cp's OEIS Frontend

Partial sums of signed sequence is shifted unsigned one: |a(n+2)| = A011655(n+1).

With interpolated zeros, a(n) = sin(5*Pi*n/6 + Pi/3)/sqrt(3) + cos(Pi*n/6 + Pi/6)/sqrt(3); this gives the diagonal sums of the Riordan array (1-x^2, x(1-x^2)). - Paul Barry, Feb 02 2005

From Tom Copeland, Nov 02 2014: (Start)

With a shift and a sign change the o.g.f. of this array becomes the compositional inverse of the shifted Motzkin or Riordan numbers A005043,

(x - x^2) / (1 - x + x^2) = x*(1-x) / (1 - x*(1-x)) = x*(1-x) + [x*(1-x)]^2 + ... . Expanding each term of this series and arranging like powers of x in columns gives skewed rows of the Pascal triangle and reading along the columns gives (mod-signs and indexing) A011973, A169803, and A115139 (see also A091867, A092865, A098925, and A102426 for these term-by-term expansions and A030528). (End)

Number of ordered trees with n+1 edges, having root of even degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002

The connection to Motzkin numbers comes from the Lagrange inversion formula. - Michael Somos, Oct 10 2003

Number of horizontal steps in all Motzkin paths of length n. - Emeric Deutsch, Nov 09 2003

Number of UHD's in all Motzkin paths of length n+2 (here U=(1,1), H=(1,0) and D=(1,-1)). Example: a(2)=2 because in the nine Motzkin paths of length 4, HHHH, HHUD, HUDH, H(UHD), UDHH, UDUD, (UHD)H, UHHD and UUDD, we have altogether two UHD's (shown between parentheses). - Emeric Deutsch, Dec 26 2003

Number of ordered trees with n+1 edges, having exactly one leaf at even height. Number of Dyck path of semilength n+1, having exactly one peak at even height. Example: a(3)=6 because we have uuu(ud)ddd, u(ud)dudud, udu(ud)dud, ududu(ud)d, u(ud)uuddd and uuudd(ud)d (here u=(1,1),d=(1,-1) and the unique peak at even height is shown between parentheses). - Emeric Deutsch, Mar 10 2004

a(n) is the number of Dyck (n+1)-paths containing exactly one UDU. - David Callan, Jul 15 2004

Number of peaks in all Motzkin paths of length n+1. - Emeric Deutsch, Sep 01 2004

This is a kind of Motzkin transform of A059841 because the substitution x -> x*A001006(x) in the independent variable of the g.f. of A059841 generates 1,0,1,2,6,16,... that is 1,0 followed by this sequence here. - R. J. Mathar, Nov 08 2008

a(n) is the number of lattice paths avoiding N^(>=3) from (0,0) to (n,n). - Shanzhen Gao, Apr 20 2010

a(n+1) is the number of binary strings having n 0's and n 1's and no appearance of 000. For example, for n = 1, there 2 strings: 01 and 10. For n = 2, there are 6: 0011, 0101, 0110, 1001, 1010, 1100. - Toby Gottfried, Sep 12 2011

a(n) is the number of paths in the half-plane x>=0, from (0,0) to (n,1), and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). For example, for n=3, we have the 6 paths HHU, HUH, UDU, UUD, UHH, DUU. - José Luis Ramírez Ramírez, Apr 19 2015

a(n) is the number of ways to tile a strip of length 2*n+1 with squares, dominos, and trominos, where the number of trominos is always one more than the number of squares. - Greg Dresden and Anna Kalynchuk, Jul 30 2025

Sum of entries in row n is A000108(n) (the Catalan numbers).

From Gus Wiseman, Nov 16 2022: (Start)

Also the number of unlabeled ordered rooted trees with n nodes and height k. For example, row n = 5 counts the following trees:

(oooo) ((o)oo) (((o))o) ((((o))))

((oo)o) (((o)o))

((ooo)) (((oo)))

(o(o)o) ((o(o)))

(o(oo)) (o((o)))

(oo(o))

((o)(o))

(End)

Also, n such that A010060(n) = A010060(n+1) where A010060 is the Thue-Morse sequence.

Sequence of n such that a(n) = 3n begins 7, 23, 27, 29, 31, 39, 71, 87, 91, 93, 95, ...

Values of k such that the Motzkin number A001006(2k) is even. Values of k such that the number of restricted hexagonal polyominoes with 2k+1 cells is even (see A002212). Values of k such that the number of directed animals of size k+1 is even (see A005773). Values of k such that the Riordan number A005043(k) is even. - Emeric Deutsch and Bruce E. Sagan, Apr 02 2003

a(n) = A036554(n)-1 = A072939(n)-2. - Ralf Stephan, Jun 09 2003

Odious and evil terms alternate. - Vladimir Shevelev, Jun 22 2009

The sequence has the following fractal property: remove terms of the form 4k+1 from the sequence, and the remaining terms are of the form 4k+3: 7, 23, 31, 39, 55, 71, 87, ...; then subtract 3 from each of these terms and divide by 4 and you get the original sequence: 1, 5, 7, 9, 13, ... - Benoit Cloitre, Apr 06 2010

A035263(a(n)) = 0. - Reinhard Zumkeller, Mar 01 2012

T(n+3, k) is also the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's cluster algebra of finite type A_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of the triangle of Narayana numbers A001263. For example, x^2 + 5*x + 5 = y^2 + 3*y + 1. - Paul Boddington, Mar 07 2003

Number of standard Young tableaux of shape (k+1,k+1,1^(n-k-3)), where 1^(n-k-3) denotes a sequence of n-k-3 1's (see the Stanley reference).

Number of k-dimensional 'faces' of the n-dimensional associahedron (see Simion, p. 168). - Mitch Harris, Jan 16 2007

Mirror image of triangle A126216. - Philippe Deléham, Oct 19 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

Row generating polynomials 1/(n+1)*Jacobi_P(n,1,1,2*x+1). Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p. 60]. See A001263 for the corresponding array of h-vectors for associahedra of type A_n. See A063007 and A080721 for the f-vectors for associahedra of type B and type D respectively. - Peter Bala, Oct 28 2008

f-vectors of secondary polytopes for Grobner bases for optimization and integer programming (see De Loera et al. and Thomas). - Tom Copeland, Oct 11 2011

From Devadoss and O'Rourke's book: The Fulton-MacPherson compactification of the configuration space of n free particles on a line segment with a fixed particle at each end is the n-Dim Stasheff associahedron whose refined f-vector is given in A133437 which reduces to A033282. - Tom Copeland, Nov 29 2011

Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012

The general results on the convolution of the refined partition polynomials of A133437, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016

The signed triangle t(n, k) =(-1)^k* T(n+2, k-1), n >= 1, k = 1..n, seems to be obtainable from the partition array A111785 (in Abramowitz-Stegun order) by adding the entries corresponding to the partitions of n with the number of parts k. E.g., triangle t, row n=4: -1, (6+3) = 9, -21, 14. - Wolfdieter Lang, Mar 17 2017

The preceding conjecture by Lang is true. It is implicit in Copeland's 2011 comments in A086810 on the relations among a gf and its compositional inverse for that entry and inversion through A133437 (a differently normalized version of A111785), whose integer partitions are the same as those for A134685. (An inversion pair in Copeland's 2008 formulas below can also be used to prove the conjecture.) In addition, it follows from the relation between the inversion formula of A111785/A133437 and the enumeration of distinct faces of associahedra. See the MathOverflow link concernimg Loday and the Aguiar and Ardila reference in A133437 for proofs of the relations between the partition polynomials for inversion and enumeration of the distinct faces of the A_n associahedra, or Stasheff polytopes. - Tom Copeland, Dec 21 2017

The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/(n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 07 2022

Chapoton's observation above is correct: the precise expansion is (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/ (n!*(n+1)!) = Sum_{k = 0..n-1} (-1)^k*T(n+2,n-k-1)*binomial(x+2*n-k,2*n-k), as can be verified using the WZ algorithm. For example, n = 4 gives (x+1)*(x+2)^2*(x+3)^2*(x+4)^2*(x+5)/(4!*5!) = 14*binomial(x+8,8) - 21*binomial(x+7,7) + 9*binomial(x+6,6) - binomial(x+5,5). - Peter Bala, Jun 24 2023

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

Number of ordered trees with n+1 edges, having root of degree 2 and nonroot nodes of outdegree at most 2.

Sequence without the initial 0 is the convolution of the sequence of Motzkin numbers (A001006) with itself.

Number of horizontal steps at level zero in all Motzkin paths of length n. Example: a(3)=5 because in the four Motzkin paths of length 3, (HHH), (H)UD, UD(H) and UHD, where H=(1,0), U=(1,1), D=(1,-1), we have altogether five horizontal steps H at level zero (shown in parentheses).

Number of peaks at level 1 in all Motzkin paths of length n+1. Example: a(3)=5 because in the nine Motzkin paths of length 4, HHHH, HH(UD), H(UD)H, HUHD, (UD)HH, (UD)(UD), UHDH, UHHD and UUDD (where H=(1,0), U=(1,1), D=(1,-1)), we have five peaks at level 1 (shown between parentheses).

a(n) = number of Motzkin paths of length n+1 that start with an up step. - David Callan, Jul 19 2004

Could be called a Motzkin transform of A130716 because the g.f. is obtained from the g.f. x*A130716(x)= x(1+x+x^2) (offset changed to 1) by the substitution x -> x*A001006(x) of the independent variable. - R. J. Mathar, Nov 08 2008

For n >= 1, a(n) is the number of length n permutations sorted to the identity by a consecutive-123-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020

Also the number of distinct orderless Mathematica expressions with one atom and n positions.

Number of partitions of n into distinct powers of 3.

Trajectory of 1 under the morphism: 1 -> 110, 0 -> 000. Thus 1 -> 110 ->110110000 -> 110110000110110000000000000 -> ... - Philippe Deléham, Jul 09 2005

Also, an example of a d-perfect sequence.

This is a composite of two earlier sequences contributed at different times by N. J. A. Sloane and by Reinhard Zumkeller, Mar 05 2005. Christian G. Bower extended them and found that they agreed for at least 512 terms. The proof that they were identical was found by Ralf Stephan, Jun 13 2005, based on the fact that they were both 3-regular sequences.

Reverse of A071947 - related to lattice paths. First column is A005043.

Triangle T(n,k), 0 <= k <= n, defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Feb 27 2007

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(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise from 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

Riordan array (f(x),x*g(x)), where f(x)is the o.g.f. of A005043 and g(x)is the o.g.f. of A001006. - Philippe Deléham, Nov 22 2009

Riordan array ((1+x-sqrt(1-2x-3x^2))/(2x(1+x)), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse of Riordan array ((1+x)/(1+x+x^2),x/(1+x+x^2)). E.g.f. of column k is exp(x)*(Bessel_I(k,2x)-Bessel_I(k+1,2x)).

Diagonal sums are A187306.

Simultaneous equations using the first n rows solve for diagonal lengths of odd N = (2n+1) regular polygons, with constants c^0, c^1, c^2, ...; where c = 1 + 2*cos( 2*Pi/N) = sin(3*Pi/N)/sin(Pi/N) = the third longest diagonal of N>5. By way of example, take the first 4 rows relating to the 9-gon (nonagon), N=(2*4 + 1), with c = 1 + 2*cos(2*Pi/9) = 2.5320888.... The simultaneous equations are (1,0,0,0) = 1; (0,1,0,0) = c; (1,1,1,0) = c^2, (1,3,2,1) = c^3. The answers are 1, 2.532..., 2.879..., and 1.879...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. - Gary W. Adamson, Sep 07 2011

Number of linearly independent irreducible representations of a given weight j in a tensor of given rank n. - Mikkel N. Schmidt, Aug 20 2025