + 21 (0')^5 (2')^2 (3') + 35 (0')^4 (1')^3 (4') + 105 (0)^4 (1')^2 (2') (3') + 35 (0')^4 (1') (2')^3 + 35 (0')^3 (1')^4 (3') + 70 (0')^3 (1')^3 (2')^2 + 21 (0')^2 (1')^5 (2') + (0') (1')^7 ]
+ ..., where from the formula section, for example, T(8,1',2',...,7') = 7! / ((8 - (1'+ 2' + ... + 7'))! * 1'! * 2'! * ... * 7'!) are the coefficients of the integer partitions (1')^1' (2')^2' ... (7')^7' in the t^8 term.
A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.
If the coefficients of partitions with the same exponent for h_0 are summed within rows, A001263 is obtained, omitting the leading 1.
From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014
From Tom Copeland, Oct 28-29 2014: (Start)
Operating with d/d(1') = d/d(h_1) on the n-th partition polynomial Prt(n;h_0,h_1,..,h_n) in square brackets above associated with t^(n+1) generates n * Prt(n-1;h_0,h_1,..,h_(n-1)); therefore, the polynomials are an Appell sequence of polynomials in the indeterminate h_1 when h_0=1 (a special type of Sheffer sequence).
Consequently, umbrally, [Prt(.;1,x,h_2,..) + y]^n = Prt(n;1,x+y,h_2,..); that is, Sum_{k=0..n} binomial(n,k) * Prt(k;1,x,h_2,..) * y^(n-k) = Prt(n;1,x+y,h_2,..).
Or, e^(x*z) * exp[Prt(.;1,0,h_2,..) * z] = exp[Prt(.;1,x,h_2,..) * z]. Then with x = h_1 = -(1/2) * d^2[f(t)]/dt^2 evaluated at t=0, the formal Laplace transform from z to 1/t of this expression generates g(t), the comp. inverse of f(t), when h_0 = 1 = df(t)/dt eval. at t=0.
I.e., t / (1 - t*(x + Prt(.;1,0,h_2,..))) = t / (1 - t*Prt(.;1,x,h_2,..)) = g(t), interpreted umbrally, when h_0 = 1.
(End)
Connections to and between arrays associated to the Catalan (A000108 and A007317), Riordan (A005043), Fibonacci (A000045), and Fine (A000957) numbers and to lattice paths, e.g., the Motzkin, Dyck, and Łukasiewicz, can be made explicit by considering the inverse in x of the o.g.f. of A104597(x,-t), i.e., f(x) = P(Cinv(x),t-1) = Cinv(x) / (1 + (t-1)*Cinv(x)) = x*(1-x) / (1 + (t-1)*x*(1-x)) = (x-x^2) / (1 + (t-1)*(x-x^2)), where Cinv(x) = x*(1-x) is the inverse of C(x) = (1 - sqrt(1-4*x)) / 2, a shifted o.g.f. for the Catalan numbers, and P(x,t) = x / (1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x / (1-t*x). Then h(x,t) = x / f(x,t) = x * (1+(t-1)Cinv(x)) / Cinv(x) = 1 + t*x + x^2 + x^3 + ..., i.e., h_1=t and all other coefficients are 1, so the inverse of f(x,t) in x, which is explicitly in closed form finv(x,t) = C(Pinv(x,t-1)), is given by A091867, whose coefficients are sums of the refined Narayana numbers above obtained by setting h_1=(1')=t in the partition polynomials and all other coefficients to one. The group generators C(x) and P(x,t) and their inverses allow associations to be easily made between these classic number arrays. - Tom Copeland, Nov 03 2014
From Tom Copeland, Nov 10 2014: (Start)
Inverting in x with t a parameter, let F(x;t,n) = x - t*x^(n+1). Then h(x) = x / F(x;t,n) = 1 / (1-t*x^n) = 1 + t*x^n + t^2*x^(2n) + t^3*x^(3n) + ..., so h_k vanishes unless k = m*n with m an integer in which case h_k = t^m.
Finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). [Added braces to disambiguate the formula. - N. J. A. Sloane, Oct 20 2015]
This relation reveals properties of the partitions and sums of the coefficients of the array. For n=1, h_k = t^k for all k, implying that the row sums are the Catalan numbers. For n = 2, h_k for k odd vanishes, implying that there are no blocks with only even-indexed h_k on the even-numbered rows and that only the blocks containing only even-sized bins contribute to the odd-row sums giving the Fuss-Catalan numbers for n=2. And so on, for n > 2.
These relations are reflected in any combinatorial structures enumerated by this array and the partitions, such as the noncrossing partitions depicted for a five-element set (a pentagon) in Wikipedia.
(End)
From Tom Copeland, Nov 12 2014: (Start)
An Appell sequence possesses an umbral inverse sequence (cf. A249548). The partition polynomials here, Prt(n;1,h_1,...), are an Appell sequence in the indeterminate h_1=u, so have an e.g.f. exp[Prt(.;1,u,h_2...)*t] = e^(u*t) * exp[Prt(.;1,0,h2,...)*t] with umbral inverses with an e.g.f e^(-u*t) / exp[Prt(.;1,0,h2,...)*t]. This makes contact with the formalism of A133314 (cf. also A049019 and A019538) and the signed, refined face partition polynomials of the permutahedra (or their duals), which determine the reciprocal of exp[Prt(.,0,u,h2...)*t] (cf. A249548) or exp[Prt(.;1,u,h2,...)*t], forming connections among the combinatorics of permutahedra and the noncrossing partitions, Dyck paths and trees (cf. A125181), and many other important structures isomorphic to the partitions of this entry, as well as to formal cumulants through A127671 and algebraic structures of Lie algebras. (Cf. relationship of permutahedra with the Eulerians A008292.)
(End)
From Tom Copeland, Nov 24 2014: (Start)
The n-th row multiplied by n gives the number of terms in the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n+1)]^n under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1^2], and 3 * A134264(3) = 3 *(1,1)= (3,3) the number of summands in the two homogeneous polynomials in the square brackets. For n=3, [a + b + c + d]^3 = [a^3 + b^3 + ...] + 3 [a*b^2 + a*c^2 + ...] + 6 [a*b*c + a*c*d + ...] maps to [4 * h_3] + 3 [12 * h_1 * h_2] + 6 [4 * (h_1)^3], and the number of terms in the brackets is given by 4 * A134264(4) = 4 * (1,3,1) = (4,12,4).
The further reduced expression is 4 h_3 + 36 h_1 h_2 + 24 (h_1)^3 = A248120(4) with h_0 = 1. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
Abramowitz and Stegun give combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link.
(End)
These partition polynomials are dubbed the Voiculescu polynomials on page 11 of the He and Jejjala link. - Tom Copeland, Jan 16 2015
See page 5 of the Josuat-Verges et al. reference for a refinement of these partition polynomials into a noncommutative version composed of nondecreasing parking functions. - Tom Copeland, Oct 05 2016
(Per Copeland's Oct 13 2014 comment.) The number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition, where the ordering of integer partitions is first by total, then by length, then lexicographically by the reversed sequence of parts. - Gus Wiseman, Feb 15 2019
With h_0 = 1 and the other h_n replaced by suitably signed partition polynomials of A263633, the refined face partition polynomials for the associahedra of normalized A133437 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Sep 09 2019
Number of primitive parking functions associated to each partition of n. See Lemma 3.8 on p. 28 of Rattan. - Tom Copeland, Sep 10 2019
With h_n = n + 1, the d_k (A006013) of Table 2, p. 18, of Jong et al. are obtained, counting the n-point correlation functions in a quantum field theory. - Tom Copeland, Dec 25 2019
By inspection of the diagrams on Robert Dickau's website, one can see the relationship between the monomials of this entry and the connectivity of the line segments of the noncrossing partitions. - Tom Copeland, Dec 25 2019
Speicher has examples of the first four inversion partition polynomials on pp. 22 and 23 with his k_n equivalent to h_n = (n') here with h_0 = 1. Identifying z = t, C(z) = t/f(t) = h(t), and M(z) = f^(-1)(t)/t, then statement (3), on p. 43, of Theorem 3.26, C(z M(z)) = M(z), is equivalent to substituting f^(-1)(t) for t in t/f(t), and statement (4), M(z/C(z)) = C(z), to substituting f(t) for t in f^(-1)(t)/t. - Tom Copeland, Dec 08 2021
Given a Laurent series of the form f(z) = 1/z + h_1 + h_2 z + h_3 z^2 + ..., the compositional inverse is f^(-1)(z) = 1/z + Prt(1;1,h_1)/z^2 + Prt(2;1,h_1,h_2)/z^3 + ... = 1/z + h_1/z^2 + (h_1^2 + h_2)/z^3 + (h_1^3 + 3 h_1 h_2 + h_3)/z^4 + (h_1^4 + 6 h_1^2 h_2 + 4 h_1 h_3 + 2 h_2^2 + h_4)/z^5 + ... for which the polynomials in the numerators are the partition polynomials of this entry. For example, this formula applied to the q-expansion of Klein's j-invariant / function with coefficients A000521, related to monstrous moonshine, gives the compositional inverse with the coefficients A091406 (see He and Jejjala). - Tom Copeland, Dec 18 2021
The partition polynomials of A350499 'invert' the polynomials of this entry giving the indeterminates h_n. A multinomial formula for the coefficients of the partition polynomials of this entry, equivalent to the multinomial formula presented in the first four sentences of the formula section below, is presented in the MathOverflow question referenced in A350499. - Tom Copeland, Feb 19 2022

Examples

			1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).
2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).
3) With f(t) = (e^(-a*t)-1) / (-a), h(t) = Sum_{n>=0} Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = Sum_{n>=1} a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, Sum_{permutations s with s(1)+s(2)+...+s(j)=j-1} h_s(1) * h_s(2) * ... * h_s(j) = j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) = a^(j-1). Note, in turn, Sum_{a=1..m} a^(j-1) = (Bernoulli(j,m+1) - Bernoulli(j)) / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x) = t / (x-1+1/(1-t)), then h(t,x) = x-1+1/(1-t), giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = (1-(1-x)*t-sqrt(1-2*(1+x)*t+((x-1)*t)^2)) / 2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.
5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)
6) With f(t,x) = x / ((1-t*x)*(1-(1+t)*x)), an o.g.f. for A074909, the reverse face polynomials of the simplices, h(t,x) = (1-t*x) * (1-(1+t)*x) with h_0=1, h_1=-(1+2*t), and h_2=t*(1+t), giving as the inverse in x about 0 the o.g.f. (1+(1+2*t)*x-sqrt(1+(1+2*t)*2*x+x^2)) / (2*t*(1+t)*x) for signed A033282, the reverse face polynomials of the Stasheff polytopes, or associahedra. Cf. A248727. (Added Jan 21 2015 by author.)
7) With f(x,t) = x / ((1+x)*(1+t*x)), an o.g.f. for the polynomials (-1)^n * (1 + t + ... + t^n), h(t,x) = (1+x) * (1+t*x) with h_0=1, h_1=(1+t), and h_2=t, giving as the inverse in x about 0 the o.g.f. (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2)) / (2*x*t) for the Narayana polynomials A001263. Cf. A046802. (Added Jan 24 2015 by author.)
From _Gus Wiseman_, Feb 15 2019: (Start)
Triangle begins:
   1
   1
   1   1
   1   3   1
   1   4   2   6   1
   1   5   5  10  10  10   1
   1   6   6   3  15  30   5  20  30  15   1
   1   7   7   7  21  42  21  21  35 105  35  35  70  21   1
Row 5 counts the following non-crossing set partitions:
  {{1234}}  {{1}{234}}  {{12}{34}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
            {{123}{4}}  {{14}{23}}  {{1}{23}{4}}
            {{124}{3}}              {{12}{3}{4}}
            {{134}{2}}              {{1}{24}{3}}
                                    {{13}{2}{4}}
                                    {{14}{2}{3}}
(End)
		

References

  • A. Nica and R. Speicher (editors), Lectures on the Combinatorics of Free Probability, London Mathematical Society Lecture Note Series: 335, Cambridge University Press, 2006 (see in particular, Eqn. 9.14 on p. 141, enumerating noncrossing partitions).

Crossrefs

(A001263,A119900) = (reduced array, associated g(x)). See A145271 for meaning and other examples of reduced and associated.
Other orderings are A125181 and A306438.
Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).
Cf. A248927 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A091867 and A125181, for relations to lattice paths and trees.
Cf. A249548 for use of Appell properties to generate the polynomials.
Cf. A133314, A049019, A019538, A127671, and A008292 for relations to permutahedra, Eulerians.
Cf. A006013.

Programs

  • Mathematica
    Table[Binomial[Total[y],Length[y]-1]*(Length[y]-1)!/Product[Count[y,i]!,{i,Max@@y}],{n,7},{y,Sort[Sort/@IntegerPartitions[n]]}] (* Gus Wiseman, Feb 15 2019 *)
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(Vec(p)) | p<-partitions(n-1)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.
For example from g(t) above, T(5,4) = (5! / ((5-3)! * 2!)) / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.
From Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.
Then for the partition polynomials of A134264,
Poly[n;h_0,...,h_(n-1)]=(1/n!)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t) = exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1.
(End)
From Tom Copeland, Oct 20 2011: (Start)
With exp(x* PS(.,t)) = exp(t*g(x)) = exp(x*W(y)d/dy) exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are
R = t*W(d/dt) = t*((h_0) + (h_1)d/dt + (h_2)(d/dt)^2 + ...)^2 / ((h_0) - (h_2)(d/dt)^2 - 2(h_3)(d/dt)^3 - 3(h_4)(d/dt)^4 + ...), and
L = (d/dt)/h(d/dt) = (d/dt) 1/((h_0) + (h_1)*d/dt + (h_2)*(d/dt)^2 + ...)
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
(End)
Using the formalism of A263634, the raising operator for the partition polynomials of this array with h_0 = 1 begins as R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 + 5 h_2^3 - 7 h_3^2 - 9 h_2 h_4) D^5/5! + (h_7 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... with D = d/d(h_1). - Tom Copeland, Sep 09 2016
Let h(x) = x/f^{-1}(x) = 1/[1-(c_2*x+c_3*x^2+...)], with c_n all greater than zero. Then h_n are all greater than zero and h_0 = 1. Determine P_n(t) from exp[t*f^{-1}(x)] = exp[x*P.(t)] with f^{-1}(x) = x/h(x) expressed in terms of the h_n (cf. A133314 and A263633). Then P_n(b.) = 0 gives a recursion relation for the inversion polynomials of this entry a_n = b_n/n! in terms of the lower order inversion polynomials and P_j(b.)P_k(b.) = P_j(t)P_k(t)|{t^n = b_n} = d{j,k} >= 0 is the coefficient of x^j/j!*y^k/k! in the Taylor series expansion of the formal group law FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)]. - Tom Copeland, Feb 09 2018
A raising operator for the partition polynomials with h_0 = 1 regarded as a Sheffer Appell sequence in h_1 is described in A249548. - Tom Copeland, Jul 03 2018

Extensions

Added explicit t^6, t^7, and t^8 polynomials and extended initial table to include the coefficients of t^8. - Tom Copeland, Sep 14 2016
Title modified by Tom Copeland, May 28 2018
More terms from Gus Wiseman, Feb 15 2019
Title modified by Tom Copeland, Sep 10 2019

A060693 Triangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 10, 6, 1, 14, 35, 30, 10, 1, 42, 126, 140, 70, 15, 1, 132, 462, 630, 420, 140, 21, 1, 429, 1716, 2772, 2310, 1050, 252, 28, 1, 1430, 6435, 12012, 12012, 6930, 2310, 420, 36, 1, 4862, 24310, 51480, 60060, 42042, 18018, 4620, 660, 45, 1, 16796
Offset: 0

Views

Author

F. Chapoton, Apr 20 2001

Keywords

Comments

The rows sum to A006318 (Schroeder numbers), the left column is A000108 (Catalan numbers); the next-to-left column is A001700, the alternating sum in each row but the first is 0.
T(n,k) is the number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k peaks. Example: T(2,1)=3 because we have UU*DD, U*DH and HU*D, the peaks being shown by *. E.g., T(n,k) = binomial(n,k)*binomial(2n-k,n-1)/n for n>0. - Emeric Deutsch, Dec 06 2003
A090181*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 14 2008
T(n,k) is also the number of rooted plane trees with maximal degree 3 and k vertices of degree 2 (a node may have at most 2 children, and there are exactly k nodes with 1 child). Equivalently, T(n,k) is the number of syntactically different expressions that can be formed that use a unary operation k times, a binary operation n-k times, and nothing else (sequence of operands is fixed). - Lars Hellstrom (Lars.Hellstrom(AT)residenset.net), Dec 08 2009

Examples

			Triangle begins:
00: [    1]
01: [    1,     1]
02: [    2,     3,      1]
03: [    5,    10,      6,      1]
04: [   14,    35,     30,     10,      1]
05: [   42,   126,    140,     70,     15,      1]
06: [  132,   462,    630,    420,    140,     21,     1]
07: [  429,  1716,   2772,   2310,   1050,    252,    28,    1]
08: [ 1430,  6435,  12012,  12012,   6930,   2310,   420,   36,   1]
09: [ 4862, 24310,  51480,  60060,  42042,  18018,  4620,  660,  45,  1]
10: [16796, 92378, 218790, 291720, 240240, 126126, 42042, 8580, 990, 55, 1]
...
		

Crossrefs

Triangle in A088617 transposed.
T(2n,n) gives A007004.

Programs

  • Maple
    A060693 := (n,k) -> binomial(n,k)*binomial(2*n-k,n)/(n-k+1); # Peter Luschny, May 17 2011
  • Mathematica
    t[n_, k_] := Binomial[n, k]*Binomial[2 n - k, n]/(n - k + 1); Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]] (* Robert G. Wilson v, May 30 2011 *)
  • PARI
    T(n, k) = binomial(n, k)*binomial(2*n - k, n)/(n - k + 1);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ")); print); \\ Indranil Ghosh, Jul 28 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return binomial(n, k) * binomial(2 * n - k, n) / (n - k + 1)
    for n in range(11): print([T(n, k) for k in range(n + 1)])  # Indranil Ghosh, Jul 28 2017

Formula

Triangle T(n, k) (0 <= k <= n) read by rows; given by [1, 1, 1, 1, 1, ...] DELTA [1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 12 2003
If C_n(x) is the g.f. of row n of the Narayana numbers (A001263), C_n(x) = Sum_{k=1..n} binomial(n,k-1)*(binomial(n-1,k-1)/k) * x^k and T_n(x) is the g.f. of row n of T(n,k), then T_n(x) = C_n(x+1), or T(n,k) = [x^n]Sum_{k=1..n}(A001263(n,k)*(x+1)^k). - Mitch Harris, Jan 16 2007, Jan 31 2007
G.f.: (1 - t*y - sqrt((1-y*t)^2 - 4*y)) / 2.
T(n, k) = binomial(2n-k, n)*binomial(n, k)/(n-k+1). - Philippe Deléham, Dec 07 2003
A060693(n, k) = binomial(2*n-k, k)*A000108(n-k); A000108: Catalan numbers. - Philippe Deléham, Dec 30 2003
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n), for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Apr 01 2007
T(n,k) = Sum_{j>=0} A090181(n,j)*binomial(j,k). - Philippe Deléham, May 04 2007
Sum_{k=0..n} T(n,k)*x^(n-k) = (-1)^n*A107841(n), A080243(n), A000007(n), A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Oct 18 2007
From Paul Barry, Jan 29 2009: (Start)
G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-.... (continued fraction);
G.f.: 1/(1-(x+xy)/(1-x/(1-(x+xy)/(1-x/(1-(x+xy)/(1-.... (continued fraction). (End)
T(n,k) = [k<=n]*(Sum_{j=0..n} binomial(n,j)^2*binomial(j,k))/(n-k+1). - Paul Barry, May 28 2009
T(n,k) = A104684(n,k)/(n-k+1). - Peter Luschny, May 17 2011
From Tom Copeland, Sep 21 2011: (Start)
With F(x,t) = (1-(2+t)*x-sqrt(1-2*(2+t)*x+(t*x)^2))/(2*x) an o.g.f. (nulling the n=0 term) in x for the A060693 polynomials in t,
G(x,t) = x/(1+t+(2+t)*x+x^2) is the compositional inverse in x.
Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+t+(2+t)*x+x^2)^2 / (1+t-x^2), the n-th A060693 polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n) x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/d) u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). (End)
See my 2008 formulas in A033282 to relate this entry to A088617, A001263, A086810, and other matrices. - Tom Copeland, Jan 22 2016
Rows of this entry are non-vanishing antidiagonals of A097610. See p. 14 of Agapito et al. for a bivariate generating function and its inverse. - Tom Copeland, Feb 03 2016
From Werner Schulte, Jan 09 2017: (Start)
T(n,k) = A126216(n,k-1) + A126216(n,k) for 0 < k < n;
Sum_{k=0..n} (-1)^k*(1+x*(n-k))*T(n,k) = x + (1-x)*A000007(n).
(End)
Conjecture: Sum_{k=0..n} (-1)^k*T(n,k)*(n+1-k)^2 = 1+n+n^2. - Werner Schulte, Jan 11 2017

Extensions

More terms from Vladeta Jovovic, Apr 21 2001
New description from Philippe Deléham, Aug 12 2003
New name using a comment by Emeric Deutsch from Peter Luschny, Jul 26 2017

A133437 Irregular triangle of coefficients of a partition transform for direct Lagrange inversion of an o.g.f., complementary to A134685 for an e.g.f.; normalized by the factorials, these are signed, refined face polynomials of the associahedra.

Original entry on oeis.org

1, -2, 12, -6, -120, 120, -24, 1680, -2520, 360, 720, -120, -30240, 60480, -20160, -20160, 5040, 5040, -720, 665280, -1663200, 907200, 604800, -60480, -362880, -181440, 20160, 40320, 40320, -5040, -17297280, 51891840, -39916800, -19958400, 6652800, 19958400, 6652800, -1814400, -1814400, -3628800, -1814400, 362880, 362880, 362880, -40320
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Sum_{n>=1} u_n * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -2 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 12 (2')^2 - 6 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -120 (2')^3 + 120 (1')(2')(3') - 24 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 1680 (2')^4 - 2520 (1') (2')^2 (3') + 360 (1')^2 (3')^2 + 720 (1')^2 (2') (4') - 120 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -30240 (2')^5 + 60480 (1') (2')^3 (3') - 20160 (1')^2 (2') (3')^2 - 20160 (1')^2 (2')^2 (4') + 5040 (1')^3 (3')(4') + 5040 (1')^3 (2')(5') - 720 (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 665280 (2')^6 - 1663200 (1')(2')^4(3') + (1')^2 [907200 (2')^2(3')^2 + 604800 (2')^3(4')] - (1')^3 [60480 (3')^3 + 362880 (2')(3')(4') + 181440 (2')^2(5')] + (1')^4 [20160 (4')^2 + 40320 (3')(5') + 40320 (2')(6')] - 5040 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -17297280 (2')^7 + 51891840 (1')(2')^5(3') - (1')^2 [39916800 (2')^3(3')^2 + 19958400 (2')^4(4')] + (1')^3 [6652800 (2')(3')^3 + 19958400 (2')^2(3')(4') + 6652800 (2')^3(5')] - (1')^4 [1814400 (3')^2(4') + 1814400 (2')(4')^2 + 3628800 (2')(3')(5') + 1814400 (2')^2(6')] + (1')^5 [362880 (4')(5') + 362880 (3')(6') + 362880 (2')(7')] - 40320 (1')^6(8')] * t^8 / 8!
...
See A134685 for more information.
A111785 is obtained from A133437 by dividing through the bracketed terms of the P(n,t) by n! and unsigned A111785 is a refinement of A033282 and A126216. - Tom Copeland, Sep 28 2008
For relation to the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see the Loday and McCammond links. E.g., P(5,t) = (1')^(-9) * [ 14 (2')^4 - 21 (1') (2')^2 (3') + 6 (1')^2 (2') (4')+ 3 (1')^2 (3')^2 - 1 (1')^3 (5') ] * t^5 is related to the 3-D associahedron with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Summing over faces of the same dimension gives A033282 or A126216. - Tom Copeland, Sep 29 2008
A relation between this Lagrange inversion for an o.g.f. and partition polynomials formed from the "refined Lah numbers" A130561 is presented in the link "Lagrange a la Lah" along with umbral binary tree representations.
With f(x,t) = x + t*Sum_{n>=2} u_n*x^n, the compositional inverse in x is related to the velocity profile of particles governed by the inviscid Burgers's, or Hopf, eqn. See A001764 and A086810. - Tom Copeland, Feb 15 2014
Newton was aware of this power series expansion for series reversion. See the Ferraro reference p. 75 eqn. 52. - Tom Copeland, Jan 22 2017
The coefficients of the partition polynomials divided by the associated factorial enumerate the faces of the convex, bounded polytopes called the associahedra, and the absolute value of the sum of the renormalized coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is either n! (unnormalized) or unity (normalized). In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With u_1 = 1 and the other u_n replaced by suitably signed partition polynomials of A263633, the partition polynomials enumerating the refined noncrossing partitions of A134264 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Nov 16 2019
Relations between associahedra and oriented n-simplices are presented by Halvorson and by Street. - Tom Copeland, Dec 08 2019
Let f(x;t,n) = x - t * x^(n+1), giving u_1 = (1') = 1 and u_(n+1) = (n+1) = -t. Then inverting in x with t a parameter gives finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). Comparing this with the same result in A134264 gives relations between the faces of associahedra and noncrossing partitions (and other combinatorial constructs related to these inversion formulas and those listed in A145271). - Tom Copeland, Jan 27 2020
From Tom Copeland, Mar 24 2020: (Start)
There is a mapping between the faces of K_n, the associahedron of dimension (n-1), and polygon dissections. The dissecting noncrossing diagonals (i.e., nonintersecting in the interior) form subpolygons. Assign the indeterminate x_k to a subpolygon where k = (number of vertices of the subpolygon) - 1. Multiply the x_k together to form the monomials for the inversion formula.
For the 3-dimensional associahedron K_4, the fundamental polygon is the hexagon, which can be dissected into pentagons, associated to x_4; tetragons, to x_3; and triangles, to x_2; for example, there are six distinguished partitions of the hexagon into one triangle and one pentagon, sharing two vertices, associated to the monomial 6 * x_2 * x_4 since the unshared vertex of the triangle can be moved consecutively from one vertex of the hexagon to the next. This term corresponds to 720 (1')^2 (2') (4') / 5! in P(5,t) above, denumerating the six pentagonal facets of K_4. (End)

References

  • G. Ferraro, The Rise and Development of the Theory of Series up to the Early 1820s, Springer Science and Business Media, 2007.
  • H. Halvorson (editor), Deep Beauty: Understanding the Quantum World Through Innovation, Cambridge Univ. Press, 2011.
  • H. Turnbull (editor), The Correspondence of Isaac Newton Vol. II 1676-1687, Cambridge Univ. Press, 1960, p. 147.

Crossrefs

Cf. A145271, (A086810, A181289) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k, {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ (e(2))! * (e(3))! * ... * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)^2]
= 1/((u_1) + 2*(u_2)*t + 3*(u_3)*t^2 + 4*(u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133437,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators defined by R PS(n,t)=PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are
R = t*h(d/dt) = t* 1/[(u_1) + 2*(u_2)*d/dt + 3*(u_3)*(d/dt)^2 + ...] and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2 + (u_3)*(d/dt)^3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x + u_3 * x^2 + ... + u_n * x^(n-1)]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / t^n, i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 6 u2 u4 + 3 u3^2, b3 = -21 u2^2 u3, and b4 = 14 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. A086810) implies that, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = [(n+1)/2] * Sum_{k = 2..n-1} PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 14 u2^4 - 2 * 21 u2^2 u3 + 1 * 6 u2 u4 + 1 * 3 u3^2 - 0 * u5 = 42 u2^4 - 42 u2^2 u3 + 6 u2 u4 + 3 u3^2 = 3 * [2 * PS(2,1,u2) * PS(4,1,u2,...,u4) + PS(3,1,u2,u3)^2] = 3 * [ 2 * (-u2) (-5 u2^3 + 5 u2 u3 - u4) + (2 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
From Tom Copeland, Sep 22 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = m!*u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently multiply the diagonals of A132159 by u_m. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ...)^T * t^n/n! in agreement with A139605. (End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Lah polynomials of A130561 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018
The derivative of the partition polynomials of A350499 with respect to a distinguished indeterminate give polynomials proportional to those of this entry. The connection of this derivative relation to the inviscid Burgers-Hopf evolution equation is given in a reference for that entry. - Tom Copeland, Feb 19 2022

Extensions

Missing coefficient in P(6,t) replaced by Tom Copeland, Nov 06 2008
P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Title modified by Tom Copeland, Jan 13 2020
Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 07 2024

A126216 Triangle read by rows: T(n,k) is the number of Schroeder paths of semilength n containing exactly k peaks but no peaks at level one (n >= 1; 0 <= k <= n-1).

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 14, 21, 9, 1, 42, 84, 56, 14, 1, 132, 330, 300, 120, 20, 1, 429, 1287, 1485, 825, 225, 27, 1, 1430, 5005, 7007, 5005, 1925, 385, 35, 1, 4862, 19448, 32032, 28028, 14014, 4004, 616, 44, 1, 16796, 75582, 143208, 148512, 91728, 34398, 7644, 936, 54, 1
Offset: 1

Views

Author

Emeric Deutsch, Dec 20 2006

Keywords

Comments

A Schroeder path of semilength n is a lattice path in the first quadrant, from the origin to the point (2n,0) and consisting of steps U=(1,1), D=(1,-1) and H=(2,0).
Also number of Schroeder paths of semilength n containing exactly k doublerises but no (2,0) steps at level 0 (n >= 1; 0 <= k <= n-1). Also number of doublerise-bicolored Dyck paths (doublerises come in two colors; also called marked Dyck paths) of semilength n and having k doublerises of a given color (n >= 1; 0 <= k <= n-1). Also number of 12312- and 121323-avoiding matchings on [2n] with exactly k crossings.
Essentially the triangle given by [1,1,1,1,1,1,1,1,...] DELTA [0,1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 20 2007
Mirror image of triangle A033282. - Philippe Deléham, Oct 20 2007
For relation to Lagrange inversion, or series reversion and the geometry of associahedra, or Stasheff polytopes (and other combinatorial objects), see A133437. - Tom Copeland, Sep 29 2008
First column (k=0) gives the Catalan numbers (A000108). - Alexander Karpov, Jun 10 2018
T(n,k) is the multiplicity of the k-hook representation of the symmetric group in the (n-1)st parking space representation (see Pak and Postnikov, 1995). - Joshua Mundinger, Jul 18 2025

Examples

			T(3,1)=5 because we have HUUDD, UUDDH, UUUDDD, UHUDD and UUDHD.
Triangle starts:
   n\k  0      1      2      3      4     5    6   7  8
   1    1;
   2    2,     1;
   3    5,     5;     1;
   4   14,    21,     9,     1;
   5   42,    84,    56,    14,     1;
   6  132,   330,   300,   120,    20,    1;
   7  429,  1287,  1485,   825,   225,   27,   1;
   8 1430,  5005,  7007,  5005,  1925,  385,  35,  1;
   9 4862, 19448, 32032, 28028, 14014, 4004, 616, 44, 1;
  10 ...
Triangle [1,1,1,1,1,1,1,...] DELTA [0,1,0,1,0,1,0,1,...] begins:
   1;
   1,  0;
   2,  1,  0;
   5,  5,  1,  0;
  14, 21,  9,  1,  0;
  42, 84, 56, 14,  1,  0;
  ...
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->binomial(n,k)*binomial(2*n-k,n+1)/n: for n from 1 to 11 do seq(T(n,k),k=0..n-1) od; # yields sequence in triangular form
  • Mathematica
    Table[Binomial[n, k] Binomial[2 n - k, n + 1]/n, {n, 10}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Jan 09 2016 *)
  • PARI
    tabl(nn) = {mP = matrix(nn, nn, n, k, binomial(n-1, k-1)); mN = matrix(nn, nn, n, k, binomial(n-1, k-1) * binomial(n, k-1) / k); mprod = mN*mP; for (n=1, nn, for (k=1, n, print1(mprod[n, k], ", ");); print(););} \\ Michel Marcus, Apr 16 2015
    
  • PARI
    t(n,k) = binomial(n,k)*binomial(2*n-k,n+1)/n;
    concat(vector(10, n, vector(n, k, t(n,k-1))))  \\ Gheorghe Coserea, Apr 24 2016

Formula

T(n,k) = C(n,k)*C(2*n-k,n+1)/n (0 <= k <= n-1).
G.f.: G(t,z) = (1-2*z-t*z-sqrt(1-4*z-2*t*z+t^2*z^2))/(2*(1+t)*z).
Equals N * P, where N = the Narayana triangle (A001263) and P = Pascal's triangle, as infinite lower triangular matrices. A126182 = P * N. - Gary W. Adamson, Nov 30 2007
G.f.: 1/(1-x-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A126216 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Oct 09 2011
From Tom Copeland, Oct 10 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = 1
P(3,t) = 2 + t
P(4,t) = 5 + 5 t + t^2
P(5,t) = 14 + 21 t + 9 t^2 + t^3
The o.g.f. A(x,t) = (1+x*t-sqrt((1-x*t)^2-4x))/(2(1+t)), and
B(x,t) = x - x^2/(1-t*x) = x - x^2 - ((t*x)^3 + (t*x)^4 + ...)/t^2 is the compositional inverse in x. [series corrected by Tom Copeland, Dec 10 2019]
Let h(x,t) = 1/(dB/dx) = (1-tx)^2/(1-(t+1)(2x-tx^2)) = 1/(1 - 2x - 3tx^2 + 4t^2x^3 + ...). Then P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(u,t)*d/du) u, evaluated at u=0, and dA/dx = h(A(x,t),t). (End)
From Tom Copeland, Dec 09 2019: (Start)
The polynomials in my 2011 formula entry above evaluate to an aerated, alternating sign sequence of the Catalan numbers A000108 with t = -2. The first few are P(2,-2) = 1, P(3,-2) = 0, P(4,t) = -1, P(5,-2) = 0, P(6,-2) = 2, P(7,-2) = 0, P(8,-2) = -5, P(9,-2) = 0, P(10,-2) = 14.
Generalizing the relations between w = theta and u = phi in Mizera on pp. 32-34, modify the inverse pair above to w = i * B(-i*u,t) = u + i * u^2/(1+i*t*u), where i is the imaginary number, and u = i*A(-i*w,t) = i*(1 - i*w*t - sqrt((1 + i*w*t)^2 + i*4*w))/(2(1+t)). Then the expression for V'(w) in Mizera generalizes to V'(w) = -i*(w - u) and reduces to V'(w) = (1 - sqrt(1-4 w^2))/2 when evaluated at t = -2, which is an o.g.f. for A126120. Cf. also A086810. (End)
Sum_{k = 0..n-1} (-1)^k*T(n,k)*binomial(x + 2*n - k, 2*n - k) = ( (x + 1) * ( Product_{k = 2..n} (x + k)^2 ) * (x + n + 1) )/(n!*(n + 1)!) for n >= 1. Cf. A243660 and A243661. - Peter Bala, Oct 08 2022

A295260 Array read by antidiagonals: T(n,k) = number of nonequivalent dissections of a polygon into n k-gons by nonintersecting diagonals up to rotation and reflection (k >= 3).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 2, 5, 4, 1, 1, 3, 8, 16, 12, 1, 1, 3, 12, 33, 60, 27, 1, 1, 4, 16, 68, 194, 261, 82, 1, 1, 4, 21, 112, 483, 1196, 1243, 228, 1, 1, 5, 27, 183, 1020, 3946, 8196, 6257, 733, 1, 1, 5, 33, 266, 1918, 10222, 34485, 58140, 32721, 2282
Offset: 1

Views

Author

Andrew Howroyd, Nov 18 2017

Keywords

Comments

The polygon prior to dissection will have n*(k-2)+2 sides.
In the Harary, Palmer and Read reference these are the sequences called h.
T(n,k) is the number of unoriented polyominoes containing n k-gonal tiles of the hyperbolic regular tiling with Schläfli symbol {k,oo}. Stereographic projections of several of these tilings on the Poincaré disk can be obtained via the Christensson link. For unoriented polyominoes, chiral pairs are counted as one. T(n,2) could represent polyominoes of the Euclidean regular tiling with Schläfli symbol {2,oo}; T(n,2) = 1. - Robert A. Russell, Jan 21 2024

Examples

			Array begins:
  ===================================================
  n\k|   3     4      5       6        7        8
  ---|-----------------------------------------------
   1 |   1     1      1       1        1        1 ...
   2 |   1     1      1       1        1        1 ...
   3 |   1     2      2       3        3        4 ...
   4 |   3     5      8      12       16       21 ...
   5 |   4    16     33      68      112      183 ...
   6 |  12    60    194     483     1020     1918 ...
   7 |  27   261   1196    3946    10222    22908 ...
   8 |  82  1243   8196   34485   109947   290511 ...
   9 | 228  6257  58140  315810  1230840  3844688 ...
  10 | 733 32721 427975 2984570 14218671 52454248 ...
  ...
		

Crossrefs

Columns k=3..7 are A000207, A005036, A005040, A004127, A005419.
Polyominoes: A295224 (oriented), A070914 (rooted).

Programs

  • Mathematica
    u[n_, k_, r_] := r*Binomial[(k - 1)*n + r, n]/((k - 1)*n + r);
    T[n_, k_] := (u[n, k, 1] + If[OddQ[n], u[(n - 1)/2, k, Quotient[k, 2]], If[OddQ[k], (u[n/2 - 1, k, k - 1] + u[n/2, k, 1])/2, u[n/2, k, 1]]] + (If[EvenQ[n], u[n/2, k, 1]] - u[n, k, 2])/2 + DivisorSum[GCD[n - 1, k], EulerPhi[#]*u[(n - 1)/#, k, k/#] &]/k)/2 /. Null -> 0;
    Table[T[n - k + 2, k + 1], {n, 1, 11}, {k, n + 1, 2, -1}] // Flatten (* Jean-François Alcover, Dec 28 2017, after Andrew Howroyd *)
  • PARI
    \\ here u is Fuss-Catalan sequence with p = k+1.
    u(n,k,r) = {r*binomial((k - 1)*n + r, n)/((k - 1)*n + r)}
    T(n,k) = {(u(n,k,1) + if(n%2, u((n-1)/2,k,k\2), if(k%2, (u(n/2-1,k,(k-1)) + u(n/2,k,1))/2, u(n/2,k,1))) + (if(n%2==0, u(n/2,k,1))-u(n,k,2))/2 + sumdiv(gcd(n-1,k), d, eulerphi(d)*u((n-1)/d,k,k/d))/k)/2}
    for(n=1, 10, for(k=3, 8, print1(T(n, k), ", ")); print);

Formula

T(n,k) ~ A295222(n,k)/(2*n) for fixed k.

A295224 Array read by antidiagonals: T(n,k) = number of nonequivalent dissections of a polygon into n k-gons by nonintersecting diagonals up to rotation (k >= 3).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 1, 2, 7, 6, 1, 1, 3, 12, 25, 19, 1, 1, 3, 19, 57, 108, 49, 1, 1, 4, 26, 118, 366, 492, 150, 1, 1, 4, 35, 203, 931, 2340, 2431, 442, 1, 1, 5, 46, 332, 1989, 7756, 16252, 12371, 1424, 1, 1, 5, 57, 494, 3766, 20254, 68685, 115940, 65169, 4522
Offset: 1

Views

Author

Andrew Howroyd, Nov 17 2017

Keywords

Comments

The polygon prior to dissection will have n*(k-2)+2 sides.
In the Harary, Palmer and Read reference these are the sequences called H.
T(n,k) is the number of oriented polyominoes containing n k-gonal tiles of the hyperbolic regular tiling with Schläfli symbol {k,oo}. Stereographic projections of several of these tilings on the Poincaré disk can be obtained via the Christensson link. For oriented polyominoes, chiral pairs are counted as two. T(n,2) could represent polyominoes of the Euclidean regular tiling with Schläfli symbol {2,oo}; T(n,2) = 1. - Robert A. Russell, Jan 21 2024

Examples

			Array begins:
  =====================================================
  n\k|    3     4      5       6        7         8
  ---|-------------------------------------------------
   1 |    1     1      1       1        1         1 ...
   2 |    1     1      1       1        1         1 ...
   3 |    1     2      2       3        3         4 ...
   4 |    4     7     12      19       26        35 ...
   5 |    6    25     57     118      203       332 ...
   6 |   19   108    366     931     1989      3766 ...
   7 |   49   492   2340    7756    20254     45448 ...
   8 |  150  2431  16252   68685   219388    580203 ...
   9 |  442 12371 115940  630465  2459730   7684881 ...
  10 | 1424 65169 854981 5966610 28431861 104898024 ...
  ...
		

Crossrefs

Columns k=3..6 are A001683(n+2), A005034, A005038, A221184(n-1).
Polyominoes: A295260 (unoriented), A070914 (rooted).

Programs

  • Mathematica
    u[n_, k_, r_] := r*Binomial[(k - 1)*n + r, n]/((k - 1)*n + r);
    T[n_, k_] := u[n, k, 1] + (If[EvenQ[n], u[n/2, k, 1], 0] - u[n, k, 2])/2 + DivisorSum[GCD[n - 1, k], EulerPhi[#]*u[(n - 1)/#, k, k/#]&]/k;
    Table[T[n - k + 1, k], {n, 1, 13}, {k, n, 3, -1}] // Flatten (* Jean-François Alcover, Nov 21 2017, after Andrew Howroyd *)
  • PARI
    \\ here u is Fuss-Catalan sequence with p = k+1.
    u(n, k, r)={r*binomial((k - 1)*n + r, n)/((k - 1)*n + r)}
    T(n,k) = u(n,k,1) + (if(n%2==0, u(n/2,k,1))-u(n,k,2))/2 + sumdiv(gcd(n-1,k), d, eulerphi(d)*u((n-1)/d,k,k/d))/k;
    for(n=1, 10, for(k=3, 8, print1(T(n, k), ", ")); print);
    
  • Python
    from sympy import binomial, gcd, totient, divisors
    def u(n, k, r): return r*binomial((k - 1)*n + r, n)//((k - 1)*n + r)
    def T(n, k): return u(n, k, 1) + ((u(n//2, k, 1) if n%2==0 else 0) - u(n, k, 2))//2 + sum([totient(d)*u((n - 1)//d, k, k//d) for d in divisors(gcd(n - 1, k))])//k
    for n in range(1, 11): print([T(n, k) for k in range(3, 9)]) # Indranil Ghosh, Dec 13 2017, after PARI code

Formula

T(n,k) ~ A295222(n,k)/n for fixed k.

A181289 Triangle read by rows: T(n,k) is the number of 2-compositions of n having length k (0 <= k <= n).

Original entry on oeis.org

1, 0, 2, 0, 3, 4, 0, 4, 12, 8, 0, 5, 25, 36, 16, 0, 6, 44, 102, 96, 32, 0, 7, 70, 231, 344, 240, 64, 0, 8, 104, 456, 952, 1040, 576, 128, 0, 9, 147, 819, 2241, 3400, 2928, 1344, 256, 0, 10, 200, 1372, 4712, 9290, 11040, 7840, 3072, 512, 0, 11, 264, 2178, 9108, 22363
Offset: 0

Views

Author

Emeric Deutsch, Oct 12 2010

Keywords

Comments

A 2-composition of n is a nonnegative matrix with two rows, such that each column has at least one nonzero entry and whose entries sum up to n. The length of the 2-composition is the number of columns.
From Tom Copeland, Sep 06 2011: (Start)
R(t,z) = (1-z)^2 / ((1+t)*(1-z)^2-1) = 1/(t - (2*z + 3*z^2 + 4*z^3 + 5*z^4 + ...)) = 1/t + (1/t)^2*2*z + (1/t)^3*(4+3t)*z^2 + (1/t)^4*(8+12*t+4*t^2)*z^3 + ... gives row reversed polynomials of A181289 with G(t,z) = R(1/t,z)/t.
R(t,z) is related to generators for A033282 and A001003 (t=1) and can be umbrally extended to give a partition generator for A133437. (End)
A refined, reverse version of this array is given in A253722. - Tom Copeland, May 02 2015
The infinitesimal generator (infinigen) for the face polynomials of associahedra A086810/A033282, read as decreasing powers, (and for the dual simplicial complex read as increasing powers) can be formed from the row polynomials P(n,t) of this entry. This type of infinigen is presented in A145271 for general sets of binomial Sheffer polynomials. This specific infinigen is presented in analytic form in A086810. Given the column vector of row polynomials V = (P(0,t) = 1, P(1,y) = 2 t, P(2,y) = 3 t + 4 t^2, P(3,y) = 4 t + 12 t^2 + 8 t^3, ...), form the lower triangular matrix M(n,k) = V(n-k,n-k), i.e., diagonally multiply the matrix with all ones on the diagonal and below by the components of V. Form the matrix MD by multiplying A132440^Transpose = A218272 = D (representing derivation of o.g.f.s) by M, i.e., MD = M*D. The non-vanishing component of the first row of (MD)^n * V / (n+1)! is the n-th face polynomial. - Tom Copeland, Dec 11 2015
T is the convolution triangle of the positive integers starting at 2 (see A357368). - Peter Luschny, Oct 19 2022

Examples

			Triangle starts:
  1;
  0,  2;
  0,  3,   4;
  0,  4,  12,    8;
  0,  5,  25,   36,   16;
  0,  6,  44,  102,   96,    32;
  0,  7,  70,  231,  344,   240,    64;
  0,  8, 104,  456,  952,  1040,   576,   128;
  0,  9, 147,  819, 2241,  3400,  2928,  1344,   256;
  0, 10, 200, 1372, 4712,  9290, 11040,  7840,  3072,  512;
  0, 11, 264, 2178, 9108, 22363, 34332, 33488, 20224, 6912, 1024;
		

Crossrefs

Cf. A003480 (row sums), A181290.
Cf. A000297 (column 3), A006636 (column 4), A006637 (column 5).

Programs

  • Maple
    T := proc (n, k) if k <= n then sum((-1)^j*2^(k-j)*binomial(k, j)*binomial(n+k-j-1, 2*k-1), j = 0 .. k) else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
    # Uses function PMatrix from A357368.
    PMatrix(10, n -> n + 1); # Peter Luschny, Oct 19 2022
  • Mathematica
    Table[Sum[(-1)^j*2^(k - j) Binomial[k, j] Binomial[n + k - j - 1, 2 k - 1], {j, 0, k}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 11 2015 *)
  • PARI
    T_xt(max_row) = {my(N=max_row+1, x='x+O('x^N), h=(1-x)^2/((1-x)^2 - t*x*(2-x))); vector(N, n, Vecrev(polcoeff(h, n-1)))}
    T_xt(10) \\ John Tyler Rascoe, Apr 05 2025

Formula

T(n,k) = Sum_{j=0..k} (-1)^j*2^(k-j)*binomial(k,j)*binomial(n+k-j-1, 2*k-1) (0 <= k <= n).
G.f.: G(t,x) = (1-x)^2/((1-x)^2 - t*x*(2-x)).
G.f. of column k = x^k*(2-x)^k/(1-x)^{2k} (k>=1) (we have a Riordan array).
Recurrences satisfied by the numbers u_{n,k}=T(n,k) can be found in the Castiglione et al. reference.
Sum_{k=0..n} k*T(n,k) = A181290(n).
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=0, T(1,1)=2, T(2,0)=0, T(1,1)=3, T(2,2)=4, T(n,k)=0, if k < 0 or if k > n. - Philippe Deléham, Nov 29 2013

A248727 A046802(x,y) --> A046802(x,y+1), transform of e.g.f. for the graded number of positroids of the totally nonnegative Grassmannians G+(k,n); enumerates faces of the stellahedra.

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 16, 24, 10, 1, 65, 130, 84, 19, 1, 326, 815, 720, 265, 36, 1, 1957, 5871, 6605, 3425, 803, 69, 1, 13700, 47950, 65646, 44240, 15106, 2394, 134, 1, 109601, 438404, 707840, 589106, 267134, 63896, 7094, 263, 1
Offset: 0

Views

Author

Tom Copeland, Oct 12 2014

Keywords

Comments

This is a transform of A046802 treating it as an array of h-vectors, so y is replaced by (y+1) in the e.g.f. for A046802.
An e.g.f. for the reversed row polynomials with signs is given by exp(a.(0;t)x) = [e^{(1+t)x} [1+t(1-e^(-x))]]^(-1) = 1 - (1+2t)x + (1+5t+5t^2)x^2/2! + ... . The reciprocal is an e.g.f. for the reversed face polynomials of the simplices A074909, i.e., exp(b.(0;t)x) = e^{(1+t)x} [1+t(1-e^(-x))] = 1 + (1+2t)x +(1+3t+3t^2) x^2/2! + ... , so the relations of A133314 apply between the two sets of polynomials. In particular, umbrally [a.(0;t)+b.(0;t)]^n vanishes except for n=0 for which it's unity, implying the two sets of Appell polynomials formed from the two bases, a_n(z;t) = (a.(0;t)+z)^n and b_n(z;t) = (b.(0;t) + z)^n, are an umbral compositional inverse pair, i.e., b_n(a.(x;t);t)= x^n = a_n(b.(x;t);t). Raising operators for these Appell polynomials are related to the polynomials of A028246, whose reverse polynomials are given by A123125 * A007318. Compare: A248727 = A007318 * A123125 * A007318 and A046802 = A007318 * A123125. See A074909 for definitions and related links. - Tom Copeland, Jan 21 2015
The o.g.f. for the umbral inverses is Og(x) = x / (1 - x b.(0;t)) = x / [(1-tx)(1-(1+t)x)] = x + (1+2t) x^2 + (1+3t+3t^2) x^3 + ... . Its compositional inverse is an o.g.f for signed A033282, the reverse f-polynomials for the simplicial duals of the Stasheff polytopes, or associahedra of type A, Oginv(x) =[1+(1+2t)x-sqrt[1+2(1+2t)x+x^2]] / (2t(1+t)x) = x - (1+2t) x^2 + (1+5t+5t^2) x^3 + ... . Contrast this with the o.g.f.s related to the corresponding h-polynomials in A046802. - Tom Copeland, Jan 24 2015
Face vectors, or coefficients of the face polynomials, of the stellahedra, or stellohedra. See p. 59 of Buchstaber and Panov. - Tom Copeland, Nov 08 2016
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Examples

			The triangle T(n, k) starts:
n\k    0     1     2     3     4    5   6  7 ...
1:     1
2:     2     1
3:     5     5     1
4:    16    24    10     1
5:    65   130    84    19     1
6:   326   815   720   265    36    1
7:  1957  5871  6605  3425   803   69   1
8: 13700 47950 65646 44240 15106 2394 134  1
... reformatted, _Wolfdieter Lang_, Mar 27 2015
		

Crossrefs

Programs

  • Mathematica
    (* t = A046802 *) t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n - 1) - 1; t[n_, k_] = Sum[((i - k + 1)^i*(k - i)^(n - i - 1) - (i - k + 2)^i*(k - i - 1)^(n - i - 1))*Binomial[n - 1, i], {i, 0, k - 1}]; T[n_, j_] := Sum[Binomial[k, j]*t[n + 1, k + 1], {k, j, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 23 2015, after Tom Copeland *)

Formula

Let M(n,k)= sum{i=0,..,k-1, C(n,i)[(i-k)^i*(k-i+1)^(n-i)- (i-k+1)^i*(k-i)^(n-i)]} with M(n,0)=1. Then M(n,k)= A046802(n,k), and T(n,j)= sum(k=j,..,n, C(k,j)*M(n,k)) for j>0 with T(n,0)= 1 + sum(k=1,..,n, M(n,k)) for n>0 and T(0,0)=1.
E.g.f: y * exp[x*(y+1)]/[y+1-exp(x*y)].
Row sums are A007047. Row polynomials evaluated at -1 are unity. Row polynomials evaluated at -2 are A122045.
First column is A000522. Second column appears to be A036918/2 = (A001339-1)/2 = n*A000522(n)/2.
Second diagonal is A052944. (Changed from conjecture to fact on Nov 08 2016.)
The raising operator for the reverse row polynomials with row signs is R = x - (1+t) - t e^(-D) / [1 + t(1-e^(-D))] evaluated at x = 0, with D = d/dx. Also R = x - d/dD log[exp(a.(0;t)D], or R = - d/dz log[e^(-xz) exp(a.(0;t)z)] = - d/dz log[exp(a.(-x;t)z)] with the e.g.f. defined in the comments and z replaced by D. Note that t e ^(-D) / [1+t(1-e^(-D))] = t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... is an e.g.f. for the signed reverse row polynomials of A028246. - Tom Copeland, Jan 23 2015
Equals A007318*(padded A090582)*A007318*A097808 = A007318*(padded (A008292*A007318))*A007318*A097808 = A007318*A130850 = A007318*(mirror of A028246). Padded means in the same way that A097805 is padded A007318. - Tom Copeland, Nov 14 2016
Umbrally, the row polynomials are p_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A130850. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = x/((1+x)*exp(-x*y) - 1), the e.g.f. of A130850, so OP(x,d/dy) y^n evaluated at y = 1 is p_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A046082, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of this entry (A248727, the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020

Extensions

Title expanded by Tom Copeland, Nov 08 2016

A068763 Irregular triangle of the Fibonacci polynomials of A011973 multiplied diagonally by the Catalan numbers.

Original entry on oeis.org

1, 1, 1, 2, 2, 5, 6, 1, 14, 20, 6, 42, 70, 30, 2, 132, 252, 140, 20, 429, 924, 630, 140, 5, 1430, 3432, 2772, 840, 70, 4862, 12870, 12012, 4620, 630, 14, 16796, 48620, 51480, 24024, 4620, 252, 58786, 184756, 218790
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

The row length sequence of this array is [1,2,2,3,3,4,4,5,5,...] = A008619(n+2), n>=0.
The row polynomials p(n,x) := Sum_{m=0..floor((n+1)/2)} a(n,m)*x^m produce, for x = (b-a^2)/a^2 (not 0), the two parameter family of sequences K(a,b; n) := (a^(n+1))*p(n,(b-a^2)/a^2) with g.f. K(a,b; x) := (1-sqrt(1-4*x*(a+x*(b-a^2))))/(2*x).
Some members are: K(1,1; n)=A000108(n) (Catalan), K(1,2; n)=A025227(n-1), K(2,1; n)=A025228(n-1), K(1,3; n)=A025229(n-1), K(3,1; n)=A025230(n-1). For a=b=2..10 the sequences K(a,a; n)/a are A068764-A068772.
The column sequences (without leading 0's) are: A000108 (Catalan), A000984 (central binomial), A002457, 2*A002802, 5*A020918, 14*A020920, 42*A020922, ...
a(n,m) is the number of ways to designate exactly m cherries over all binary trees with n internal nodes. A cherry is an internal node whose descendants are both external nodes. Cf. A091894 which gives the number of binary trees with m cherries. - Geoffrey Critzer, Jul 24 2020
This irregular triangle is essentially that of A011973 with its diagonals multiplied by the Catalan numbers of A000108. The diagonals of this triangle are then rows of the Pascal matrix A007318 multiplied by the Catalan numbers. - Tom Copeland, Dec 23 2023

Examples

			The irregular triangle begins:
   n\m    0     1     2     3    4   5
   0:     1
   1:     1     1
   2:     2     2
   3:     5     6     1
   4:    14    20     6
   5:    42    70    30     2
   6:   132   252   140    20
   7:   429   924   630   140    5
   8:  1430  3432  2772   840   70
   9:  4862 12870 12012  4620  630  14
  10: 16796 48620 51480 24024 4620 252
  ...
p(3,x) = 5 + 6*x + x^2.
		

Crossrefs

Cf. A025227(n-1) (row sums).
Cf. A000007(n) (alternating row sums).

Programs

  • Mathematica
    nn = 10; b[z_] := (1 - Sqrt[1 - 4 z])/(2 z);Map[Select[#, # > 0 &] &,
    CoefficientList[Series[v b[v z] /. v -> (1 + u z ), {z, 0, nn}], {z, u}]] // Grid (* Geoffrey Critzer, Jul 24 2020 *)

Formula

a(n, m) = binomial(n+1-m, m)*C(n-m) if 0 <= m <= floor((n+1)/2), otherwise 0, with C(n) := A000108(n) (Catalan).
G.f. for column m=1, 2, ...: (x^(2*m-1))*C(m-1)/(1-4*x)^((2*m-1)/2); m=0: c(x), g.f. for A000108 (Catalan).
G.f. for row polynomials p(n, x): c(z) + x*z*c(x*(z^2)/(1-4*z))/sqrt(1-4*z) = (1-sqrt(1-4*z*(1+x*z)))/(2*z), where c(x) is the g.f. of A000108 (Catalan).
G.f. for triangle: (1 - sqrt(1 - 4*x (1 + y*x)))/(2*x). - Geoffrey Critzer, Jul 24 2020
The series expansion of f(x) = (1 + 2sx - sqrt(1 + 4sx + 4d^2x^2))/(2x) at x = 0 is (s^2 - d^2) x + (2 d^2s - 2 s^3) x^2 + (d^4 - 6 d^2 s^2 + 5 s^4) x^3 + (-6 d^4 s + 20 d^2 s^3 - 14 s^5) x^4 + ..., containing the coefficients of this array. With s = (a+b)/2 and d = (a-b)/2, then f(x)/ab = g(x) = (1 + (a+b)x - sqrt((1+(a+b)x)^2 - 4abx^2))/(2abx) = x - (a + b) x^2 + (a^2 + 3 a b + b^2) x^3 - (a^3 + 6 a^2 b + 6 a b^2 + b^3) x^4 + ..., containing the Narayana polynomials of A001263, which can be simply transformed into A033282. The compositional inverse about the origin of g(x) is g^(-1)(x) = x/((1-ax)(1-bx)) = x/((1-(s+d)x)(1-(s-d)x)) = x + (a + b) x^2 + (a^2 + a b + b^2) x^3 + (a^3 + a^2 b + a b^2 + b^3) x^4 + ..., containing the complete homogeneous symmetric polynomials h_n(a,b) = (a^n - b^n)/(a-b), which are the polynomials of A034867 when expressed in s and d, e.g., ((s + d)^7 - (s - d)^7)/(2 d) = d^6 + 21 d^4 s^2 + 35 d^2 s^4 + 7 s^6. A133437 and A134264 for compositional inversion of o.g.f.s can be used to relate the sets of polynomials above. - Tom Copeland, Nov 28 2023

Extensions

Title changed by Tom Copeland, Dec 23 2023

A108263 Triangle read by rows: T(n,k) is the number of short bushes with n edges and k branchnodes (i.e., nodes of outdegree at least two). A short bush is an ordered tree with no nodes of outdegree 1.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 5, 0, 1, 9, 5, 0, 1, 14, 21, 0, 1, 20, 56, 14, 0, 1, 27, 120, 84, 0, 1, 35, 225, 300, 42, 0, 1, 44, 385, 825, 330, 0, 1, 54, 616, 1925, 1485, 132, 0, 1, 65, 936, 4004, 5005, 1287, 0, 1, 77, 1365, 7644, 14014, 7007, 429, 0, 1, 90, 1925, 13650, 34398
Offset: 0

Views

Author

Emeric Deutsch, May 29 2005

Keywords

Comments

Row n has 1+floor(n/2) terms. Row sums are the Riordan numbers (A005043). Column 3 yields A033275; column 4 yields A033276.
Related to the number of certain non-crossing partitions for the root system A_n. Cf. p. 12, Athanasiadis and Savvidou. Diagonals are A033282/A086810. Also see A132081 and A100754.- Tom Copeland, Oct 19 2014

Examples

			T(6,3)=5 because the only short bushes with 6 edges and 3 branchnodes are the five full binary trees with 6 edges.
Triangle begins:
1;
0;
0,1;
0,1;
0,1,2;
0,1,5;
0,1,9,5
		

Crossrefs

Programs

  • Maple
    G:=(1+z-sqrt((1-z)^2-4*t*z^2))/2/z/(1+t*z): Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 16 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 16 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od; # yields sequence in triangular form
    A108263 := (n,k) -> binomial(n-k-1,n-2*k)*binomial(n,k)/(n-k+1);
    seq(print(seq(A108263(n,k),k=0..ceil((n-1)/2))),n=0..8); # Peter Luschny, Sep 25 2014
  • Mathematica
    T[n_,k_]:=Binomial[n-k-1,n-2k]*Binomial[n,k]/(n-k+1); Flatten[Table[T[n,k],{n,0,11},{k,0,Ceiling[(n-1)/2]}]] (* Indranil Ghosh, Feb 20 2017 *)

Formula

G.f. G=G(t, z) satisfies z*(1+t*z)*G^2 - (1+z)*G + 1 = 0.
T(n, k) = A086810(n-k, k). - Philippe Deléham, May 30 2005
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