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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A249548 Coefficients of reduced partition polynomials of A134264 for computing Lagrange compositional inversion.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 5, 1, 6, 3, 5, 1, 7, 7, 21, 1, 8, 8, 4, 28, 28, 14, 1, 9, 9, 9, 36, 72, 12, 84, 1, 10, 10, 10, 5, 45, 90, 45, 45, 120, 180, 42, 1, 11, 11, 11, 11, 55, 110, 110, 55, 55, 165, 495, 165, 330, 1, 12, 12, 12, 12, 6, 66, 132, 132, 66, 66, 132, 22, 220, 660, 330, 660, 55, 495, 990, 132
Offset: 0

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Author

Tom Copeland, Oct 31 2014

Keywords

Comments

Coefficients of reduced partition polynomials of A134264 for computing the complete partition polynomials for the Lagrange compositional inversion of A134264 (see Oct 2014 comment by Copeland there). Umbrally,
e^(x*t) * exp[Prt(.;1,0,h_2,..) * t] = exp[Prt(.;1,x,h_2,..) * t], where Prt(n;1,0,h_2,..,h_n) are the reduced (h_0 = 1 and h_1 = 0) partition polynomials of the complete polynomials Prt(n;h_0,h_1,h_2,..,h_n) of A134264.
Partitions are given in the order of those on page 831 of Abramowitz and Stegun. Formulas for the coefficients of the partitions are given in A134264.
Row sums are the Motzkin sums or Riordan numbers A005043. - Tom Copeland, Nov 09 2014
From Tom Copeland, Jul 03 2018: (Start)
The matrix and operator formalism for Sheffer Appell sequences leads to the following relations with D = d/dh_1.
Exp[Prt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + Prt(.;1,0,h_2,...)]^n = Prt(n;1,h_1,h_2,...), the partition polynomials of A134264 for g(t)/t with h_0 = 1.
For the umbral compositional inverses described below,
Exp[UPrt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + UPrt(.;1,0,h_2,...)]^n = UPrt(n;1,h_1,h_2,...).
The respective e.g.f.s are multiplicative inverses; that is, exp[Prt(.;1,0,h_2,..) * t] = 1/exp[UPrt(.;1,0,h_2,..) * t], so the formalism of A133314 applies.
The raising operator R such that R Prt(n;1,h_1,h_2,...) = Prt(n+1;1,h_1,h_2,...) is R = exp[Prt(.;1,0,h_2,...)*D] h_1 exp[UPrt(.;1,0,h_2,..)*D] since R Prt(n+1;1,h_1,h_2,...) = exp[Prt(.;1,0,h_2,...)*D] h_1 (h_1)^n = Prt(n+1;h_1,h_2,...) from the definition of the umbral compositional inverse. This may also be expressed as R = h_1 + d/dD log[exp[Prt(.;1,0,h_2,...) * D]], so, using A127671, 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 - 9 h_2 h_4 + 5 h_2^3 - 7 h_3^2) D^5/5! + (h_7 - 28 h_3 h_4 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... . (End)

Examples

			Prt(0) = 1
Prt(1;1,0) = 0
Prt(2;1,0,h_2) = 1 h_2
Prt(3;1,0,h_2,h_3) = 1 h_3
Prt(4;1,0,h_2,..,h_4) = 1 h_4 + 2 (h_2)^2
Prt(5;1,0,h_2,..,h_5) = 1 h_5 + 5 h_2 h_3
Prt(6;1,0,h_2,..,h_6) = 1 h_6 + 6 h_2 h_4 + 3 (h_3)^2 + 5 (h_2)^3
Prt(7;1,0,h_2,..,h_7) = 1 h_7 + 7 h_3 h_4 + 7 h_2 h_5 + 21 h_3 (h_2)^2
...
------------
With h_n denoted by (n'), the first seven partition polynomials of A134264 with h_0=1 are given by the first seven coefficients of the truncated Taylor series expansion of the Euler binomial transform
e^[(1') * t] * {1 + 1 (2') *  t^2/2! + 1 (3') *  t^3/3! + [1 (4') + 2 (2')^2] *  t^4/4! + [1 (5') + 5 (2')(3')] *  t^5/5! + [1 (6') + 6 (2')(4') + 3 (3')^2 + 5 (2')^3] *  t^6/6!}, giving the truncated expansion
1 + 1 (1') * t + [1 (2') + 1 (1')^2] * t^2/2! + ... + [1 (6') + 6 (1')(5') + 6 (2')(4') + 3 (3')^2 + 15 (1')^2(4') + 30 (1')(2')(3') + 5 (2')^3 + 20 (1')^3(3') + 30 (1')^2(2')^2 + 15 (1')^4(2') + 1 (1')^6] * t^6/6!.
Extending the number of reduced partition polynomials of the transform allows for further complete polynomials of A134264 to be computed.
		

Crossrefs

Cf. A127671.
Rows lengths are given by A002865 (except for row 1).

Programs

  • Mathematica
    rows[n_] := {{1}, {0}}~Join~Module[
        {g = 1 / D[t / (1 + Sum[h[k] t^k, {k, 2, n}] + O[t]^(n+1)), t], p = t, r},
        r = Reap[Do[p = g D[p, t]/k; Sow[Expand[Normal@p /. {t -> 0}]], {k, n+1}]][[2, 1, 2 ;;]];
        Table[Coefficient[r[[k]], Product[h[t], {t, p}]], {k, 2, n}, {p, Sort[Sort /@ IntegerPartitions[k, k, Range[2, k]]]}]];
    rows[12] // Flatten (* Andrey Zabolotskiy, Feb 18 2024 *)

Formula

From Tom Copeland, Nov 10 2014: (Start)
Terms may be computed symbolically up to order n by using an iterated derivative evaluated at t=0:
with g(t) = 1/{d/dt [t/(1 + 0 t + h_2 t^2 + h_3 t^3 + ... + h_n t^n)]},
evaluate 1/n! * [g(t) d/dt]^n t at t=0, i.e., ask a symbolic math app for the first term in a series expansion of this iterated derivative, to obtain Prt(n-1).
Alternatively, the explicit formula in A134264 for the numerical coefficients of each partition can be used. (End)
From Tom Copeland, Nov 12 2014: (Start)
The first few partitions polynomials formed by taking the reciprocal of the e.g.f. of this entry's e.g.f. (cf. A133314) are
UPrt(0) = 1
UPrt(1;1,0) = 0
UPrt(2;1,0,h_2) = -1 h_2
UPrt(3;1,0,h_2,h_3) = -1 h_3
UPrt(4;1,0,h_2,..,h_4) = -1 h_4 + 4 (h_2)^2
UPrt(5;1,0,h_2,..,h_5) = -1 h_5 + 15 h_2 h_3
UPrt(6;1,0,h_2,..,h_6) = -1 h_6 + 24 h_2 h_4 + 17 (h_3)^2 + -35 (h_2)^3
...
Therefore, umbrally, [Prt(.;1,0,h_2,...) + UPrt(.;1,0,h_2,...)]^n = 0 for n>0 and unity for n=0.
Example of the umbral operation:
(a. + b.)^2 = a.^2 + 2 a.* b. + b.^2 = a_2 + 2 a_1 * b_1 + b_2.
This implies that the umbral compositional inverses (see below) of the partition polynomials of the Lagrange inversion formula (LIF) of A134264 with h_0=1 are given by UPrt(n;1,h_1,h_2,...,h_n) = [UPrt(.;1,0,h_2,...,h_n) + h_1]^n, so that the sequence of polynomials UPrt(n;1,h_1,h_2,...,h_n) is an Appell sequence in the indeterminate h_1. So, if one calculates UPrt(n;1,h_1,...,h_n), the lower order UPrt(n-1;1,h_1,...,h_(n-1)) can be found by taking the derivative w.r.t. h_1 and dividing by n. Same applies for Prt(n;1,h_1,h_2,...,h_n).
This connects the combinatorics of the permutohedra through A133314 and A049019, or their duals, to the noncrossing partitions, Dyck lattice paths, etc. that are isomorphic with the LIF of A134264.
An Appell sequence P(.,x) with the e.g.f. e^(x*t)/w(t) possesses an umbral inverse sequence UP(.,x) with the e.g.f. w(t)e^(x*t), i.e., polynomials such that P(n,UP(.,x))= x^n = UP(n,P(.,x)) through umbral substitution, as in the binomial example. The Bernoulli polynomials with w(t) = t/(e^t - 1) are a good example with the umbral compositional inverse sequence UP(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1) (cf. A074909 and A135278). (End)

Extensions

Formula for Prt(7,..) and a(12)-a(15) added by Tom Copeland, Jul 22 2016
Rows 8-12 added by Andrey Zabolotskiy, Feb 18 2024

A338135 Irregular triangle read by rows: Row p gives number of non-overlapping clusters of 2q-plets joining 2p points on a circle, i.e., number of noncrossing partitions from A134264 with h_k for k odd replaced by zero.

Original entry on oeis.org

1, 1, 2, 1, 6, 5, 1, 8, 4, 28, 14, 1, 10, 10, 45, 45, 120, 42, 1, 12, 12, 6, 66, 132, 22, 220, 330, 495, 132, 1, 14, 14, 14, 91, 182, 91, 91, 364, 1092, 364, 1001, 2002, 2002, 429
Offset: 1

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Author

Tom Copeland, Oct 11 2020

Keywords

Comments

This combinatorial problem arises in relating connected and disconnected Green functions associated to a "zero-dimensional" quantum field theory presented by Brezin et al. in "Planar Diagrams" via Eqn. 31 on p. 42.
Appears to be a refinement of A120986 and A108767 in that summing the coefficients of partitions with the same sum of exponents gives the rows or reverse rows of the two entries; for example, row 4 here becomes x + 8 xx + 4 x^2 + 28 x^2x + 14 x^4 = x + 12 x^2 + 28 x^3 + 14 x^4, which is row 4 of A108767. In short, replace each g_k or (k) by x in the formula section here to obtain the coarser entry or its reverse from this refined entry, apparently.
This also gives the relationship between moments and free cumulants in free probability theory restricted to an even number of noncrossing partitions as given by restricting the similar enumeration formuia on p. 34 of Novak and LaCroix to b_{2n} = G_{2n} and K_{2q} = g_{2q}. This is consistent with setting h_k to zero for odd k in A134264, e.g., doing so for the coefficients of t^7 for g(t) there gives G_6 here.
A125181 is another version of A134264, providing interpretations in terms of Dyck paths and trees.

Examples

			row 1: G_2  = g_2
row 2: G_4  = g_4  +  2 g_2^2
row 3: G_6  = g_6  +  6 g_2 g_4 +  5 g_2^3
row 4: G_8  = g_8  +  8 g_2 g_6 +  4 g_4^2   +  28 g_2^2 g_4 + 14 g_2^4
row 5: G_10 = g_10 + 10 g_2 g_8 + 10 g_4 g_6 +  45 g_2^2 g_6 + 45 g_2 g_4^2
              + 120 g_2^3 g_4  + 42 g_2^5
_____________
In the notation of Abramowitz and Stegun p. 831 with indices of the partitions above divided by 2;
R_1  = (1)
R_2  = (2)  +  2 (1)^2
R_3  = (3)  +  6 (1) (2) +  5 (1)^3
R_4  = (4)  +  8 (1) (3) +  4 (2)^2   +  28 (1)^2 (2) + 14 (1)^4
R_5  = (5)  + 10 (1) (4) + 10 (2) (3) +  45 (1)^2 (3) + 45 (1) (2)^2
        + 120 (1)^3 (2) + 42 (1)^5
______________
		

Crossrefs

Programs

  • Mathematica
    Table[(2 n)!/((2 n + 1 - Length@p)! Product[r!, {r, Last /@ Tally[p]}]), {n, 5}, {p, Sort[Sort /@ IntegerPartitions[n]]}] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

Under the constraint 2p = Sum_{q} 2q r_q, then G_{2p} = Sum_{r_q >= 0} [(2p)! / (2p + 1 - Sum_{q} r_q)! ] (g_2^r_1 /(r_1)!) (g_4^r_2 / (r_2)!) ... (g_{2q}^r_q / (r_q)!) where g_{2k} are the connected Green functions.
With R_p = G_{2p} and N_q = g_{2q}, then R_p = Sum_{r_q >= 0} [(2p)! / (2p + 1 - Sum_{q} r_q)! ] (N_1^r_1 /(r_1)!) (N_2^r_2 / (r_2)!) ... (N_{q}^r_q / (r_q)!) where N_q are the partitions in Abramowitz and Stegun on p. 831.
Coefficients of the final terms g_{2}^p = (1)^p are the Catalan numbers A000108.

Extensions

Rows 6-7 from Andrey Zabolotskiy, Mar 07 2024

A354622 Irregular triangle read by rows: Refined 3-Narayana triangle. Coefficients of partition polynomials of A134264, a refined Narayana triangle enumerating noncrossing partitions, with all h_k other than h_0, h_3, h_6, ..., h_(3n), ... replaced by zero.

Original entry on oeis.org

1, 1, 3, 1, 9, 12, 1, 12, 6, 66, 55, 1, 15, 15, 105, 105, 455, 273, 1, 18, 18, 9, 153, 306, 51, 816, 1224, 3060, 1428, 1, 21, 21, 21, 210, 420, 210, 210, 1330, 3990, 1330, 5985, 11970, 20349, 7752, 1, 24, 24, 24, 12, 276, 552, 552, 276, 276, 2024, 6072, 3036, 6072, 506, 10626, 42504, 21252, 42504, 106260, 134596, 43263
Offset: 1

Views

Author

Tom Copeland, Jul 08 2022

Keywords

Comments

A set of partition polynomials with these coefficients and the polynomials of A338135 can be generated by substitution of the refined Narayana, or noncrossing partition, polynomials N_n[h_1,...,h_n] of A134264 (h_0=1) into themselves--once for A338135 and twice for this entry--or by setting the indeterminates h_n of A134264 to zero except for h_0, h_3, h_6, ..., h_(3n), ... with h_0 = 1 and ultimately re-indexing. This is equivalent to recursive use of the Lagrange inversion formula on f(x) = x / h(x) = x / (1 + h_1 x + h_2 x^2 + ...) since its compositional inverse is f^{(-1)}(x) = x + N_1(h_1) x + N_2(h_1,h_2) x^2 + .... The equivalence of the two methods of generation--the substitution and the zeroing out--follows from the general theorems stated by Peter Bala in his presentation of formulas for A108767 in 2008, which stem from a fixed point-iteration formalism of a basic identity for a compositional inverse pair, x* h(f^{(-1)}(x)) = f^{(-1)}(x), where, as above, h(x) = x / f(x).
The sets of refined m-Narayana polynomials are used by Cachazo and Umbert to characterize the scattering amplitudes of a class of quantum fields (see, e.g., section 7.3).
These could also be called the refined 3-Dyck path polynomials. From the interpretation of A134264 as Dyck paths in A125181, or staircases whose steps never rise above the diagonal of a square grid (see illustrations in Weisstein), the monomials of the partition polynomial N_4 = 1 (4') + 4 (1') (3') + 2 (2')^2 + 6 (1')^2 (2') + 1 (1')^4 of A134264 have the following correspondences:
1 (4') --> 1 staircase of one step of height 4,
4 (1') (3') --> 4 staircases of 1 step of height 1 and 1 step of height 3,
2 (2')^2 --> 2 staircases of 2 steps of height 2,
6 (1')^2 (2') --> 6 staircases of 2 steps of height 1 and 1 step of height 2,
1 (1')^4 --> 1 staircase of 4 steps of height 1.
Consequently, the partition polynomials G_{3n} of this entry enumerate staircases of height 3n with steps of heights 3, 6, 9, ..., 3k, ... only.
Diverse combinatorial models of the refined m-Narayana, or m-Dyck, polynomials are inherited from those presented for the refined Narayana, or noncrossing partition, polynomials in A134264 and A125181 and in the references therein.
A127537 gives a combinatorial model (see title and Domb and Barret therein, Table 2, p. 355) that contains the coefficients of the monomials h_1^n and h_1^(n-2) h_2, i.e., A001764 and A003408.

Examples

			Triangle begins:
  1;
  1,  3;
  1,  9, 12;
  1, 12,  6, 66,  55;
  1, 15, 15, 105, 105, 455, 273;
  ...
Row 1: G_3  = g_3
row 2: G_6  = g_6 + 3 g_3^2
row 3: G_9  = g_9 + 9 g_3 g_6 + 12 g_3^3
row 4: G_12 = g_12 + 12 g_3 g_9 + 6 g_6^2 + 66 g_3^2 g_6 + 55 g_3^4
row 5: G_15 = g_15 + 15 g_3 g_12 + 15 g_6 g_9 + 105 g_3^2 g_9 + 105 g_3 g_6^2
              + 455 g_3^3 g_6 + 273 g_3^5.
.
In the notation of Abramowitz and Stegun p. 831 with indices of the partitions above divided by 3 and partition indeterminates h_n denoted (n):
R_1 = (1);
R_2 = (2) + 3 (1)^2;
R_3 = (3) + 9 (1) (2) + 12 (1)^3;
R_4 = (4) + 12 (1) (3) + 6 (2)^2 + 66 (1)^2 (2) + 55 (1)^4;
R_5 = (5) + 15 (1) (4) + 15 (2) (3) + 105 (1)^2 (3) + 105 (1) (2)^2 + 455 (1)^3(2)
          + 273 (1)^5.
		

Crossrefs

The length of row n is equal to A000041(n).
Row sums give A002293, n >= 1.

Programs

  • Mathematica
    Table[Binomial[Total[y], Length[y]-1] (Length[y]-1)! / Product[Count[y, i]!, {i, Max@@y}], {n, 8}, {y, Sort[Sort /@ IntegerPartitions[3n, n, Range[3, 3n, 3]]]}] // Flatten (* Andrey Zabolotskiy, Feb 19 2024, using Gus Wiseman's code for A134264 *)
  • PARI
    \\ Compare with A134264
    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(3*Vec(p)) | p<-partitions(n)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 19 2024

Formula

Coefficients of the monomials are those of the surviving monomials of the partition polynomials of A134264 after zeroing all indeterminates other than h_0, h_3, h_6, h_9, ..., h_(3n), .... The multinomial coefficients of A125181 also apply for G_n, giving the coefficient of the monomial h_1^e_1 h_2^e_2 ... h_n^n of R_n with se := e_1 + e_2 + ... + e_n as (3n)! / ((3n-se+1)! (e_1)! (e_2)! ... (e_n)!).
1*e_1 + 2*e_2 + ... + n*e_n = n for each monomial of R_n.
The partition polynomials R_n = N_n^3 of this entry can be determined from those of A338135, N_n^2, by substituting the partition polynomials of A134264, N_n, for the indeterminate h_n = (n) of N_n^2 or by doing the same for A134264 twice. E.g., N_1(h_1) = h_1, N_2(h_1,h_2) = h_2 + h_1^2, so N_2^2(h_1,h_2) = N_2(N_1,N_2) = N_2 + N_1 = h_2 + h_1^2 + h_1^2 = h_2 + 2 h_1^2 and N_2^3(h_1,h_2) = N_2^2(N_1,N_2) = N_2 + 2 N_1^2 = h_2 + h_1^2 + 2 h_1^2 = h_2 + 3 h_1^2.
Reduces with all indeterminates h_n = (n) = t to A173020.
The coefficient of the monomial h_1^n is (3*n)! / ((3*n-n+1)! n!) = A001764(n) (see also A179848 and A235534). In general, the coefficients of these monomials of the refined (m+1)-Narayana polynomials are the Fuss-Catalan sequence associated to the row sums of the refined m-Narayana polynomials.
The coefficient of the monomial h_1^(n-2) h_2 is (3n)! / ((3n-n+2)! (n-2)!) = A003408(n-2) for n > 1.
The coefficient of the monomial h_1^(n-3) h_3 is (3n)! / ((3n-n+3)! (n-3)!) = A004321(n) for n > 2.

Extensions

Rows 6-8 added by Andrey Zabolotskiy, Feb 19 2024

A000108 Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!).

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
Offset: 0

Views

Author

Keywords

Comments

These were formerly sometimes called Segner numbers.
A very large number of combinatorial interpretations are known - see references, esp. R. P. Stanley, "Catalan Numbers", Cambridge University Press, 2015. This is probably the longest entry in the OEIS, and rightly so.
The solution to Schröder's first problem: number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways: ((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0, 0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or a (+1,-1) step. Then the number of such paths that never go below the x-axis (Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14 noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011
Noncrossing partitions are partitions of genus 0. - Robert Coquereaux, Feb 13 2024
a(n-1) is the number of ways of expressing an n-cycle (123...n) in the symmetric group S_n as a product of n-1 transpositions (u_1,v_1)*(u_2,v_2)*...*(u_{n-1},v_{n-1}) where u_iA000272. - Joerg Arndt and Greg Stevenson, Jul 11 2011
a(n) is the number of ordered rooted trees with n nodes, not including the root. See the Conway-Guy reference where these rooted ordered trees are called plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. - Wolfdieter Lang, Aug 07 2007
As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the number of labeled dissections of a disk, related to the number R(n)=A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22 2012
Shifts one place left when convolved with itself.
For n >= 1, a(n) is also the number of rooted bicolored unicellular maps of genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Number of ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then the number of ways of forming n chords is given by (2n-1)!! = (2n)!/(n!*2^n) = A001147(n).)
Arises in Schubert calculus - see Sottile reference.
Inverse Euler transform of sequence is A022553.
With interpolated zeros, the inverse binomial transform of the Motzkin numbers A001006. - Paul Barry, Jul 18 2003
The Hankel transforms of this sequence or of this sequence with the first term omitted give A000012 = 1, 1, 1, 1, 1, 1, ...; example: Det([1, 1, 2, 5; 1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132]) = 1 and Det([1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132; 14, 42, 132, 429]) = 1. - Philippe Deléham, Mar 04 2004
a(n) equals the sum of squares of terms in row n of triangle A053121, which is formed from successive self-convolutions of the Catalan sequence. - Paul D. Hanna, Apr 23 2005
Also coefficients of the Mandelbrot polynomial M iterated an infinite number of times. Examples: M(0) = 0 = 0*c^0 = [0], M(1) = c = c^1 + 0*c^0 = [1 0], M(2) = c^2 + c = c^2 + c^1 + 0*c^0 = [1 1 0], M(3) = (c^2 + c)^2 + c = [0 1 1 2 1], ... ... M(5) = [0 1 1 2 5 14 26 44 69 94 114 116 94 60 28 8 1], ... - Donald D. Cross (cosinekitty(AT)hotmail.com), Feb 04 2005
The multiplicity with which a prime p divides C_n can be determined by first expressing n+1 in base p. For p=2, the multiplicity is the number of 1 digits minus 1. For p an odd prime, count all digits greater than (p+1)/2; also count digits equal to (p+1)/2 unless final; and count digits equal to (p-1)/2 if not final and the next digit is counted. For example, n=62, n+1 = 223_5, so C_62 is not divisible by 5. n=63, n+1 = 224_5, so 5^3 | C_63. - Franklin T. Adams-Watters, Feb 08 2006
Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are a(2) = 2 and a(3) = 5. Is the only semiprime Catalan number a(4) = 14? - Jonathan Vos Post, Mar 06 2006
The answer is yes. Using the formula C_n = binomial(2n,n)/(n+1), it is immediately clear that C_n can have no prime factor greater than 2n. For n >= 7, C_n > (2n)^2, so it cannot be a semiprime. Given that the Catalan numbers grow exponentially, the above consideration implies that the number of prime divisors of C_n, counted with multiplicity, must grow without limit. The number of distinct prime divisors must also grow without limit, but this is more difficult. Any prime between n+1 and 2n (exclusive) must divide C_n. That the number of such primes grows without limit follows from the prime number theorem. - Franklin T. Adams-Watters, Apr 14 2006
The number of ways to place n indistinguishable balls in n numbered boxes B1,...,Bn such that at most a total of k balls are placed in boxes B1,...,Bk for k=1,...,n. For example, a(3)=5 since there are 5 ways to distribute 3 balls among 3 boxes such that (i) box 1 gets at most 1 ball and (ii) box 1 and box 2 together get at most 2 balls:(O)(O)(O), (O)()(OO), ()(OO)(O), ()(O)(OO), ()()(OOO). - Dennis P. Walsh, Dec 04 2006
a(n) is also the order of the semigroup of order-decreasing and order-preserving full transformations (of an n-element chain) - now known as the Catalan monoid. - Abdullahi Umar, Aug 25 2008
a(n) is the number of trivial representations in the direct product of 2n spinor (the smallest) representations of the group SU(2) (A(1)). - Rutger Boels (boels(AT)nbi.dk), Aug 26 2008
The invert transform appears to converge to the Catalan numbers when applied infinitely many times to any starting sequence. - Mats Granvik, Gary W. Adamson and Roger L. Bagula, Sep 09 2008, Sep 12 2008
Limit_{n->oo} a(n)/a(n-1) = 4. - Francesco Antoni (francesco_antoni(AT)yahoo.com), Nov 24 2008
Starting with offset 1 = row sums of triangle A154559. - Gary W. Adamson, Jan 11 2009
C(n) is the degree of the Grassmannian G(1,n+1): the set of lines in (n+1)-dimensional projective space, or the set of planes through the origin in (n+2)-dimensional affine space. The Grassmannian is considered a subset of N-dimensional projective space, N = binomial(n+2,2) - 1. If we choose 2n general (n-1)-planes in projective (n+1)-space, then there are C(n) lines that meet all of them. - Benji Fisher (benji(AT)FisherFam.org), Mar 05 2009
Starting with offset 1 = A068875: (1, 2, 4, 10, 18, 84, ...) convolved with Fine numbers, A000957: (1, 0, 1, 2, 6, 18, ...). a(6) = 132 = (1, 2, 4, 10, 28, 84) dot (18, 6, 2, 1, 0, 1) = (18 + 12 + 8 + 10 + 0 + 84) = 132. - Gary W. Adamson, May 01 2009
Convolved with A032443: (1, 3, 11, 42, 163, ...) = powers of 4, A000302: (1, 4, 16, ...). - Gary W. Adamson, May 15 2009
Sum_{k>=1} C(k-1)/2^(2k-1) = 1. The k-th term in the summation is the probability that a random walk on the integers (beginning at the origin) will arrive at positive one (for the first time) in exactly (2k-1) steps. - Geoffrey Critzer, Sep 12 2009
C(p+q)-C(p)*C(q) = Sum_{i=0..p-1, j=0..q-1} C(i)*C(j)*C(p+q-i-j-1). - Groux Roland, Nov 13 2009
Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.) - Peter Luschny, Mar 13 2010
Let A179277 = A(x). Then C(x) is satisfied by A(x)/A(x^2). - Gary W. Adamson, Jul 07 2010
a(n) is also the number of quivers in the mutation class of type B_n or of type C_n. - Christian Stump, Nov 02 2010
From Matthew Vandermast, Nov 22 2010: (Start)
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color while satisfying the following conditions: 1. No two colors are chosen the same positive number of times. 2. For any two colors (c, d) that are chosen at least once, color c is chosen more times than color d iff color c appears more times in the original set than color d.
If the second requirement is lifted, the number of acceptable ways equals A000110(n+1). See related comments for A016098, A085082. (End)
Deutsch and Sagan prove the Catalan number C_n is odd if and only if n = 2^a - 1 for some nonnegative integer a. Lin proves for every odd Catalan number C_n, we have C_n == 1 (mod 4). - Jonathan Vos Post, Dec 09 2010
a(n) is the number of functions f:{1,2,...,n}->{1,2,...,n} such that f(1)=1 and for all n >= 1 f(n+1) <= f(n)+1. For a nice bijection between this set of functions and the set of length 2n Dyck words, see page 333 of the Fxtbook (see link below). - Geoffrey Critzer, Dec 16 2010
Postnikov (2005) defines "generalized Catalan numbers" associated with buildings (e.g., Catalan numbers of Type B, see A000984). - N. J. A. Sloane, Dec 10 2011
Number of permutations in S(n) for which length equals depth. - Bridget Tenner, Feb 22 2012
a(n) is also the number of standard Young tableau of shape (n,n). - Thotsaporn Thanatipanonda, Feb 25 2012
a(n) is the number of binary sequences of length 2n+1 in which the number of ones first exceed the number of zeros at entry 2n+1. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012
Number of binary necklaces of length 2*n+1 containing n 1's (or, by symmetry, 0's). All these are Lyndon words and their representatives (as cyclic maxima) are the binary Dyck words. - Joerg Arndt, Nov 12 2012
Number of sequences consisting of n 'x' letters and n 'y' letters such that (counting from the left) the 'x' count >= 'y' count. For example, for n=3 we have xxxyyy, xxyxyy, xxyyxy, xyxxyy and xyxyxy. - Jon Perry, Nov 16 2012
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps come in 2 colors. Example: a(4)=14 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 8 paths of shape HHH, 2 paths of shape UHD, 2 paths of shape UDH, and 2 paths of shape HUD. - José Luis Ramírez Ramírez, Jan 16 2013
If p is an odd prime, then (-1)^((p-1)/2)*a((p-1)/2) mod p = 2. - Gary Detlefs, Feb 20 2013
Conjecture: For any positive integer n, the polynomial Sum_{k=0..n} a(k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
a(n) is the size of the Jones monoid on 2n points (cf. A225798). - James Mitchell, Jul 28 2013
For 0 < p < 1, define f(p) = Sum_{n>=0} a(n)*(p*(1-p))^n, then f(p) = min{1/p, 1/(1-p)}, so f(p) reaches its maximum value 2 at p = 0.5, and p*f(p) is constant 1 for 0.5 <= p < 1. - Bob Selcoe, Nov 16 2013 [Corrected by Jianing Song, May 21 2021]
No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
From Alexander Adamchuk, Dec 27 2013: (Start)
Prime p divides a((p+1)/2) for p > 3. See A120303(n) = Largest prime factor of Catalan number.
Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27 = 1.80613.. = A121839.
Log(Phi) = (125*C - 55) / (24*sqrt(5)), where C = Sum_{k>=1} (-1)^(k+1)*1/a(k). See A002390 = Decimal expansion of natural logarithm of golden ratio.
3-d analog of the Catalan numbers: (3n)!/(n!(n+1)!(n+2)!) = A161581(n) = A006480(n) / ((n+1)^2*(n+2)), where A006480(n) = (3n)!/(n!)^3 De Bruijn's S(3,n). (End)
For a relation to the inviscid Burgers's, or Hopf, equation, see A001764. - Tom Copeland, Feb 15 2014
From Fung Lam, May 01 2014: (Start)
One class of generalized Catalan numbers can be defined by g.f. A(x) = (1-sqrt(1-q*4*x*(1-(q-1)*x)))/(2*q*x) with nonzero parameter q. Recurrence: (n+3)*a(n+2) -2*q*(2*n+3)*a(n+1) +4*q*(q-1)*n*a(n) = 0 with a(0)=1, a(1)=1.
Asymptotic approximation for q >= 1: a(n) ~ (2*q+2*sqrt(q))^n*sqrt(2*q*(1+sqrt(q))) /sqrt(4*q^2*Pi*n^3).
For q <= -1, the g.f. defines signed sequences with asymptotic approximation: a(n) ~ Re(sqrt(2*q*(1+sqrt(q)))*(2*q+2*sqrt(q))^n) / sqrt(q^2*Pi*n^3), where Re denotes the real part. Due to Stokes' phenomena, accuracy of the asymptotic approximation deteriorates at/near certain values of n.
Special cases are A000108 (q=1), A068764 to A068772 (q=2 to 10), A240880 (q=-3).
(End)
Number of sequences [s(0), s(1), ..., s(n)] with s(n)=0, Sum_{j=0..n} s(j) = n, and Sum_{j=0..k} s(j)-1 >= 0 for k < n-1 (and necessarily Sum_{j=0..n-1} s(j)-1 = 0). These are the branching sequences of the (ordered) trees with n non-root nodes, see example. - Joerg Arndt, Jun 30 2014
Number of stack-sortable permutations of [n], these are the 231-avoiding permutations; see the Bousquet-Mélou reference. - Joerg Arndt, Jul 01 2014
a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 132. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
In a one-dimensional medium with elastic scattering (zig-zag walk), first recurrence after 2n+1 scattering events has the probability C(n)/2^(2n+1). - Joachim Wuttke, Sep 11 2014
The o.g.f. C(x) = (1 - sqrt(1-4x))/2, for the Catalan numbers, with comp. inverse Cinv(x) = x*(1-x) and the functions P(x) = x / (1 + t*x) and its inverse Pinv(x,t) = -P(-x,t) = x / (1 - t*x) form a group under composition that generates or interpolates among many classic arrays, such as the Motzkin (Riordan, A005043), Fibonacci (A000045), and Fine (A000957) numbers and polynomials (A030528), and enumerating arrays for Motzkin, Dyck, and Łukasiewicz lattice paths and different types of trees and non-crossing partitions (A091867, connected to sums of the refined Narayana numbers A134264). - Tom Copeland, Nov 04 2014
Conjecture: All the rational numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015
The Catalan number series A000108(n+3), offset n=0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset n=0 (empirical observation). - Tony Foster III, Sep 05 2016
Hankel transforms of the Catalan numbers with the first 2, 4, and 5 terms omitted give A001477, A006858, and A091962, respectively, without the first 2 terms in all cases. More generally, the Hankel transform of the Catalan numbers with the first k terms omitted is H_k(n) = Product_{j=1..k-1} Product_{i=1..j} (2*n+j+i)/(j+i) [see Cigler (2011), Eq. (1.14) and references therein]; together they form the array A078920/A123352/A368025. - Andrey Zabolotskiy, Oct 13 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. See S. J. Miller, ed., 2015, p. 5. - N. J. A. Sloane, Feb 09 2017
Coefficients of the generating series associated to the Magmatic and Dendriform operadic algebras. Cf. p. 422 and 435 of the Loday et al. paper. - Tom Copeland, Jul 08 2018
Let M_n be the n X n matrix with M_n(i,j) = binomial(i+j-1,2j-2); then det(M_n) = a(n). - Tony Foster III, Aug 30 2018
Also the number of Catalan trees, or planted plane trees (Bona, 2015, p. 299, Theorem 4.6.3). - N. J. A. Sloane, Dec 25 2018
Number of coalescent histories for a caterpillar species tree and a matching caterpillar gene tree with n+1 leaves (Rosenberg 2007, Corollary 3.5). - Noah A Rosenberg, Jan 28 2019
Finding solutions of eps*x^2+x-1 = 0 for eps small, that is, writing x = Sum_{n>=0} x_{n}*eps^n and expanding, one finds x = 1 - eps + 2*eps^2 - 5*eps^3 + 14*eps^3 - 42*eps^4 + ... with x_{n} = (-1)^n*C(n). Further, letting x = 1/y and expanding y about 0 to find large roots, that is, y = Sum_{n>=1} y_{n}*eps^n, one finds y = 0 - eps + eps^2 - 2*eps^3 + 5*eps^3 - ... with y_{n} = (-1)^n*C(n-1). - Derek Orr, Mar 15 2019
Permutations of length n that produce a bipartite permutation graph of order n [see Knuth (1973), Busch (2006), Golumbic and Trenk (2004)]. - Elise Anderson, R. M. Argus, Caitlin Owens, Tessa Stevens, Jun 27 2019
For n > 0, a random selection of n + 1 objects (the minimum number ensuring one pair by the pigeonhole principle) from n distinct pairs of indistinguishable objects contains only one pair with probability 2^(n-1)/a(n) = b(n-1)/A098597(n), where b is the 0-offset sequence with the terms of A120777 repeated (1,1,4,4,8,8,64,64,128,128,...). E.g., randomly selecting 6 socks from 5 pairs that are black, blue, brown, green, and white, results in only one pair of the same color with probability 2^(5-1)/a(5) = 16/42 = 8/21 = b(4)/A098597(5). - Rick L. Shepherd, Sep 02 2019
See Haran & Tabachnikov link for a video discussing Conway-Coxeter friezes. The Conway-Coxeter friezes with n nontrivial rows are generated by the counts of triangles at each vertex in the triangulations of regular n-gons, of which there are a(n). - Charles R Greathouse IV, Sep 28 2019
For connections to knot theory and scattering amplitudes from Feynman diagrams, see Broadhurst and Kreimer, and Todorov. Eqn. 6.12 on p. 130 of Bessis et al. becomes, after scaling, -12g * r_0(-y/(12g)) = (1-sqrt(1-4y))/2, the o.g.f. (expressed as a Taylor series in Eqn. 7.22 in 12gx) given for the Catalan numbers in Copeland's (Sep 30 2011) formula below. (See also Mizera p. 34, Balduf pp. 79-80, Keitel and Bartosch.) - Tom Copeland, Nov 17 2019
Number of permutations in S_n whose principal order ideals in the weak order are modular lattices. - Bridget Tenner, Jan 16 2020
Number of permutations in S_n whose principal order ideals in the weak order are distributive lattices. - Bridget Tenner, Jan 16 2020
Legendre gives the following formula for computing the square root modulo 2^m:
sqrt(1 + 8*a) mod 2^m = (1 + 4*a*Sum_{i=0..m-4} C(i)*(-2*a)^i) mod 2^m
as cited by L. D. Dickson, History of the Theory of Numbers, Vol. 1, 207-208. - Peter Schorn, Feb 11 2020
a(n) is the number of length n permutations sorted to the identity by a consecutive-132-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020
Number of non-crossing partitions of a 2*n-set with n blocks of size 2. Also number of non-crossing partitions of a 2*n-set with n+1 blocks of size at most 3, and without cyclical adjacencies. The two partitions can be mapped by rotated Kreweras bijection. - Yuchun Ji, Jan 18 2021
Named by Riordan (1968, and earlier in Mathematical Reviews, 1948 and 1964) after the French and Belgian mathematician Eugène Charles Catalan (1814-1894) (see Pak, 2014). - Amiram Eldar, Apr 15 2021
For n >= 1, a(n-1) is the number of interpretations of x^n is an algebra where power-associativity is not assumed. For example, for n = 4 there are a(3) = 5 interpretations: x(x(xx)), x((xx)x), (xx)(xx), (x(xx))x, ((xx)x)x. See the link "Non-associate powers and a functional equation" from I. M. H. Etherington and the page "Nonassociative Product" from Eric Weisstein's World of Mathematics for detailed information. See also A001190 for the case where multiplication is commutative. - Jianing Song, Apr 29 2022
Number of states in the transition diagram associated with the Laplacian system over the complete graph K_N, corresponding to ordered initial conditions x_1 < x_2 < ... < x_N. - Andrea Arlette España, Nov 06 2022
a(n) is the number of 132-avoiding stabilized-interval-free permutations of size n+1. - Juan B. Gil, Jun 22 2023
Number of rooted polyominoes composed of n triangular cells of the hyperbolic regular tiling with Schläfli symbol {3,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
a(n) is the number of extremely lucky Stirling permutations of order n; i.e., the number of Stirling permutations of order n that have exactly n lucky cars. (see Colmenarejo et al. reference) - Bridget Tenner, Apr 16 2024

Examples

			From _Joerg Arndt_ and Greg Stevenson, Jul 11 2011: (Start)
The following products of 3 transpositions lead to a 4-cycle in S_4:
(1,2)*(1,3)*(1,4);
(1,2)*(1,4)*(3,4);
(1,3)*(1,4)*(2,3);
(1,4)*(2,3)*(2,4);
(1,4)*(2,4)*(3,4). (End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
For n=3, a(3)=5 since there are exactly 5 binary sequences of length 7 in which the number of ones first exceed the number of zeros at entry 7, namely, 0001111, 0010111, 0011011, 0100111, and 0101011. - _Dennis P. Walsh_, Apr 11 2012
From _Joerg Arndt_, Jun 30 2014: (Start)
The a(4) = 14 branching sequences of the (ordered) trees with 4 non-root nodes are (dots denote zeros):
01:  [ 1 1 1 1 . ]
02:  [ 1 1 2 . . ]
03:  [ 1 2 . 1 . ]
04:  [ 1 2 1 . . ]
05:  [ 1 3 . . . ]
06:  [ 2 . 1 1 . ]
07:  [ 2 . 2 . . ]
08:  [ 2 1 . 1 . ]
09:  [ 2 1 1 . . ]
10:  [ 2 2 . . . ]
11:  [ 3 . . 1 . ]
12:  [ 3 . 1 . . ]
13:  [ 3 1 . . . ]
14:  [ 4 . . . . ]
(End)
		

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  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., 14 (1922), 55-62, 122-138 and 143-146.
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  • P. J. Larcombe and P. D. C. Wilson, On the trail of the Catalan sequence, Mathematics Today, 34 (1998), 114-117.
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  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 41.
  • J. Wuttke, The zig-zag walk with scattering and absorption on the real half line and in a lattice model, J. Phys. A 47 (2014), 215203, 1-9.

Crossrefs

A row of A060854.
See A001003, A001190, A001699, A000081 for other ways to count parentheses.
Enumerates objects encoded by A014486.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A051168 (diagonal of the square array described).
Cf. A033552, A176137 (partitions into Catalan numbers).
Cf. A000753, A000736 (Boustrophedon transforms).
Cf. A120303 (largest prime factor of Catalan number).
Cf. A121839 (reciprocal Catalan constant), A268813.
Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).
Cf. A002390 (decimal expansion of natural logarithm of golden ratio).
Coefficients of square root of the g.f. are A001795/A046161.
For a(n) mod 6 see A259667.
For a(n) in base 2 see A264663.
Hankel transforms with first terms omitted: A001477, A006858, A091962, A078920, A123352, A368025.
Cf. A332602 (conjectured production matrix).
Polyominoes: A001683(n+2) (oriented), A000207 (unoriented), A369314 (chiral), A208355(n-1) (achiral), A001764 {4,oo}.

Programs

  • GAP
    A000108:=List([0..30],n->Binomial(2*n,n)/(n+1)); # Muniru A Asiru, Feb 17 2018
  • Haskell
    import Data.List (genericIndex)
    a000108 n = genericIndex a000108_list n
    a000108_list = 1 : catalan [1] where
       catalan cs = c : catalan (c:cs) where
          c = sum $ zipWith (*) cs $ reverse cs
    -- Reinhard Zumkeller, Nov 12 2011
    a000108 = map last $ iterate (scanl1 (+) . (++ [0])) [1]
    -- David Spies, Aug 23 2015
    
  • Magma
    C:= func< n | Binomial(2*n,n)/(n+1) >; [ C(n) : n in [0..60]];
    
  • Magma
    [Catalan(n): n in [0..40]]; // Vincenzo Librandi, Apr 02 2011
    
  • Maple
    A000108 := n->binomial(2*n,n)/(n+1);
    G000108 := (1 - sqrt(1 - 4*x)) / (2*x);
    spec := [ A, {A=Prod(Z,Sequence(A))}, unlabeled ]: [ seq(combstruct[count](spec, size=n+1), n=0..42) ];
    with(combstruct): bin := {B=Union(Z,Prod(B,B))}: seq(count([B,bin,unlabeled],size=n+1), n=0..25); # Zerinvary Lajos, Dec 05 2007
    gser := series(G000108, x=0, 42): seq(coeff(gser, x, n), n=0..41); # Zerinvary Lajos, May 21 2008
    seq((2*n)!*coeff(series(hypergeom([],[2],x^2),x,2*n+2),x,2*n),n=0..30); # Peter Luschny, Jan 31 2015
    A000108List := proc(m) local A, P, n; A := [1, 1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
    A := [op(A), P[-1]] od; A end: A000108List(31); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[(2 n)!/n!/(n + 1)!, {n, 0, 20}]
    Table[4^n Gamma[n + 1/2]/(Sqrt[Pi] Gamma[n + 2]), {n, 0, 20}] (* Eric W. Weisstein, Oct 31 2024 *)
    Table[Hypergeometric2F1[1 - n, -n, 2, 1], {n, 0, 20}] (* Richard L. Ollerton, Sep 13 2006 *)
    Table[CatalanNumber @ n, {n, 0, 20}] (* Robert G. Wilson v, Feb 15 2011 *)
    CatalanNumber[Range[0, 20]] (* Eric W. Weisstein, Oct 31 2024 *)
    CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *)
    CoefficientList[Series[(1 - Sqrt[1 - 4 x])/(2 x), {x, 0, 20}], x] (* Stefano Spezia, Aug 31 2018 *)
  • Maxima
    A000108(n):=binomial(2*n,n)/(n+1)$ makelist(A000108(n),n,0,30); /* Martin Ettl, Oct 24 2012 */
    
  • MuPAD
    combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007
    
  • PARI
    a(n)=binomial(2*n,n)/(n+1) \\ M. F. Hasler, Aug 25 2012
    
  • PARI
    a(n) = (2*n)! / n! / (n+1)!
    
  • PARI
    a(n) = my(A, m); if( n<0, 0, m=1; A = 1 + x + O(x^2); while(m<=n, m*=2; A = sqrt(subst(A, x, 4*x^2)); A += (A - 1) / (2*x*A)); polcoeff(A, n));
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */
    
  • PARI
    (recur(a,b)=if(b<=2,(a==2)+(a==b)+(a!=b)*(1+a/2), (1+a/b)*recur(a,b-1))); a(n)=recur(n,n); \\ R. J. Cano, Nov 22 2012
    
  • PARI
    x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015
    
  • Python
    from gmpy2 import divexact
    A000108 = [1, 1]
    for n in range(1, 10**3):
        A000108.append(divexact(A000108[-1]*(4*n+2),(n+2))) # Chai Wah Wu, Aug 31 2014
    
  • Python
    # Works in Sage also.
    A000108 = [1]
    for n in range(1000):
        A000108.append(A000108[-1]*(4*n+2)//(n+2)) # Günter Rote, Nov 08 2023
    
  • Sage
    [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    # Generalized algorithm of L. Seidel
    def A000108_list(n) :
        D = [0]*(n+1); D[1] = 1
        b = True; h = 1; R = []
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1; R.append(D[1])
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            b = not b
        return R
    A000108_list(31) # Peter Luschny, Jun 02 2012
    

Formula

a(n) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!) = A000984(n)/(n+1).
Recurrence: a(n) = 2*(2*n-1)*a(n-1)/(n+1) with a(0) = 1.
Recurrence: a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x), and satisfies A(x) = 1 + x*A(x)^2.
a(n) = Product_{k=2..n} (1 + n/k).
a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... - Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015
Using the Stirling approximation in A000142 we get the asymptotic expansion a(n) ~ 4^n / (sqrt(Pi * n) * (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001
Integral representation: a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*sqrt((4-x)/x). - Karol A. Penson, Apr 12 2001
E.g.f.: exp(2*x)*(I_0(2*x)-I_1(2*x)), where I_n is Bessel function. - Karol A. Penson, Oct 07 2001
a(n) = polygorial(n, 6)/polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com), Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun 27 2003
G.f. A(x) satisfies Sum_{k>=1} k(A(x)-1)^k = Sum_{n>=1} 4^{n-1}*x^n. - Shapiro, Woan, Getu
a(n+m) = Sum_{k} A039599(n, k)*A039599(m, k). - Philippe Deléham, Dec 22 2003
a(n+1) = (1/(n+1))*Sum_{k=0..n} a(n-k)*binomial(2k+1, k+1). - Philippe Deléham, Jan 24 2004
a(n) = Sum_{k>=0} A008313(n, k)^2. - Philippe Deléham, Feb 14 2004
a(m+n+1) = Sum_{k>=0} A039598(m, k)*A039598(n, k). - Philippe Deléham, Feb 15 2004
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
Sum_{n>=0} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = A268813 = 2.806133050770763... (see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13 2005
a(n) = Sum_{k=0..floor(n/2)} ((n-2*k+1)*binomial(n, n-k)/(n-k+1))^2, which is equivalent to: a(n) = Sum_{k=0..n} A053121(n, k)^2, for n >= 0. - Paul D. Hanna, Apr 23 2005
a((m+n)/2) = Sum_{k>=0} A053121(m, k)*A053121(n, k) if m+n is even. - Philippe Deléham, May 26 2005
E.g.f. Sum_{n>=0} a(n) * x^(2*n) / (2*n)! = BesselI(1, 2*x) / x. - Michael Somos, Jun 22 2005
Given g.f. A(x), then B(x) = x * A(x^3) satisfies 0 = f(x, B(X)) where f(u, v) = u - v + (u*v)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005
a(n) = a(n-1)*(4-6/(n+1)). a(n) = 2a(n-1)*(8a(n-2)+a(n-1))/(10a(n-2)-a(n-1)). - Franklin T. Adams-Watters, Feb 08 2006
Sum_{k>=1} a(k)/4^k = 1. - Franklin T. Adams-Watters, Jun 28 2006
a(n) = A047996(2*n+1, n). - Philippe Deléham, Jul 25 2006
Binomial transform of A005043. - Philippe Deléham, Oct 20 2006
a(n) = Sum_{k=0..n} (-1)^k*A116395(n,k). - Philippe Deléham, Nov 07 2006
a(n) = (1/(s-n))*Sum_{k=0..n} (-1)^k (k+s-n)*binomial(s-n,k) * binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
a(k) = Sum_{i=1..k} |A008276(i,k)| * (k-1)^(k-i) / k!. - André F. Labossière, May 29 2007
a(n) = Sum_{k=0..n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20 2007
a(n) = Sum_{k=0..n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20 2007
Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007
Limit_{n->oo} (1 + Sum_{k=0..n} a(k)/A004171(k)) = 4/Pi. - Reinhard Zumkeller, Aug 26 2008
a(n) = Sum_{k=0..n} A120730(n,k)^2 and a(k+1) = Sum_{n>=k} A120730(n,k). - Philippe Deléham, Oct 18 2008
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i < l_(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(n) = A000680(n)/A006472(n+1). - Mark Dols, Jul 14 2010; corrected by M. F. Hasler, Nov 08 2015
Let A(x) be the g.f., then B(x)=x*A(x) satisfies the differential equation B'(x)-2*B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011
Complement of A092459; A010058(a(n)) = 1. - Reinhard Zumkeller, Mar 29 2011
G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18 2011
With F(x) = (1-2*x-sqrt(1-4*x))/(2*x) an o.g.f. in x for the Catalan series, G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). - Tom Copeland, Sep 04 2011
With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for A115291. - Tom Copeland, Sep 04 2011
From Tom Copeland, Sep 30 2011: (Start)
With F(x) = (1-sqrt(1-4*x))/2 an o.g.f. in x for the Catalan series, G(x)= x*(1-x) is the compositional inverse and this relates the Catalan numbers to the row sums of A125181.
With H(x) = 1/(dG(x)/dx) = 1/(1-2x), the n-th Catalan number (offset 1) is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
G.f.: (1-sqrt(1-4*x))/(2*x) = G(0) where G(k) = 1 + (4*k+1)*x/(k+1-2*x*(k+1)*(4*k+3)/(2*x*(4*k+3)+(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(1,2*x)) = G(0) where G(k) = 1 + (4*k+1)*x/((k+1)*(2*k+1)-x*(k+1)*(2*k+1)*(4*k+3)/(x*(4*k+3)+(k+1)*(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given just above, and also by Karol A. Penson further above. - Wolfdieter Lang, Jan 13 2012
A076050(a(n)) = n + 1 for n > 0. - Reinhard Zumkeller, Feb 17 2012
a(n) = A208355(2*n-1) = A208355(2*n) for n > 0. - Reinhard Zumkeller, Mar 04 2012
a(n+1) = A214292(2*n+1,n) = A214292(2*n+2,n). - Reinhard Zumkeller, Jul 12 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = k*(4*x+1) + 2*x + 2 - x*(2*k+3)*(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013
For n > 0: a(n) = sum of row n in triangle A001263. - Reinhard Zumkeller, Oct 10 2013
a(n) = binomial(2n,n-1)/n and a(n) mod n = binomial(2n,n) mod n = A059288(n). - Jonathan Sondow, Dec 14 2013
a(n-1) = Sum_{t1+2*t2+...+n*tn=n} (-1)^(1+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*a(1)^t1*a(2)^t2*...*a(n)^tn. - Mircea Merca, Feb 27 2014
a(n) = Sum_{k=1..n} binomial(n+k-1,n)/n if n > 0. Alexander Adamchuk, Mar 25 2014
a(n) = -2^(2*n+1) * binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014
a(n) = (4*A000984(n) - A000984(n+1))/2. - Stanislav Sykora, Aug 09 2014
a(n) = A246458(n) * A246466(n). - Tom Edgar, Sep 02 2014
a(n) = (2*n)!*[x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015
a(2n) = 2*A000150(2n); a(2n+1) = 2*A000150(2n+1) + a(n). - John Bodeen, Jun 24 2015
a(n) = Sum_{t=1..n+1} n^(t-1)*abs(Stirling1(n+1, t)) / Sum_{t=1..n+1} abs(Stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel Marcus, Oct 06 2015
a(n) ~ 4^(n-2)*(128 + 160/N^2 + 84/N^4 + 715/N^6 - 10180/N^8)/(N^(3/2)*Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*binomial(n+1-k,k)*a(n-k) if n > 0; and a(0) = 1. - David Pasino, Jun 29 2016
Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24*arccsch(2)/(25*sqrt(5)) = 14/25 - 24*A002390/(25*sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun 30 2016
C(n) = (1/n) * Sum_{i+j+k=n-1} C(i)*C(j)*C(k)*(k+1), n >= 1. - Yuchun Ji, Feb 21 2016
C(n) = 1 + Sum_{i+j+kYuchun Ji, Sep 01 2016
a(n) = A001700(n) - A162551(n) = binomial(2*n+1,n+1). - 2*binomial(2*n,n-1). - Taras Goy, Aug 09 2018
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x) = 2F1(1/2,1;2;4*x). G.f. A(x) satisfies A = 1 + x*A^2. - R. J. Mathar, Nov 17 2018
C(n) = 1 + Sum_{i=0..n-1} A000245(i). - Yuchun Ji, Jan 10 2019
From A.H.M. Smeets, Apr 11 2020: (Start)
(1+sqrt(1+4*x))/2 = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4; and sqrt(x+sqrt(x+sqrt(x+...))) = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4 and x <> 0. (End)
a(3n+1)*a(5n+4)*a(15n+10) = a(3n+2)*a(5n+2)*a(15n+11). The first case of Catalan product equation of a triple partition of 23n+15. - Yuchun Ji, Sep 27 2020
a(n) = 4^n * (-1)^(n+1) * 3F2[{n + 1,n + 1/2,n}, {3/2,1}, -1], n >= 1. - Sergii Voloshyn, Oct 22 2020
a(n) = 2^(1 + 2 n) * (-1)^(n)/(1 + n) * 3F2[{n, 1/2 + n, 1 + n}, {1/2, 1}, -1], n >= 1. - Sergii Voloshyn, Nov 08 2020
a(n) = (1/Pi)*4^(n+1)*Integral_{x=0..Pi/2} cos(x)^(2*n)*sin(x)^2 dx. - Greg Dresden, May 30 2021
From Peter Bala, Aug 17 2021: (Start)
G.f. A(x) satisfies A(x) = 1/sqrt(1 - 4*x) * A( -x/(1 - 4*x) ) and (A(x) + A(-x))/2 = 1/sqrt(1 - 4*x) * A( -2*x/(1 - 4*x) ); these are the cases k = 0 and k = -1 of the general formula 1/sqrt(1 - 4*x) * A( (k-1)*x/(1 - 4*x) ) = Sum_{n >= 0} ((k^(n+1) - 1)/(k - 1))*Catalan(n)*x^n.
2 - sqrt(1 - 4*x)/A( k*x/(1 - 4*x) ) = 1 + Sum_{n >= 1} (1 + (k + 1)^n) * Catalan(n-1)*x^n. (End)
Sum_{n>=0} a(n)*(-1/4)^n = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Mar 22 2022
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n>=0. - Michael Somos, Dec 12 2022
G.f.: (offset 1) 1/G(x), with G(x) = 1 - 2*x - x^2/G(x) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 01 2023
a(n) = K^(2n+1, n, 1) for all n >= 0, where K^(n, s, x) is the Krawtchouk polynomial defined to be Sum_{k=0..s} (-1)^k * binomial(n-x, s-k) * binomial(x, k). - Vladislav Shubin, Aug 17 2023
From Peter Bala, Feb 03 2024: (Start)
The g.f. A(x) satisfies the following functional equations:
A(x) = 1 + x/(1 - 4*x) * A(-x/(1 - 4*x))^2,
A(x^2) = 1/(1 - 2*x) * A(- x/(1 - 2*x))^2 and, for arbitrary k,
1/(1 - k*x) * A(x/(1 - k*x))^2 = 1/(1 - (k+4)*x) * A(-x/(1 - (k+4)*x))^2. (End)
a(n) = A363448(n) + A363449(n). - Julien Rouyer, Jun 28 2024

A001147 Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).

Original entry on oeis.org

1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
Offset: 0

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Comments

The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085).
a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022]
a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012
For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number (A008517). - Ross La Haye, Feb 13 2005
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/2)/sqrt(2*Pi*x) dx, n >= 0. - Karol A. Penson, Oct 10 2005
a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for iDavid Callan, Sep 25 2006
a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006
Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006
a(n) = total weight of all Dyck n-paths (A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006
a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials (A000142) and increasing quaternary trees by triple factorials (A007559). - David Callan, Mar 30 2007
From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start)
a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations.
This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels.
For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End)
A139541(n) = a(n) * a(2*n). - Reinhard Zumkeller, Apr 25 2008
a(n+1) = Sum_{j=0..n} A074060(n,j) * 2^j. - Tom Copeland, Sep 01 2008
From Emeric Deutsch, Jun 05 2009: (Start)
a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
a(n) = Sum_{k>=0} k*A079267(n,k).
(End)
Hankel transform is A137592. - Paul Barry, Sep 18 2009
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009
a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009
The Hankel transform of a(n+1) is A168467. - Paul Barry, Dec 04 2009
Partial products of odd numbers. - Juri-Stepan Gerasimov, Oct 17 2010
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011
a(n) is the leading coefficient of the Bessel polynomial y_n(x) (cf. A001498). - Leonid Bedratyuk, Jun 01 2012
For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013
a(n) is the size of the Brauer monoid on 2n points (see A227545). - James Mitchell, Jul 28 2013
For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014
a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014
For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014
Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014
a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017
Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
From Christopher J. Smyth, Jan 26 2018: (Start)
a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021
Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021
From Tom Copeland, Jun 14 2021: (Start)
Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial.
These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991.
a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End)
Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021
b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022

Examples

			a(3) = 1*3*5 = 15.
From _Joerg Arndt_, Sep 10 2013: (Start)
There are a(3)=15 involutions of 6 elements without fixed points:
  #:    permutation           transpositions
  01:  [ 1 0 3 2 5 4 ]      (0, 1) (2, 3) (4, 5)
  02:  [ 1 0 4 5 2 3 ]      (0, 1) (2, 4) (3, 5)
  03:  [ 1 0 5 4 3 2 ]      (0, 1) (2, 5) (3, 4)
  04:  [ 2 3 0 1 5 4 ]      (0, 2) (1, 3) (4, 5)
  05:  [ 2 4 0 5 1 3 ]      (0, 2) (1, 4) (3, 5)
  06:  [ 2 5 0 4 3 1 ]      (0, 2) (1, 5) (3, 4)
  07:  [ 3 2 1 0 5 4 ]      (0, 3) (1, 2) (4, 5)
  08:  [ 3 4 5 0 1 2 ]      (0, 3) (1, 4) (2, 5)
  09:  [ 3 5 4 0 2 1 ]      (0, 3) (1, 5) (2, 4)
  10:  [ 4 2 1 5 0 3 ]      (0, 4) (1, 2) (3, 5)
  11:  [ 4 3 5 1 0 2 ]      (0, 4) (1, 3) (2, 5)
  12:  [ 4 5 3 2 0 1 ]      (0, 4) (1, 5) (2, 3)
  13:  [ 5 2 1 4 3 0 ]      (0, 5) (1, 2) (3, 4)
  14:  [ 5 3 4 1 2 0 ]      (0, 5) (1, 3) (2, 4)
  15:  [ 5 4 3 2 1 0 ]      (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
  • Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
  • F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
  • C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
  • G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.

Crossrefs

Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables).
Constant terms of polynomials in A098503. First row of array A099020.
Subsequence of A248652.
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A053871 (binomial transform).

Programs

  • GAP
    A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018
    
  • Haskell
    a001147 n = product [1, 3 .. 2 * n - 1]
    a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
    -- Reinhard Zumkeller, Feb 15 2015, Dec 03 2011
    
  • Magma
    A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ A001147(n): n in [0..20] ]; // Klaus Brockhaus, Jun 22 2011
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    f := n->(2*n)!/(n!*2^n);
    A001147 := proc(n) doublefactorial(2*n-1); end: # R. J. Mathar, Jul 04 2009
    A001147 := n -> 2^n*pochhammer(1/2, n); # Peter Luschny, Aug 09 2009
    G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009
    series(hypergeom([1,1/2],[],2*x),x=0,20); # Mark van Hoeij, Apr 07 2013
  • Mathematica
    Table[(2 n - 1)!!, {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
    a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *)
    Join[{1}, Range[1, 41, 2]!!] (* Harvey P. Dale, Jan 28 2017 *)
    a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
    (2 Range[0, 20] - 1)!! (* Eric W. Weisstein, Jul 22 2017 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(sum(binomial(n-1,i)*binomial(n-i-1,j)*a(i)*a(j)*a(n-i-j-1),j,0,n-i-1),i,0,n-1); /* Vladimir Kruchinin, May 06 2020 */
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011
    
  • Python
    from sympy import factorial2
    def a(n): return factorial2(2 * n - 1)
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 22 2017
    
  • Sage
    [rising_factorial(n+1,n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/A000165(n).
a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003
The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=0..n} (-2)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006
1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006
a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008
a(n) = A006882(2n-1). - R. J. Mathar, Jul 04 2009
G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009
a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009
a(n) = 2^n*gamma(n+1/2)/gamma(1/2). - Jaume Oliver Lafont, Nov 09 2009
G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = Sum_{i=1..n} binomial(n,i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010
E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011
a(n) = A123023(2*n). - Michael Somos, Jul 24 2011
a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle).
a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle.
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix:
1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011
a(n) = A009445(n) / A014481(n). - Reinhard Zumkeller, Dec 03 2011
a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = (2*n)4! = Gauss_factorial(2*n,4) = Product{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013
G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013
G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014
0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017
Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017
a(n) = A028338(n, 0), n >= 0. - Wolfdieter Lang, May 27 2017
a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020
From Amiram Eldar, Jun 29 2020: (Start)
Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024

Extensions

Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009

A001764 a(n) = binomial(3*n,n)/(2*n+1) (enumerates ternary trees and also noncrossing trees).

Original entry on oeis.org

1, 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, 8414640, 50067108, 300830572, 1822766520, 11124755664, 68328754959, 422030545335, 2619631042665, 16332922290300, 102240109897695, 642312451217745, 4048514844039120, 25594403741131680, 162250238001816900
Offset: 0

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Keywords

Comments

Smallest number of straight line crossing-free spanning trees on n points in the plane.
Number of dissections of some convex polygon by nonintersecting diagonals into polygons with an odd number of sides and having a total number of 2n+1 edges (sides and diagonals). - Emeric Deutsch, Mar 06 2002
Number of lattice paths of n East steps and 2n North steps from (0,0) to (n,2n) and lying weakly below the line y=2x. - David Callan, Mar 14 2004
With interpolated zeros, this has g.f. 2*sqrt(3)*sin(arcsin(3*sqrt(3)*x/2)/3)/(3*x) and a(n) = C(n+floor(n/2),floor(n/2))*C(floor(n/2),n-floor(n/2))/(n+1). This is the first column of the inverse of the Riordan array (1-x^2,x(1-x^2)) (essentially reversion of y-y^3). - Paul Barry, Feb 02 2005
Number of 12312-avoiding matchings on [2n].
Number of complete ternary trees with n internal nodes, or 3n edges.
Number of rooted plane trees with 2n edges, where every vertex has even outdegree ("even trees").
a(n) is the number of noncrossing partitions of [2n] with all blocks of even size. E.g.: a(2)=3 counts 12-34, 14-23, 1234. - David Callan, Mar 30 2007
Pfaff-Fuss-Catalan sequence C^{m}_n for m=3, see the Graham et al. reference, p. 347. eq. 7.66.
Also 3-Raney sequence, see the Graham et al. reference, p. 346-7.
The number of lattice paths from (0,0) to (2n,0) using an Up-step=(1,1) and a Down-step=(0,-2) and staying above the x-axis. E.g., a(2) = 3; UUUUDD, UUUDUD, UUDUUD. - Charles Moore (chamoore(AT)howard.edu), Jan 09 2008
a(n) is (conjecturally) the number of permutations of [n+1] that avoid the patterns 4-2-3-1 and 4-2-5-1-3 and end with an ascent. For example, a(4)=55 counts all 60 permutations of [5] that end with an ascent except 42315, 52314, 52413, 53412, all of which contain a 4-2-3-1 pattern and 42513. - David Callan, Jul 22 2008
Central terms of pendular triangle A167763. - Philippe Deléham, Nov 12 2009
With B(x,t)=x+t*x^3, the comp. inverse in x about 0 is A(x,t) = Sum_{j>=0} a(j) (-t)^j x^(2j+1). Let U(x,t)=(x-A(x,t))/t. Then DU(x,t)/Dt=dU/dt+U*dU/dx=0 and U(x,0)=x^3, i.e., U is a solution of the inviscid Burgers's, or Hopf, equation. Also U(x,t)=U(x-t*U(x,t),0) and dB(x,t)/dt = U(B(x,t),t) = x^3 = U(x,0). The characteristics for the Hopf equation are x(t) = x(0) + t*U(x(t),t) = x(0) + t*U(x(0),0) = x(0) + t*x(0)^3 = B(x(0),t). These results apply to all the Fuss-Catalan sequences with 3 replaced by n>0 and 2 by n-1 (e.g., A000108 with n=2 and A002293 with n=4), see also A086810, which can be generalized to A133437, for associahedra. - Tom Copeland, Feb 15 2014
Number of intervals (i.e., ordered pairs (x,y) such that x<=y) in the Kreweras lattice (noncrossing partitions ordered by refinement) of size n, see the Bernardi & Bonichon (2009) and Kreweras (1972) references. - Noam Zeilberger, Jun 01 2016
Number of sum-indecomposable (4231,42513)-avoiding permutations. Conjecturally, number of sum-indecomposable (2431,45231)-avoiding permutations. - Alexander Burstein, Oct 19 2017
a(n) is the number of topologically distinct endstates for the game Planted Brussels Sprouts on n vertices, see Ji and Propp link. - Caleb Ji, May 14 2018
Number of complete quadrillages of 2n+2-gons. See Baryshnikov p. 12. See also Nov 10 2014 comments in A134264. - Tom Copeland, Jun 04 2018
a(n) is the number of 2-regular words on the alphabet [n] that avoid the patterns 231 and 221. Equivalently, this is the number of 2-regular tortoise-sortable words on the alphabet [n] (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
a(n) is the number of Motzkin paths of length 3n with n steps of each type, with the condition that (1, 0) and (1, 1) steps alternate (starting with (1, 0)). - Helmut Prodinger, Apr 08 2019
a(n) is the number of uniquely sorted permutations of length 2n+1 that avoid the patterns 312 and 1342. - Colin Defant, Jun 08 2019
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. - Tom Copeland, Dec 13 2019
The sequences of Fuss-Catalan numbers, of which this is the first after the Catalan numbers A000108 (the next is A002293), appear in articles on random matrices and quantum physics. See Banica et al., Collins et al., and Mlotkowski et al. Interpretations of these sequences in terms of the cardinality of specific sets of noncrossing partitions are provided by A134264. - Tom Copeland, Dec 21 2019
Call C(p, [alpha], g) the number of partitions of a cyclically ordered set with p elements, of cyclic type [alpha], and of genus g (the genus g Faa di Bruno coefficients of type [alpha]). This sequence counts the genus 0 partitions (non-crossing, or planar, partitions) of p = 3n into n parts of length 3: a(n) = C(3n, [3^n], 0). For genus 1 see A371250, for genus 2 see A371251. - Robert Coquereaux, Mar 16 2024
a(n) is the total number of down steps before the first up step in all 2_1-Dyck paths of length 3*n for n > 0. A 2_1-Dyck path is a lattice path with steps (1,2), (1,-1) that starts and ends at y = 0 and does not go below the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [n]. - Francesca Aicardi, May 28 2022
a(n) is the number of parking functions of size n avoiding the patterns 231 and 321. - Lara Pudwell, Apr 10 2023
Number of rooted polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {4,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 3 of the family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 05 2024
The number of Apollonian networks (planar 3-trees) with n+3 vertices with a given base triangle. - Allan Bickle, Feb 20 2024
Number of rooted polyominoes composed of n tetrahedral cells of the hyperbolic regular tiling with Schläfli symbol {3,3,oo}. A rooted polyomino has one external face identified, and chiral pairs are counted as two. a(n) = T(n) in the second Beineke and Pippert link. - Robert A. Russell, Mar 20 2024

Examples

			a(2) = 3 because the only dissections with 5 edges are given by a square dissected by any of the two diagonals and the pentagon with no dissecting diagonal.
G.f. = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + 43263*x^8 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347. See also the Pólya-Szegő reference.
  • W. Kuich, Languages and the enumeration of planted plane trees. Nederl. Akad. Wetensch. Proc. Ser. A 73 = Indag. Math. 32, (1970), 268-280.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, p. 98.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001762, A001763, A002294 - A002296, A006013, A025174, A063548, A064017, A072247, A072248, A134264, A143603, A258708, A256311, A188687 (binomial transform), A346628 (inverse binomial transform).
A column of triangle A102537.
Bisection of A047749 and A047761.
Row sums of triangles A108410 and A108767.
Second column of triangle A062993.
Mod 3 = A113047.
2D Polyominoes: A005034 (oriented), A005036 (unoriented), A369315 (chiral), A047749 (achiral), A000108 {3,oo}, A002293 {5,oo}.
3D Polyominoes: A007173 (oriented), A027610 (unoriented), A371350 (chiral), A371351 (achiral).
Cf. A130564 (for C(k, n) cases).

Programs

  • GAP
    List([0..25],n->Binomial(3*n,n)/(2*n+1)); # Muniru A Asiru, Oct 31 2018
    
  • Haskell
    a001764 n = a001764_list !! n
    a001764_list = 1 : [a258708 (2 * n) n | n <- [1..]]
    -- Reinhard Zumkeller, Jun 23 2015
    
  • Magma
    [Binomial(3*n,n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Sep 04 2014
    
  • Maple
    A001764 := n->binomial(3*n,n)/(2*n+1): seq(A001764(n), n=0..25);
    with(combstruct): BB:=[T,{T=Prod(Z,F),F=Sequence(B),B=Prod(F,Z,F)}, unlabeled]:seq(count(BB,size=i),i=0..22); # Zerinvary Lajos, Apr 22 2007
    with(combstruct):BB:=[S, {B = Prod(S,S,Z), S = Sequence(B)}, labelled]: seq(count(BB, size=n)/n!, n=0..21); # Zerinvary Lajos, Apr 25 2008
    n:=30:G:=series(RootOf(g = 1+x*g^3, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015
    alias(PS=ListTools:-PartialSums): A001764List := proc(m) local A, P, n;
    A := [1,1]; P := [1]; for n from 1 to m - 2 do P := PS(PS([op(P), P[-1]]));
    A := [op(A), P[-1]] od; A end: A001764List(25); # Peter Luschny, Mar 26 2022
  • Mathematica
    InverseSeries[Series[y-y^3, {y, 0, 24}], x] (* then a(n)=y(2n+1)=ways to place non-crossing diagonals in convex (2n+4)-gon so as to create only quadrilateral tiles *) (* Len Smiley, Apr 08 2000 *)
    Table[Binomial[3n,n]/(2n+1),{n,0,25}] (* Harvey P. Dale, Jul 24 2011 *)
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n! / (2*n + 1)!)};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( serreverse( x - x^3 + O(x^(2*n + 2))), 2*n + 1))};
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( m=1, n, A = 1 + x * A^3); polcoeff(A, n))};
    
  • PARI
    b=vector(22);b[1]=1;for(n=2,22,for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if((i-1)+(j-1)+(k-1)-(n-2),NULL,b[n]=b[n]+b[i]*b[j]*b[k])))));a(n)=b[n+1]; print1(a(0));for(n=1,21,print1(", ",a(n))) \\ Gerald McGarvey, Oct 08 2008
    
  • PARI
    Vec(1 + serreverse(x / (1+x)^3 + O(x^30))) \\ Gheorghe Coserea, Aug 05 2015
    
  • Python
    from math import comb
    def A001764(n): return comb(3*n,n)//(2*n+1) # Chai Wah Wu, Nov 10 2022
  • Sage
    def A001764_list(n) :
        D = [0]*(n+1); D[1] = 1
        R = []; b = false; h = 1
        for i in range(2*n) :
            for k in (1..h) : D[k] += D[k-1]
            if not b : R.append(D[h])
            else : h += 1
            b = not b
        return R
    A001764_list(22) # Peter Luschny, May 03 2012
    

Formula

From Karol A. Penson, Nov 08 2001: (Start)
G.f.: (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))).
E.g.f.: hypergeom([1/3, 2/3], [1, 3/2], 27/4*x).
Integral representation as n-th moment of a positive function on [0, 27/4]: a(n) = Integral_{x=0..27/4} (x^n*((1/12) * 3^(1/2) * 2^(1/3) * (2^(1/3)*(27 + 3 * sqrt(81 - 12*x))^(2/3) - 6 * x^(1/3))/(Pi * x^(2/3)*(27 + 3 * sqrt(81 - 12*x))^(1/3)))), n >= 0. This representation is unique. (End)
G.f. A(x) satisfies A(x) = 1+x*A(x)^3 = 1/(1-x*A(x)^2) [Cyvin (1998)]. - Ralf Stephan, Jun 30 2003
a(n) = n-th coefficient in expansion of power series P(n), where P(0) = 1, P(k+1) = 1/(1 - x*P(k)^2).
G.f. Rev(x/c(x))/x, where c(x) is the g.f. of A000108 (Rev=reversion of). - Paul Barry, Mar 26 2010
From Gary W. Adamson, Jul 07 2011: (Start)
Let M = the production matrix:
1, 1
2, 2, 1
3, 3, 2, 1
4, 4, 3, 2, 1
5, 5, 4, 3, 2, 1
...
a(n) = upper left term in M^n. Top row terms of M^n = (n+1)-th row of triangle A143603, with top row sums generating A006013: (1, 2, 7, 30, 143, 728, ...). (End)
Recurrence: a(0)=1; a(n) = Sum_{i=0..n-1, j=0..n-1-i} a(i)a(j)a(n-1-i-j) for n >= 1 (counts ternary trees by subtrees of the root). - David Callan, Nov 21 2011
G.f.: 1 + 6*x/(Q(0) - 6*x); Q(k) = 3*x*(3*k + 1)*(3*k + 2) + 2*(2*(k^2) + 5*k +3) - 6*x*(2*(k^2) + 5*k + 3)*(3*k + 4)*(3*k + 5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2011
D-finite with recurrence: 2*n*(2n+1)*a(n) - 3*(3n-1)*(3n-2)*a(n-1) = 0. - R. J. Mathar, Dec 14 2011
REVERT transform of A115140. BINOMIAL transform is A188687. SUMADJ transform of A188678. HANKEL transform is A051255. INVERT transform of A023053. INVERT transform is A098746. - Michael Somos, Apr 07 2012
(n + 1) * a(n) = A174687(n).
G.f.: F([2/3,4/3], [3/2], 27/4*x) / F([2/3,1/3], [1/2], (27/4)*x) where F() is the hypergeometric function. - Joerg Arndt, Sep 01 2012
a(n) = binomial(3*n+1, n)/(3*n+1) = A062993(n+1,1). - Robert FERREOL, Apr 03 2015
a(n) = A258708(2*n,n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
0 = a(n)*(-3188646*a(n+2) + 20312856*a(n+3) - 11379609*a(n+4) + 1437501*a(n+5)) + a(n+1)*(177147*a(n+2) - 2247831*a(n+3) + 1638648*a(n+4) - 238604*a(n+5)) + a(n+2)*(243*a(n+2) + 31497*a(n+3) - 43732*a(n+4) + 8288*a(n+5)) for all integer n. - Michael Somos, Jun 03 2016
a(n) ~ 3^(3*n + 1/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). - Ilya Gutkovskiy, Nov 21 2016
Given g.f. A(x), then A(1/8) = -1 + sqrt(5), A(2/27) = (-1 + sqrt(3))*3/2, A(4/27) = 3/2, A(3/64) = -2 + 2*sqrt(7/3), A(5/64) = (-1 + sqrt(5))*2/sqrt(5), etc. A(n^2/(n+1)^3) = (n+1)/n if n > 1. - Michael Somos, Jul 17 2018
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/3)*binomial(3*n,n)*x^n/n ).
The sequence defined by b(n) := [x^n] A(x)^n = A224274(n) for n >= 1 and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. Cf. A060941. (End)
G.f.: 1/sqrt(B(x)+(1-6*x)/(9*B(x))+1/3), with B(x):=((27*x^2-18*x+2)/54-(x*sqrt((-(4-27*x))*x))/(2*3^(3/2)))^(1/3). - Vladimir Kruchinin, Sep 28 2021
x*A'(x)/A(x) = (A(x) - 1)/(- 2*A(x) + 3) = x + 5*x^2 + 28*x^3 + 165*x^4 + ... is the o.g.f. of A025174. Cf. A002293 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -2*n], [2], 1). Row sums of A108767. - Peter Bala, Aug 30 2023
G.f.: z*exp(3*z*hypergeom([1, 1, 4/3, 5/3], [3/2, 2, 2], (27*z)/4)) + 1.
- Karol A. Penson, Dec 19 2023
G.f.: hypergeometric([1/3, 2/3], [3/2], (3^3/2^2)*x). See the e.g.f. above. - Wolfdieter Lang, Feb 04 2024
a(n) = (3*n)! / (n!*(2*n+1)!). - Allan Bickle, Feb 20 2024
Sum_{n >= 0} a(n)*x^n/(1 + x)^(3*n+1) = 1. See A316371 and A346627. - Peter Bala, Jun 02 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 16 2025

A001263 Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
Offset: 1

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Comments

Number of antichains (or order ideals) in the poset 2*(k-1)*(n-k) or plane partitions with rows <= k-1, columns <= n-k and entries <= 2. - Mitch Harris, Jul 15 2000
T(n,k) is the number of Dyck n-paths with exactly k peaks. a(n,k) = number of pairs (P,Q) of lattice paths from (0,0) to (k,n+1-k), each consisting of unit steps East or North, such that P lies strictly above Q except at the endpoints. - David Callan, Mar 23 2004
Number of permutations of [n] which avoid-132 and have k-1 descents. - Mike Zabrocki, Aug 26 2004
T(n,k) is the number of paths through n panes of glass, entering and leaving from one side, of length 2n with k reflections (where traversing one pane of glass is the unit length). - Mitch Harris, Jul 06 2006
Antidiagonal sums given by A004148 (without first term).
T(n,k) is the number of full binary trees with n internal nodes and k-1 jumps. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007
From Gary W. Adamson, Oct 22 2007: (Start)
The n-th row can be generated by the following operation using an ascending row of (n-1) triangular terms, (A) and a descending row, (B); e.g., row 6:
A: 1....3....6....10....15
B: 15...10....6.....3.....1
C: 1...15...50....50....15....1 = row 6.
Leftmost column of A,B -> first two terms of C; then followed by the operation B*C/A of current column = next term of row C, (e.g., 10*15/3 = 50). Continuing with the operation, we get row 6: (1, 15, 50, 50, 15, 1). (End)
The previous comment can be upgraded to: The ConvOffsStoT transform of the triangular series; and by rows, row 6 is the ConvOffs transform of (1, 3, 6, 10, 15). Refer to triangle A117401 as another example of the ConvOffsStoT transform, and OEIS under Maple Transforms. - Gary W. Adamson, Jul 09 2012
For a connection to Lagrange inversion, see A134264. - Tom Copeland, Aug 15 2008
T(n,k) is also the number of order-decreasing and order-preserving mappings (of an n-element set) of height k (height of a mapping is the cardinal of its image set). - Abdullahi Umar, Aug 21 2008
Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p.60]. See A033282 for the corresponding array of f-vectors for associahedra of type A_n. See A008459 and A145903 for the h-vectors for associahedra of type B and type D respectively. The Hilbert transform of this triangle (see A145905 for the definition of this transform) is A145904. - Peter Bala, Oct 27 2008
T(n,k) is also the number of noncrossing set partitions of [n] into k blocks. Given a partition P of the set {1,2,...,n}, a crossing in P are four integers [a, b, c, d] with 1 <= a < b < c < d <= n for which a, c are together in a block, and b, d are together in a different block. A noncrossing partition is a partition with no crossings. - Peter Luschny, Apr 29 2011
Noncrossing set partitions are also called genus 0 partitions. In terms of genus-dependent Stirling numbers of the second kind S2(n,k,g) that count partitions of genus g of an n-set into k nonempty subsets, one has T(n,k) = S2(n,k,0). - Robert Coquereaux, Feb 15 2024
Diagonals of A089732 are rows of A001263. - Tom Copeland, May 14 2012
From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Generating function E(y)*E(x*y) = 1 + (1 + x)*y/(1!*2!) + (1 + 3*x + x^2)*y^2/(2!*3!) + (1 + 6*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*E(x*y).
The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse A103364 has a generating function E(x*y)/E(y). (End)
T(n,k) is the number of nonintersecting n arches above the x axis, starting and ending on vertices 1 to 2n, with k being the number of arches starting on an odd vertice and ending on a higher even vertice. Example: T(3,2)=3 [16,25,34] [14,23,56] [12,36,45]. - Roger Ford, Jun 14 2014
Fomin and Reading on p. 31 state that the rows of the Narayana matrix are the h-vectors of the associahedra as well as its dual. - Tom Copeland, Jun 27 2017
The row polynomials P(n, x) = Sum_{k=1..n} T(n, k)*x^(k-1), together with P(0, x) = 1, multiplied by (n+1) are the numerator polynomials of the o.g.f.s of the diagonal sequences of the triangle A103371: G(n, x) = (n+1)*P(n, x)/(1 - x)^{2*n+1}, for n >= 0. This is proved with Lagrange's theorem applied to the Riordan triangle A135278 = (1/(1 - x)^2, x/(1 - x)). See an example below. - Wolfdieter Lang, Jul 31 2017
T(n,k) is the number of Dyck paths of semilength n with k-1 uu-blocks (pairs of consecutive up-steps). - Alexander Burstein, Jun 22 2020
In case you were searching for Narayama numbers, the correct spelling is Narayana. - N. J. A. Sloane, Nov 11 2020
Named after the Canadian mathematician Tadepalli Venkata Narayana (1930-1987). They were also called "Runyon numbers" after John P. Runyon (1922-2013) of Bell Telephone Laboratories, who used them in a study of a telephone traffic system. - Amiram Eldar, Apr 15 2021 The Narayana numbers were first studied by Percy Alexander MacMahon (see reference, Article 495) as pointed out by Bóna and Sagan (see link). - Peter Luschny, Apr 28 2022
From Andrea Arlette España, Nov 14 2022: (Start)
T(n,k) is the degree distribution of the paths towards synchronization in the transition diagram associated with the Laplacian system over the complete graph K_n, corresponding to ordered initial conditions x_1 < x_2 < ... < x_n.
T(n,k) for n=2N+1 and k=N+1 is the number of states in the transition diagram associated with the Laplacian system over the complete bipartite graph K_{N,N}, corresponding to ordered (x_1 < x_2 < ... < x_N and x_{N+1} < x_{N+2} < ... < x_{2N}) and balanced (Sum_{i=1..N} x_i/N = Sum_{i=N+1..2N} x_i/N) initial conditions. (End)
From Gus Wiseman, Jan 23 2023: (Start)
Also the number of unlabeled ordered rooted trees with n nodes and k leaves. See the link by Marko Riedel. For example, row n = 5 counts the following trees:
((((o)))) (((o))o) ((o)oo) (oooo)
(((o)o)) ((oo)o)
(((oo))) ((ooo))
((o)(o)) (o(o)o)
((o(o))) (o(oo))
(o((o))) (oo(o))
The unordered version is A055277. Leaves in standard ordered trees are counted by A358371. (End)

Examples

			The initial rows of the triangle are:
  [1] 1
  [2] 1,  1
  [3] 1,  3,   1
  [4] 1,  6,   6,    1
  [5] 1, 10,  20,   10,    1
  [6] 1, 15,  50,   50,   15,    1
  [7] 1, 21, 105,  175,  105,   21,   1
  [8] 1, 28, 196,  490,  490,  196,  28,  1
  [9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
  ...
For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions.
Example of umbral representation:
  A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1),
  so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1}
  = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1).
  First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - _Tom Copeland_, Sep 21 2011
Row polynomials and diagonal sequences of A103371: n = 4,  P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), pp. 103-124.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 196.
  • P. A. MacMahon, Combinatory Analysis, Vols. 1 and 2, Cambridge University Press, 1915, 1916; reprinted by Chelsea, 1960, Sect. 495.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.
  • T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 2.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 17.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.36(a) and (b).

Crossrefs

Other versions are in A090181 and A131198. - Philippe Deléham, Nov 18 2007
Cf. variants: A181143, A181144. - Paul D. Hanna, Oct 13 2010
Row sums give A000108 (Catalan numbers), n>0.
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A016098 and A189232 for numbers of crossing set partitions.
Cf. A243752.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • GAP
    Flat(List([1..11],n->List([1..n],k->Binomial(n-1,k-1)*Binomial(n,k-1)/k))); # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001263 n k = a001263_tabl !! (n-1) !! (k-1)
    a001263_row n = a001263_tabl !! (n-1)
    a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where
       dt us vs = zipWith (-) (zipWith (*) us (tail vs))
                              (zipWith (*) (tail us ++ [0]) (init vs))
    -- Reinhard Zumkeller, Oct 10 2013
    
  • Magma
    /* triangle */ [[Binomial(n-1,k-1)*Binomial(n,k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    A001263 := (n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k;
    a:=proc(n,k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1,i),i=1..k-1); fi; end:
    # Alternatively, as a (0,0)-based triangle:
    R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n,x),x,j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
  • Mathematica
    T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k];
    Flatten[Table[Binomial[n-1,k-1] Binomial[n,k-1]/k,{n,15},{k,n}]] (* Harvey P. Dale, Feb 29 2012 *)
    TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x];
    Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *)
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Length[Position[#,{}]]==k&]],{n,2,9},{k,1,n-1}] (* Gus Wiseman, Jan 23 2023 *)
    T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = (2n/k-1) T[n-1,k-1] + T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    {a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j)*x^m/m) +O(x^(n+1))),n,x),k,y)} \\ Paul D. Hanna, Oct 13 2010
    
  • Sage
    @CachedFunction
    def T(n, k):
        if k == n or k == 1: return 1
        if k <= 0 or k > n: return 0
        return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1)))
    for n in (1..9): print([T(n, k) for k in (1..n)])  # Peter Luschny, Oct 28 2014
    

Formula

a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938.
0Mike Zabrocki, Aug 26 2004
T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005
T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005
a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006
Central column = A000891, (2n)!*(2n+1)! / (n!*(n+1)!)^2. - Zerinvary Lajos, Oct 29 2006
G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
From Peter Bala, Oct 22 2008: (Start)
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459.
T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599.
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End)
G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010
E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010
With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011
With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)=A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011
With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana 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/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011
T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011
A166360(n-k) = T(n,k) mod 2. - Reinhard Zumkeller, Oct 10 2013
Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014
Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016
Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017
The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018
Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020
T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
From Sergii Voloshyn, Nov 25 2024: (Start)
G.f.: F(x,y) = (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) is the solution of the differential equation x^3 * d^2(x*F(x,y))/dx^2 = y * d^2(x*F(x,y))/dy^2.
Let E be the operator x*D*D, where D denotes the derivative operator d/dx. Then (1/(n! (1 + n)!)) * E^n(x/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} C(n-1, k-1)*C(n, k-1)/k*x^k. For example, when n = 4 we have (1/4!/5!)*E^3(x/(1 - x)) = x (1 + 6 x + 6 x^2 + x^3)/(1 - x)^9. (End)

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A036038 Triangle of multinomial coefficients.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 6, 12, 24, 1, 5, 10, 20, 30, 60, 120, 1, 6, 15, 20, 30, 60, 90, 120, 180, 360, 720, 1, 7, 21, 35, 42, 105, 140, 210, 210, 420, 630, 840, 1260, 2520, 5040, 1, 8, 28, 56, 70, 56, 168, 280, 420, 560, 336, 840, 1120, 1680, 2520, 1680, 3360, 5040, 6720
Offset: 1

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Comments

The number of terms in the n-th row is the number of partitions of n, A000041(n). - Amarnath Murthy, Sep 21 2002
For each n, the partitions are ordered according to A-St: first by length and then lexicographically (arranging the parts in nondecreasing order), which is different from the usual practice of ordering all partitions lexicographically. - T. D. Noe, Nov 03 2006
For this ordering of the partitions, for n >= 1, see the remarks and the C. F. Hindenburg link given in A036036. - Wolfdieter Lang, Jun 15 2012
The relation (n+1) * A134264(n+1) = A248120(n+1) / a(n) where the arithmetic is performed for matching partitions in each row n connects the combinatorial interpretations of this array to some topological and algebraic constructs of the two other entries. Also, these seem (cf. MOPS reference, Table 2) to be the coefficients of the Jack polynomial J(x;k,alpha=0). - Tom Copeland, Nov 24 2014
The conjecture on the Jack polynomials of zero order is true as evident from equation a) on p. 80 of the Stanley reference, suggested to me by Steve Kass. The conventions for denoting the more general Jack polynomials J(n,alpha) vary. Using Stanley's convention, these Jack polynomials are the umbral extensions of the multinomial expansion of (s_1*x_1 + s_2*x_2 + ... + s_(n+1)*x_(n+1))^n in which the subscripts of the (s_k)^j in the symmetric monomial expansions are finally ignored and the exponent dropped to give s_j(alpha) = j-th row polynomial of A094638 or |A008276| in ascending powers of alpha. (The MOPS table has some inconsistency between n = 3 and n = 4.) - Tom Copeland, Nov 26 2016

Examples

			1;
1, 2;
1, 3,  6;
1, 4,  6, 12, 24;
1, 5, 10, 20, 30, 60, 120;
1, 6, 15, 20, 30, 60,  90, 120, 180, 360, 720;
		

References

  • Abramowitz and Stegun, Handbook, p. 831, column labeled "M_1".

Crossrefs

Cf. A036036-A036040. Different from A078760. Row sums give A005651.
Cf. A183610 is a table of sums of powers of terms in rows.
Cf. A134264 and A248120.
Cf. A096162 for connections to A130561.

Programs

  • Maple
    nmax:=7: with(combinat): for n from 1 to nmax do P(n):=sort(partition(n)): for r from 1 to numbpart(n) do B(r):=P(n)[r] od: for m from 1 to numbpart(n) do s:=0: j:=0: while sA036038(n, m) := n!/ (mul((t!)^q(t), t=1..n)); od: od: seq(seq(A036038(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jul 14 2016
  • Mathematica
    Flatten[Table[Apply[Multinomial, Reverse[Sort[IntegerPartitions[i],  Length[ #1]>Length[ #2]&]], {1}], {i,9}]] (* T. D. Noe, Nov 03 2006 *)
  • Sage
    def ASPartitions(n, k):
        Q = [p.to_list() for p in Partitions(n, length=k)]
        for q in Q: q.reverse()
        return sorted(Q)
    def A036038_row(n):
        return [multinomial(p) for k in (0..n) for p in ASPartitions(n, k)]
    for n in (1..10): print(A036038_row(n))
    # Peter Luschny, Dec 18 2016, corrected Apr 30 2022

Formula

The n-th row is the expansion of (x_1 + x_2 + ... + x_(n+1))^n in the basis of the monomial symmetric polynomials (m.s.p.). E.g., (x_1 + x_2 + x_3 + x_4)^3 = m[3](x_1,..,x_4) + 3*m[1,2](x_1,..,x_4) + 6*m[1,1,1](x_1,..,x_4) = (Sum_{i=1..4} x_i^3) + 3*(Sum_{i,j=1..4;i != j} x_i^2 x_j) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k). The number of indeterminates can be increased indefinitely, extending each m.s.p., yet the expansion coefficients remain the same. In each m.s.p., unique combinations of exponents and subscripts appear only once with a coefficient of unity. Umbral reduction by replacing x_k^j with x_j in the expansions gives the partition polynomials of A248120. - Tom Copeland, Nov 25 2016
From Tom Copeland, Nov 26 2016: (Start)
As an example of the umbral connection to the Jack polynomials: J(3,alpha) = (Sum_{i=1..4} x_i^3)*s_3(alpha) + 3*(Sum_{i,j=1..4;i!=j} x_i^2 x_j)*s_2(alpha)*s_1(alpha)+ 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k)*s_1(alpha)*s_1(alpha)*s_1(alpha) = (Sum_{i=1..4} x_i^3)*(1+alpha)*(1+2*alpha)+ 3*(sum_{i,j=1..4;i!=j} x_i^2 x_j)*(1+alpha) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k).
See the Copeland link for more relations between the multinomial coefficients and the Jack symmetric functions. (End)

Extensions

More terms from David W. Wilson and Wouter Meeussen

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

Examples

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 117.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182-183.

Crossrefs

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.

Programs

  • Haskell
    a011973 n k = a011973_tabf !! n !! k
    a011973_row n = a011973_tabf !! n
    a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
    -- Reinhard Zumkeller, Jul 14 2015
  • Maple
    a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
    T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013
  • Mathematica
    (* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
    (* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
    (* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
    Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
    CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
    CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};
    
  • Sage
    # Prints the table; cf. A145574.
    for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012
    

Formula

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Wouter Meeussen, Oct 01 2002

Keywords

Comments

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
B) = [St2]*[dP]*[St2]^(-1)
C) = [St1]^(-1)*[dP]*[St1],
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

Examples

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018
    
  • Haskell
    a074909 n k = a074909_tabl !! n !! k
    a074909_row n = a074909_tabl !! n
    a074909_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Feb 25 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018
    
  • Maple
    A074909 := proc(n,k)
        if k > n or k < 0 then
            0;
        else
            binomial(n+1,k) ;
        end if;
    end proc: # Zerinvary Lajos, Nov 09 2006
  • Mathematica
    Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]
  • PARI
    print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    from math import comb, isqrt
    def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

Extensions

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003
Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 2014
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