A115728 -id:A115728 - cp's OEIS frontend

A006013 a(n) = binomial(3*n+1,n)/(n+1).

1, 2, 7, 30, 143, 728, 3876, 21318, 120175, 690690, 4032015, 23841480, 142498692, 859515920, 5225264024, 31983672534, 196947587823, 1219199353190, 7583142491925, 47365474641870, 296983176369495, 1868545312633440, 11793499763070480

Offset: 0

N. J. A. Sloane

Enumerates pairs of ternary trees [Knuth, 2014]. - N. J. A. Sloane, Dec 09 2014
G.f. (offset 1) is series reversion of x - 2x^2 + x^3.
Hankel transform is A005156(n+1). - Paul Barry, Jan 20 2007
a(n) = number of ways to connect 2*n - 2 points labeled 1, 2, ..., 2*n-2 in a line with 0 or more noncrossing arcs above the line such that each maximal contiguous sequence of isolated points has even length. For example, with arcs separated by dashes, a(3) = 7 counts {} (no arcs), 12, 14, 23, 34, 12-34, 14-23. It does not count 13 because 2 is an isolated point. - David Callan, Sep 18 2007
In my 2003 paper I introduced L-algebras. These are K-vector spaces equipped with two binary operations > and < satisfying (x > y) < z = x > (y < z). In my arXiv paper math-ph/0709.3453 I show that the free L-algebra on one generator is related to symmetric ternary trees with odd degrees, so the dimensions of the homogeneous components are 1, 2, 7, 30, 143, .... These L-algebras are closely related to the so-called triplicial-algebras, 3 associative operations and 3 relations whose free object is related to even trees. - Philippe Leroux (ph_ler_math(AT)yahoo.com), Oct 05 2007
a(n-1) is also the number of projective dependency trees with n nodes. - Marco Kuhlmann (marco.kuhlmann(AT)lingfil.uu.se), Apr 06 2010
Number of subpartitions of [1^2, 2^2, ..., n^2]. - Franklin T. Adams-Watters, Apr 13 2011
a(n) = sum of (n+1)-th row terms of triangle A143603. - Gary W. Adamson, Jul 07 2011
Also the number of Dyck n-paths with up steps colored in two ways (N or A) and avoiding NA. The 7 Dyck 2-paths are NDND, ADND, NDAD, ADAD, NNDD, ANDD, and AADD. - David Scambler, Jun 24 2013
a(n) is also the number of permutations avoiding 213 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014
With offset 1, a(n) is the number of ordered trees (A000108) with n non-leaf vertices and n leaf vertices such that every non-leaf vertex has a leaf child (and hence exactly one leaf child). - David Callan, Aug 20 2014
a(n) is the number of paths in the plane with unit east and north steps, never going above the line x=2y, from (0,0) to (2n+1,n). - Ira M. Gessel, Jan 04 2018
a(n) is the number of words on the alphabet [n+1] that avoid the patterns 231 and 221 and contain exactly one 1 and exactly two occurrences of every other letter. - Colin Defant, Sep 26 2018
a(n) is the number of Motzkin paths of length 3n with n of each type of step, such that (1, 1) and (1, 0) alternate (ignoring (-1, 1) steps). All paths start with a (1, 1) step. - Helmut Prodinger, Apr 08 2019
Hankel transform omitting a(0) is A051255(n+1). - Michael Somos, May 15 2022
If f(x) is the generating function for (-1)^n*a(n), a real solution of the equation y^3 - y^2 - x = 0 is given by y = 1 + x*f(x). In particular 1 + f(1) is Narayana's cow constant, A092526, aka the "supergolden" ratio. - R. James Evans, Aug 09 2023
This is instance k = 2 of the family {c(k, n+1)}A130564.%20_Wolfdieter%20Lang">{n>=0} described in A130564. _Wolfdieter Lang, Feb 04 2024
Also the number of quadrangulations of a (2n+4)-gon which do not contain any diagonals incident to a fixed vertex. - Esther Banaian, Mar 12 2025

			a(3) = 30 since the top row of Q^3 = (12, 12, 5, 1).
G.f. = 1 + 2*x + 7*x^2 + 30*x^3 + 143*x^4 + 728*x^5 + 3876*x^6 + 21318*x^7 + ... - _Michael Somos_, May 15 2022

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms n = 0..100 from T. D. Noe)
A. Aggarwal, Armstrong's Conjecture for (k, mk+1)-Core Partitions, arXiv:1407.5134 [math.CO], 2014.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, (2016), #16.3.5.
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
W. G. Brown, Enumeration of non-separable planar maps, Canad. J. Math., 15 (1963), 526-545.
W. G. Brown, Enumeration of non-separable planar maps. [Annotated scanned copy]
Naiomi Cameron and J. E. McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, 19 (2016), #16.6.1.
F. Cazals, Combinatorics of Non-Crossing Configurations, Studies in Automatic Combinatorics, Volume II (1997).
F. Chapoton, F. Hivert, and J.-C. Novelli, A set-operad of formal fractions and dendriform-like sub-operads, arXiv:1307.0092 [math.CO], 2013.
F. Chapoton and S. Giraudo, Enveloping operads and bicoloured noncrossing configurations, arXiv:1310.4521 [math.CO], 2013.
Jins de Jong, Alexander Hock, and Raimar Wulkenhaar, Catalan tables and a recursion relation in noncommutative quantum field theory, arXiv:1904.11231 [math-ph], 2019.
C. Defant and N. Kravitz, Stack-sorting for words, arXiv:1809.09158 [math.CO], 2018.
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Emeric Deutsch, S. Feretic and M. Noy, Diagonally convex directed polyominoes and even trees: a bijection and related issues, Discrete Math., 256 (2002), 645-654.
I. Gessel and G. Xin, The generating function of ternary trees and continued fractions, arXiv:math/0505217 [math.CO], 2005.
Samuele Giraudo, Tree series and pattern avoidance in syntax trees, arXiv:1903.00677 [math.CO], 2019.
Clemens Heuberger, Sarah J. Selkirk, and Stephan Wagner, Enumeration of Generalized Dyck Paths Based on the Height of Down-Steps Modulo k, arXiv:2204.14023 [math.CO], 2022.
Hsien-Kuei Hwang, Mihyun Kang, and Guan-Huei Duh, Asymptotic Expansions for Sub-Critical Lagrangean Forms, LIPIcs Proceedings of Analysis of Algorithms 2018, Vol. 110. Schloss Dagstuhl-Leibniz-Zentrum für Informatik, 2018.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 432 [broken link].
Pakawut Jiradilok, Large-scale Rook Placements, arXiv:2204.00615 [math.CO], 2022.
S. Kitaev and A. de Mier, Enumeration of fixed points of an involution on beta(1, 0)-trees, arXiv:1210.2618 [math.CO], 2012. - From _N. J. A. Sloane_, Dec 31 2012
Sergey Kitaev, Anna de Mier, and Marc Noy, On the number of self-dual rooted maps, European J. Combin. 35 (2014), 377-387. MR3090510. See Theorem 1. - _N. J. A. Sloane_, May 19 2014
Don Knuth, 20th Anniversary Christmas Tree Lecture.
Philippe Leroux, An algebraic framework of weighted directed graphs, Int. J. Math. Math. Sci. 58. (2003).
Philippe Leroux, L-algebras, triplicial-algebras, within an equivalence of categories motivated by graphs, arXiv:0709.3453 [math-ph], 2008.
Ho-Hon Leung and Thotsaporn "Aek" Thanatipanonda, A Probabilistic Two-Pile Game, arXiv:1903.03274 [math.CO], 2019.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Hugo Mlodecki, Decompositions of packed words and self duality of Word Quasisymmetric Functions, arXiv:2205.13949 [math.CO], 2022. See Table 4 p. 20.
W. Mlotkowski and K. A. Penson, The probability measure corresponding to 2-plane trees, arXiv:1304.6544 [math.PR], 2013.
Henri Muehle and Philippe Nadeau, A Poset Structure on the Alternating Group Generated by 3-Cycles, arXiv:1803.00540 [math.CO], 2018.
Liviu I. Nicolaescu, Counting Morse functions on the 2-sphere, arXiv:math/0512496 [math.GT], 2005.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
M. Noy, Enumeration of noncrossing trees on a circle, Discrete Math., 180, 301-313, 1998.
Isaac Owino Okoth, Bijections of k-plane trees, Open J. Discret. Appl. Math. (2022) Vol. 5, No. 1, 29-35.
J.-B. Priez and A. Virmaux, Non-commutative Frobenius characteristic of generalized parking functions: Application to enumeration, arXiv:1411.4161 [math.CO], 2014-2015.
Helmut Prodinger, On some questions by Cameron about ternary paths --- a linear algebra approach, arXiv:1910.02320 [math.CO], 2019.
Helmut Prodinger, Sarah J. Selkirk, and Stephan Wagner, On two subclasses of Motzkin paths and their relation to ternary trees, arXiv:1902.01681 [math.CO], 2019.
Jocelyn Quaintance, Combinatoric Enumeration of Two-Dimensional Proper Arrays, Discrete Math., 307 (2007), 1844-1864.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.
D. G. Rogers, Comments on A111160, A055113 and A006013.
Joe Sawada, Jackson Sears, Andrew Trautrim, and Aaron Williams, Demystifying our Grandparent's De Bruijn Sequences with Concatenation Trees, arXiv:2308.12405 [math.CO], 2023.
Michael Somos, Number Walls in Combinatorics.
Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 7.
Zhujun Zhang, A Note on Counting Dependency Trees, arXiv:1708.08789 [math.GM], 2017. See p. 3.
S.-n. Zheng and S.-l. Yang, On the-Shifted Central Coefficients of Riordan Matrices, Journal of Applied Mathematics, Volume 2014, Article ID 848374, 8 pages.

These are the odd indices of A047749.
Cf. A121645, A115728, A143603, A236194.
Cf. A000260, A007318, A024485, A051255, A071948, A092526, A110616, A258708.
Cf. A305574 (the same with offset 1 and the initial 1 replaced with 5).
Cf. A130564 (comment on c(k, n+1)).

a006013 n = a007318 (3 * n + 1) n `div` (n + 1)
a006013' n = a258708 (2 * n + 1) n
-- Reinhard Zumkeller, Jun 22 2015

[Binomial(3*n+1,n)/(n+1): n in [0..30]]; // Vincenzo Librandi, Mar 29 2015

Binomial[3#+1,#]/(#+1)&/@Range[0,30]  (* Harvey P. Dale, Mar 16 2011 *)

A006013(n) = binomial(3*n+1,n)/(n+1) \\ M. F. Hasler, Jan 08 2024

from math import comb
def A006013(n): return comb(3*n+1,n)//(n+1) # Chai Wah Wu, Jul 30 2022

def A006013_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 b : R.append(D[h]); h += 1
        b = not b
    return R
A006013_list(23) # Peter Luschny, May 03 2012

G.f. is square of g.f. for ternary trees, A001764 [Knuth, 2014]. - N. J. A. Sloane, Dec 09 2014
Convolution of A001764 with itself: 2*C(3*n + 2, n)/(3*n + 2), or C(3*n + 2, n+1)/(3*n + 2).
G.f.: (4/(3*x)) * sin((1/3)*arcsin(sqrt(27*x/4)))^2.
G.f.: A(x)/x with A(x)=x/(1-A(x))^2. - Vladimir Kruchinin, Dec 26 2010
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) is the top left term in M^n, where M is the infinite square production matrix:
   2, 1, 0, 0, 0, ...
   3, 2, 1, 0, 0, ...
   4, 3, 2, 1, 0, ...
   5, 4, 3, 2, 1, ...
   ... (End)
From Gary W. Adamson, Aug 11 2011: (Start)
a(n) is the sum of top row terms in Q^n, where Q is the infinite square production matrix as follows:
   1, 1, 0, 0, 0, ...
   2, 2, 1, 0, 0, ...
   3, 3, 2, 1, 0, ...
   4, 4, 3, 2, 1, ...
   ... (End)
D-finite with recurrence: 2*(n+1)*(2n+1)*a(n) = 3*(3n-1)*(3n+1)*a(n-1). - R. J. Mathar, Dec 17 2011
a(n) = 2*A236194(n)/n for n > 0. - Bruno Berselli, Jan 20 2014
a(n) = A258708(2*n+1, n). - Reinhard Zumkeller, Jun 22 2015
From Ilya Gutkovskiy, Dec 29 2016: (Start)
E.g.f.: 2F2([2/3, 4/3]; [3/2,2]; 27*x/4).
a(n) ~ 3^(3*n+3/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). (End)
a(n) = A110616(n+1, 1). - Ira M. Gessel, Jan 04 2018
0 = v0*(+98415*v2 -122472*v3 +32340*v4) +v1*(+444*v3 -2968*v4) +v2*(-60*v2 +56*v3 +64*v4) where v0=a(n)^2, v1=a(n)*a(n+1), v2=a(n+1)^2, v3=a(n+1)*a(n+2), v4=a(n+2)^2 for all n in Z if a(-1)=-2/3 and a(n)=0 for n<-1. - Michael Somos, May 15 2022
a(n) = (1/4^n) * Product_{1 <= i <= j <= 2*n} (2*i + j + 2)/(2*i + j - 1). Cf. A000260. - Peter Bala, Feb 21 2023
From Karol A. Penson, Jun 02 2023: (Start)
a(n) = Integral_{x=0..27/4} x^n*W(x) dx, where
W(x) = (((9 + sqrt(81 - 12*x))^(2/3) - (9 - sqrt(81 - 12*x))^(2/3)) * 2^(1/3) * 3^(1/6)) / (12 * Pi * x^(1/3)), for x in (0, 27/4).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem. Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with the singularity x^(-1/3), and for x > 0 is monotonically decreasing to zero at x = 27/4. At x = 27/4 the first derivative of W(x) is infinite. (End)
G.f.: hypergeometric([2/3,1,4/3], [3/2,2], (3^3/2^2)*x). See the e.g.f. above. - Wolfdieter Lang, Feb 04 2024
a(n) = A024485(n+1)/3. - Michael Somos, Oct 14 2024
G.f.: (Sum_{n >= 0} binomial(3*n+2, n)*x^n) / (Sum_{n >= 0} binomial(3*n, n)*x^n) = (B(x) - 1)/(x*B(x)), where B(x) = Sum_{n >= 0} binomial(3*n, n)/(2*n+1) * x^n is the g.f. of A001764. - Peter Bala, Dec 13 2024
The g.f. A(x) is uniquely determined by the conditions A(0) = 1 and [x^n] A(x)^(-n) = -2 for all n >= 1. Cf. A006632. - Peter Bala, Jul 24 2025

Edited by N. J. A. Sloane, Feb 21 2008

A121412 Triangular matrix T, read by rows, where row n of T equals row (n-1) of T^(n+1) with an appended '1'.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 18, 4, 1, 1, 170, 30, 5, 1, 1, 2220, 335, 45, 6, 1, 1, 37149, 4984, 581, 63, 7, 1, 1, 758814, 92652, 9730, 924, 84, 8, 1, 1, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1, 508907970, 53636520, 4843125, 387567, 28365, 1965, 135, 10, 1, 1

Offset: 0

Paul D. Hanna, Jul 30 2006

Related to the number of subpartitions of a partition as defined in A115728; for examples involving column k of successive matrix powers, see A121430, A121431, A121432 and A121433. Essentially the same as triangle A101479, but this form best illustrates the nice properties of this triangle.

			Triangle T begins:
1;
1, 1;
3, 1, 1;
18, 4, 1, 1;
170, 30, 5, 1, 1;
2220, 335, 45, 6, 1, 1;
37149, 4984, 581, 63, 7, 1, 1;
758814, 92652, 9730, 924, 84, 8, 1, 1;
18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1;
508907970, 53636520, 4843125, 387567, 28365, 1965, 135, 10, 1, 1;
To get row 4 of T, append '1' to row 3 of matrix power T^5:
1;
5, 1;
25, 5, 1;
170, 30, 5, 1; ...
To get row 5 of T, append '1' to row 4 of matrix power T^6:
1;
6, 1;
33, 6, 1;
233, 39, 6, 1;
2220, 335, 45, 6, 1; ...
Likewise, get row n of T by appending '1' to row (n-1) of T^(n+1).

Alois P. Heinz, Rows n = 0..45, flattened

Cf. A121416 (T^2), A121420 (T^3), columns: A121413, A121414, A121415; related tables: A121424, A121426, A121428; related subpartitions: A121430, A121431, A121432, A121433.

T[n_, k_] := Module[{A = {{1}}, B}, Do[B = Array[0&, {m, m}]; Do[Do[B[[i, j]] = If[j == i, 1, MatrixPower[A, i][[i-1, j]]], {j, 1, i}], {i, 1, m}]; A = B, {m, 1, n+1}]; A[[n+1, k+1]]];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 03 2019 *)

{T(n, k) = my(A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, B[i, j]=(A^i)[i-1, j]); )); A=B); return((A^1)[n+1, k+1])}
for(n=0,12, for(k=0,n, print1( T(n,k),", "));print(""))

G.f.: Column k of successive powers of T satisfy the amazing relation given by: 1 = Sum_{n>=0} (1-x)^(n+1) * x^(n(n+1)/2 + k*n) * Sum_{j=0..n+k} [T^(j+1)](n+k,k) * x^j.

A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 7, 7, 3, 1, 1, 37, 37, 15, 4, 1, 1, 268, 268, 106, 26, 5, 1, 1, 2496, 2496, 975, 230, 40, 6, 1, 1, 28612, 28612, 11100, 2565, 425, 57, 7, 1, 1, 391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1, 6230646, 6230646, 2401365, 544423

Offset: 0

Paul D. Hanna, Jun 04 2005

Remarkably, T equals the product of these triangular matrices: T = A107862^-1*A107867 = A107867^-1*A107870 = A107870^-1*A107873; reversing the order of these products yields triangle A101479.
Column m of T^k is the number of subpartitions of the initial terms of the sequence (k-1)+n(m-1)+n(n-1)/2 (ignoring 0's above the diagonal). E.g., column 4 of T^3 is 1,3,15,106,975,.... The sequence above is 2,5,9,14,20,.... subp([]) = 1, subp([2]) = 3, subp([2,5]) = 15, subp([2,5,9]) = 106, etc. The matrix product of T^(k-1) * T computes the number of such subpartitions by looking at the first part index where the subpartition is maxed - for [2,5,9,14,20] the third term (9 maxed) has subp([1,4]) for the first two values (not maxed), times subp([5,11]) for the last two values (possibly maxed). - Franklin T. Adams-Watters, Jun 26 2006
T(n,k) is the number of Dyck paths whose sequence of ascent lengths is exactly k,k+1,...,n, for example the T(4,3) = 3 paths are UUUdUUUUd^6, UUUddUUUUd^5 and UUUdddUUUUd^4. - David Scambler, May 30 2012

			G.f. for column 1:
1 = T(1,1)*(1-x)^1 + T(2,1)*x*(1-x)^2 + T(3,1)*x^2*(1-x)^4 + T(4,1)*x^3*(1-x)^7 + T(5,1)*x^4*(1-x)^11 + T(6,1)*x^5*(1-x)^16 +...
= 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 +...
G.f. for column 2:
1 = T(2,2)*(1-x)^1 + T(3,2)*x*(1-x)^3 + T(4,2)*x^2*(1-x)^6 + T(5,2)*x^3*(1-x)^10 + T(6,2)*x^4*(1-x)^15 + T(7,2)*x^5*(1-x)^21 +...
= 1*(1-x)^1 + 1*x*(1-x)^3 + 3*x^2*(1-x)^6 + 15*x^3*(1-x)^10 + 106*x^4*(1-x)^15 + 975*x^5*(1-x)^21 +...
Triangle T begins:
       1;
       1,      1;
       1,      1,      1;
       2,      2,      1,     1;
       7,      7,      3,     1,    1;
      37,     37,     15,     4,    1,   1;
     268,    268,    106,    26,    5,   1,  1;
    2496,   2496,    975,   230,   40,   6,  1, 1;
   28612,  28612,  11100,  2565,  425,  57,  7, 1, 1;
  391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1; ...
where column 1 of T = SHIFT_LEFT(column 0 of T).
Matrix square T^2 begins:
     1;
     2,   1;
     3,   2,   1;
     7,   5,   2,  1;
    26,  19,   7,  2,  1;
   141, 104,  37,  9,  2, 1;
  1034, 766, 268, 61, 11, 2, 1; ...
Compare column 2 of T^2 with column 1 of T.
Matrix inverse begins:
   1;
  -1,    1;
   0,   -1,   1;
   0,   -1,  -1,   1;
   0,   -3,  -2,  -1,  1;
   0,  -15,  -9,  -3, -1,  1;
   0, -106, -61, -18, -4, -1, 1; ...
Compare column 1 of T^-1 with column 2 of T and
compare column 2 of T^-1 with column 3 of T^2.

Alois P. Heinz, Rows n = 0..50, flattened

Cf. A107862, A107865, A107867, A107870, A107877 (column 1), A107878 (column 2), A107879 (column 3), A107880 (matrix square), A107884 (matrix cube), A107889 (matrix inverse).
Cf. A115728, A115729, A101479 (dual triangle).
T(2n,n) gives A300954.

max = 10;
A107862 = Table[Binomial[If[nA107867 = Table[Binomial[If[nA107862].A107867;
Table[t[[n, k]], {n, 1, max+1}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012, after first comment, fixed by Vaclav Kotesovec, Jun 13 2018 *)

{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(1+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))

/* Print the Triangular Matrix to the Power p: */
{T(n,k,p)=polcoeff(1- sum(j=0,n-k-1,T(j+k,k,p)*x^j*(1-x+x*O(x^n))^(j*(j-1)/2+j*k+p)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k,1),", ")); print(""))

G.f. for column k of T^m, the m-th matrix power of this triangle T:
(1) 1 = Sum_{j>=0} T(k+j, k) * x^j * (1-x)^(1+(k+j)*(k+j-1)/2-k*(k-1)/2) for m=1.
(2) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j*(1-x)^(m+(k+j)*(k+j-1)/2-k*(k-1)/2) for all m and k>=0.
(3) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j / C(x)^(m-j+(k+j)*(k+j-1)/2-k*(k-1)/2) where C(x)=2/(1+sqrt(1-4*x)) is g.f. for A000108 (Catalan numbers).
Matrix inverse of this triangle T satisfies:
(4) [T^-1](n,k) = -[T^k](n,k+1) for n>k>=0.

A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)prime_m(2)...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 4, 6, 7, 6, 7, 7, 9, 9, 5, 8, 9, 9, 10, 12, 11, 10, 9, 10, 13, 10, 13, 11, 14, 12, 6, 15, 15, 14, 12, 13, 17, 18, 13, 14, 19, 15, 16, 16, 19, 16, 11, 15, 16, 21, 19, 17, 14, 18, 17, 24, 21, 18, 19, 19, 23, 22, 7, 22, 24, 20, 22, 27, 23, 21

Offset: 1

Matthew Vandermast, Apr 28 2014

nonn

Equivalently, a(n) equals the number of values of m such that each value of A238689 T(m,k) <= A238689 T(n,k). (Since the prime factorization of 1 is the empty factorization, we consider each prime_1(i) not to be greater than prime_n(i) for all positive integers n.)
Suppose we say that n "covers" m iff both m and n are factorized as described in the sequence definition and each prime_m(i) <= prime_n(i). At least three sequences (A037019, A108951 and A181821) have the property that a(m) divides a(n) iff n "covers" m.  These sequences are also divisibility sequences (i.e., sequences with the property that a(m) divides a(n) if m divides n), since any positive integer "covers" each of its divisors.
For any positive integers m and k, the following integer sequences (with n >= 0) are arithmetic progressions:
1. The sequence b(n) = a(m*(2^n)).
2. The sequence b(n) = a(m*(prime(n+k))) if prime(k) >= A006530(m).
Also, a(n) = the number of distinct prime signatures that occur among the divisors of any integer m such that A181819(m) = n and/or A238745(m) = n.
Number of skew partitions whose numerator has Heinz number n, where a skew partition is a pair y/v of integer partitions such that the diagram of v fits inside the diagram of y. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). - Gus Wiseman, Feb 24 2018

			The prime factorizations of integers 1 through 9, with prime factors sorted from largest to smallest:
1 - the empty factorization (no prime factors)
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 3*2
7 = 7
8 = 2*2*2
9 = 3*3
To find a(9), we consider 9 = 3*3. There are 6 positive integers (1, 2, 3, 4, 6 and 9) which satisfy the following criteria:
1) The largest prime factor, if one exists, is not greater than 3;
2) The second-largest prime factor, if one exists, is not greater than 3;
3) The total number of prime factors (counting repeated factors) does not exceed 2.
Therefore, a(9) = 6.
From _Gus Wiseman_, Feb 24 2018: (Start)
Heinz numbers of the a(15) = 9 partitions contained within the partition (32) are 1, 2, 3, 4, 5, 6, 9, 10, 15. The a(15) = 9 skew partitions are (32)/(), (32)/(1), (32)/(11), (32)/(2), (32)/(21), (32)/(22), (32)/(3), (32)/(31), (32)/(32).
Corresponding diagrams are:
  o o o   . o o   . o o   . . o   . . o   . . o   . . .   . . .   . . .
  o o     o o     . o     o o     . o     . .     o o     . o     . .    (End)

Amiram Eldar, Table of n, a(n) for n = 1..10000

Rearrangement of A115728, A115729 and A238746. A116473(n) is the number of times n appears in the sequence.

Cf. A000041, A000085, A000720, A056239, A063834, A112798, A122111, A153452, A215366, A238689, A259478, A259480, A296150, A296188, A296561, A297388, A299925, A299926, A299966, A299967.

undptns[y_]:=Select[Tuples[Range[0,#]&/@y],OrderedQ[#,GreaterEqual]&];
primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Length[undptns[Reverse[primeMS[n]]]],{n,100}] (* Gus Wiseman, Feb 24 2018 *)

a(n) = A085082(A108951(n)) = A085082(A181821(n)).
a(n) = a(A122111(n)).
a(prime(n)) = a(2^n) = n+1.
a((prime(n))^m) = a((prime(m))^n) = binomial(n+m, n).
a(A002110(n)) = A000108(n+1).
A000005(n) <= a(n) <= n.

A001192 Number of full sets of size n.

Original entry on oeis.org

1, 1, 1, 2, 9, 88, 1802, 75598, 6421599, 1097780312, 376516036188, 258683018091900, 355735062429124915, 978786413996934006272, 5387230452634185460127166, 59308424712939278997978128490, 1305926814154452720947815884466579

Offset: 0

N. J. A. Sloane

A set x is full if every element of x is also a subset of x.
Equals the subpartitions of Eulerian numbers (A000295(n)=2^n-n-1); see A115728 for the definition of subpartitions of a partition. - Paul D. Hanna, Jul 03 2006
Also number of transitive rooted identity trees with n branches. - Gus Wiseman, Dec 21 2016

			Examples of full sets are 0 := {}, 1 := {0}, 2 := {1,0}, 3a := {2,1,0}, 3b := { {1}, 1, 0}, 4a := { 3a, 2, 1, 0 }.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 123, Problem 20.
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).

T. D. Noe, Table of n, a(n) for n = 0..50
Bojan Bašić, Paul Ellis, Dana C. Ernst, Danijela Popović, and Nándor Sieben, Categories of impartial rulegraphs and gamegraphs, arXiv:2312.00650 [math.CO], 2023. See p. 20.
Alberto Casagrande, Carla Piazza, and Alberto Policriti, Is hyper-extensionality preservable under deletions of graph elements?, Elec. Notes Theor. Comp. Sci. (2016) Vol. 322, 103-118.
Richard Peddicord, The number of full sets with n elements, Proc. Amer. Math. Soc., 13 (1962), 825-828.
John Riordan, Letter to N. J. A. Sloane, Jul. 1970
Alexandru Ioan Tomescu, Sets as Graphs, Ph. D. Thesis, Università degli Studi di Udine, Dipartimento di Matematica e Informatica, Dottorato di Ricerca in Informatica, Dec. 2011.
Stephan Wagner, Asymptotic enumeration of extensional acyclic digraphs, in Proceedings of the SIAM Meeting on Analytic Algorithmics and Combinatorics (ANALCO12).
Gus Wiseman, Transitive rooted identity trees with n=5 branches

Cf. A000295, A004111, A115728, A182161, A279861, A279863.

A001192 := proc(n) option remember: if(n=0)then return 1: fi: return add((-1)^(n-k-1)*binomial(2^k-k,n-k)*procname(k), k=0..n-1); end: seq(A001192(n), n=0..16); # Nathaniel Johnston, Apr 18 2012

max = 16; f[x_] := Sum[a[n]*(x^n/(1+x)^2^n), {n, 0, max}] - 1; cc = CoefficientList[ Series[f[x], {x, 0, max}], x]; Table[a[n], {n, 0, max}] /. First[ Solve[ Thread[cc == 0]]] (* Jean-François Alcover, Nov 02 2011, after Vladeta Jovovic *)

{a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-k-1) ), n)} \\ Paul D. Hanna, Jul 03 2006

1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n). E.g., 1 = 1/(1+x) + 1*x/(1+x)^2 + 1*x^2/(1+x)^4 + 2*x^3/(1+x)^8 + 9*x^4/(1+x)^16 + 88*x^5/(1+x)^32 + 1802*x^6/(1+x)^64 + ... . - Vladeta Jovovic, May 26 2005
Equivalently, a(n) = (-1)^n*C(2^n+n-1, n) - Sum_{k=0..n-1} a(k)*(-1)^(n-k)*C(2^n+2^k+n-k-1, n-k). - Paul D. Hanna, May 26 2005
G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-n-1) = 1*(1-x)^0 + 1*x*(1-x)^0 + 1*x^2*(1-x)^1 + 2*x^3*(1-x)^4 + 9*x^3*(1-x)^11 + ... + a(n)*x^n*(1-x)^(2^n-n-1) + ... . - Paul D. Hanna, Jul 03 2006

More terms from Ryan Propper, Jun 13 2005

A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 2, 7, 44, 516, 11622, 512022, 44588536, 7718806044, 2664170119608, 1836214076324153, 2529135272371085496, 6964321029630556852944, 38346813253279804426846032, 422247020982575523983378003936, 9298487213328788062025571134762096

Offset: 0

Max Alekseyev, May 24 2005

Number of subpartitions of partition [1,3,7,...,2^n-1]. - Franklin T. Adams-Watters, Mar 11 2006
Can also be computed summing forwards:
  1
  1,1
  1,2,2, 2
  1,3,5, 7, 7, 7, 7, 7
  1,4,9,16,23,30,37,44,44,44,44,44,44,44,44,44

			For n=4, the array looks like this:
  1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
  ........................1..2..3..4..5..6..7..8
  ....................................5.11.18.26
  .........................................18.44
  ............................................44
  Therefore a(4)=44.
For n=5, we can illustrate the recurrence by:
a(5) = 516 = C(19, 4) - ( 1*C(17, 4) + 2*C(14, 3) + 7*C(9, 2) ) = C(16+4-1, 4) - ( 1*C(16-2+4-1, 4) + 2*C(16-4+3-1, 3) + 7*C(16-8+2-1, 2) ).

Alois P. Heinz, Table of n, a(n) for n = 0..82 (first 26 terms from Reinhard Zumkeller)

Cf. A105996; variants: A109055 - A109061; subpartitions defined: A115728, A115729.

Column k=2 of A355576.

a107354 n = head $ snd $ until ((== 1) . fst)
                               f (2^n, replicate (2^n) 1) where
   f (len, xs) = (len', scanl1 (+) $ drop len' xs) where
      len' = len `div` 2
-- Feasible only for small n.
-- Reinhard Zumkeller, Nov 20 2011

a:= proc(n) option remember; `if`(n=0, 1, -add(
      a(j)*(-1)^(n-j)*binomial(2^j, n-j), j=0..n-1))
    end:
seq(a(n), n=0..16);  # Alois P. Heinz, Jul 08 2022

f[n_] := If[n == 0, 1, Binomial[2^(n - 1) + n - 2, n - 1] - Sum[ f[k]*Binomial[2^(n - 1) - 2^k + n - k - 1, n - k], {k, n - 2}]]; Table[ f[n], {n, 0, 15}] (* Robert G. Wilson v, May 25 2005 *)
Table[NestWhile[Accumulate[Drop[#,Ceiling[Length[#]/2]]]&,PadRight[{},2^n+1,1], Length[ #]> 1&],{n,0,16}]//Flatten (* Harvey P. Dale, Jun 24 2018 *)

{a(n)=if(n==0,1,binomial(2^(n-1)+n-2,n-1)- sum(k=1,n-2,a(k)*binomial(2^(n-1)-2^k+n-k-1,n-k)))} \\ Paul D. Hanna, May 24 2005

{a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-1) ), n)} \\ Paul D. Hanna, May 24 2005

a(n) = C(2^(n-1)+n-2,n-1) - Sum_{k=1..n-2} a(k)*C(2^(n-1)-2^k+n-k-1,n-k) for n>=2, with a(0)=1, a(1)=1, where C = binomial. - Paul D. Hanna, May 24 2005
The first number in row 3 is 2^(n-2)+1. - Ralf Stephan, May 24 2005
G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-1) (g.f. of subpartitions). - Paul D. Hanna, Jul 03 2006
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n+n). - Paul D. Hanna, Jul 03 2006

Edited by Paul D. Hanna, Jul 03 2006

A115729 Number of subpartitions of partitions in Mathematica order.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 7, 6, 7, 5, 6, 9, 9, 10, 9, 9, 6, 7, 11, 12, 10, 13, 14, 10, 13, 12, 11, 7, 8, 13, 15, 14, 16, 19, 16, 16, 17, 19, 14, 16, 15, 13, 8, 9, 15, 18, 18, 15, 19, 24, 23, 22, 19, 21, 26, 22, 23, 15, 21, 24, 18, 19, 18, 15, 9, 10, 17, 21, 22, 20, 22

Offset: 0

Franklin T. Adams-Watters, Mar 11 2006

nonn

subpart([n^k]) = C(n+k,k); subpart([n,n-1,n-2,...,1]) = C_n = A000108(n).

			Partition 5 in Mathematica order is [2,1]; it has 5
subpartitions: [], [1], [2], [1^2] and [2,1] itself.

Cf. A115728, A080577, A000108, A007318.

/* Expects input as vector in decreasing order - e.g. [3,2,1,1] */ subpart2(p)=local(i,j,v,n,k);n=matsize(p)[2];if(n==0,1,v=vector(p[1]+1,i, 1);for(i=1,n,k=p[i];for(j=1,k,v[k+1-j]+=v[k+2-j]));v[1])

For a partition P = [p_1,...,p_n] with the p_i in decreasing order, define b(i,j) to be the number of subpartitions of [p_1,...,p_i] with the i-th part = j (b(i,0) is subpartitions with less than i parts). Then b(1,j)=1 for j<=p_1, b(i+1,j) = Sum_{k=j}^{p_i} b(i,k) for 0<=k<=p_{i+1}; and the total number of subpartitions is sum_{k=1}^{p_n} b(n,k).

A107877 Column 1 of triangle A107876.

Original entry on oeis.org

1, 1, 2, 7, 37, 268, 2496, 28612, 391189, 6230646, 113521387, 2332049710, 53384167192, 1348601249480, 37291381915789, 1120914133433121, 36406578669907180, 1271084987848923282, 47487293697623885913, 1890771531272515677250, 79947079338974990793060

Offset: 0

Paul D. Hanna, Jun 04 2005, Apr 10 2007

nonn

Also number of subpartitions of partition consisting of first n-1 triangular numbers; e.g., a(4) = subp([1,3,6]) = 37. - Franklin T. Adams-Watters, Jun 26 2006
Number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= s(k-1)+k, see Fxtbook link and example. - Joerg Arndt, Apr 30 2011
Number of Dyck paths whose ascent lengths are exactly {1,2,...,n+1}; for example, the a(2) = 2 paths are uduuduuudddd and uduudduuuddd. - David Scambler, May 30 2012
Number of types of cells of a fine mixed subdivision of the Tesler flow polytope. - Alejandro H. Morales, Oct 11 2017

			1 = 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 + 2496*x^6*(1-x)^22 + ...
Also equals the final term in rows of the triangle where row n+1 equals the partial sums of row n with the final term repeated n+1 times, starting with a '1' in row 0, as illustrated by:
1;
1, 1;
1, 2,  2,  2;
1, 3,  5,  7,  7,  7,   7;
1, 4,  9, 16, 23, 30,  37,  37,  37,  37,  37;
1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268; ...
Restricted growth strings: a(0)=1 corresponds to the empty string; a(1)=1 to [0];
a(2) = 2 to [00] and [01]; a(3)=7 to
  1:  [ 0 0 0 ],
  2:  [ 0 0 1 ],
  3:  [ 0 0 2 ],
  4:  [ 0 1 0 ],
  5:  [ 0 1 1 ],
  6:  [ 0 1 2 ],
  7:  [ 0 1 3 ].
[_Joerg Arndt_, Apr 30 2011]

R. P. Stanley, Enumerative Combinatorics volume 1, 2nd edition, Cambridge University Press, 2011, Ch. 3

Alois P. Heinz, Table of n, a(n) for n = 0..390
Joerg Arndt, Matters Computational (The Fxtbook), section 17.3.6, pp. 368-369
K. Mészáros, A. H. Morales, Volumes and Ehrhart polynomials of flow polytopes, arXiv:1710.00701 [math.CO], 2017, sections 6.1 and 7.

Cf. A107876, A107878, A107879, A115728, A115729.

Cf. A305605, A305601.

b:= proc(n, y) option remember; `if`(n=0, 1, add(
      b(n-1, y+i-n), i=max(1, n-y)..n*(n-1)/2+1-y))
    end:
a:= n-> b(n+1, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 26 2016
# second Maple program:
a:= n-> LinearAlgebra:-Determinant(Matrix(n,(i,j)->
        binomial(binomial(n+1-i,2)+1,i-j+1))):
seq(a(n), n=0..25); # Alejandro H. Morales, Aug 31 2017

a[ n_, k_: 1, j_: 1] := If[ n < 2, Boole[n >= 0], a[n, k, j] = Sum[a[n - 1, i, j + 1], {i, k + j}]]; (* Michael Somos, Nov 26 2016 *)

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^(1+k*(k+1)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^(1 + k*(k+1)/2).
G.f.: 1 = Sum_{k>=0} a(k)*x^k/(1+x)^((k+1)*(k+2)/2).
From Benedict W. J. Irwin, Nov 26 2016: (Start)
Conjecture: a(n) can be expressed with a series of nested sums,
a(3) = Sum_{i=1..2} i+2,
a(4) = Sum_{i=1..2} Sum_{j=1..i+2} j+3,
a(5) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} k+4,
a(6) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} Sum_{l=1..k+4} l+5. (End)
Determinantal formula: a(n) = Det(A) where A is the n X n matrix with entries A(i,j) = binomial(binomial(n+1-i,2)+1,i-j+1). This follows by the formula by MacMahon (see EC1 Ex 3.63) for the number of such subpartitions. -  Alejandro H. Morales, Aug 31 2017

A126460 Triangle T, read by rows, where column k of matrix power T^( k(k+1)/2 ) equals left-shifted column (k-1) of T for k>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 21, 21, 6, 1, 1, 274, 274, 75, 10, 1, 1, 5806, 5806, 1565, 195, 15, 1, 1, 182766, 182766, 48950, 5940, 420, 21, 1, 1, 8034916, 8034916, 2145626, 257300, 17570, 798, 28, 1, 1, 471517614, 471517614, 125727238, 14989472, 1006880

Offset: 0

Paul D. Hanna, Dec 27 2006

Amazingly, A126460 = A126445^-1*A126450 = A126450^-1*A126454 = A126454^-1*A126457; and also A126465 = A126450*A126445^-1 = A126454*A126450^-1 = A126457*A126454^-1. Also, column k equals unsigned column k of the matrix inverse of triangle P_k defined by P_k(m,j) = C( C(j+2,3) - C(k+2,3) + m-j, m-j) for m>=j>=0.

			Triangle T begins:
1;
1, 1;
1, 1, 1;
3, 3, 1, 1;
21, 21, 6, 1, 1;
274, 274, 75, 10, 1, 1;
5806, 5806, 1565, 195, 15, 1, 1;
182766, 182766, 48950, 5940, 420, 21, 1, 1;
8034916, 8034916, 2145626, 257300, 17570, 798, 28, 1, 1; ...
where column 1 of T^1 equals left-shifted column 0 of T.
Matrix cube T^3 begins:
1;
3, 1;
6, 3, (1);
22, 12, (3), 1;
163, 91, (21), 3, 1;
2167, 1219, (274), 33, 3, 1;
46248, 26091, (5806), 661, 48, 3, 1;
1460301, 824853, (182766), 20341, 1369, 66, 3, 1; ...
where column 2 of T^3 equals left-shifted column 1 of T.
Matrix power T^6 begins:
1;
6, 1;
21, 6, 1;
98, 33, 6, (1);
791, 281, 51, (6), 1;
10850, 3929, 710, (75), 6, 1;
234472, 85557, 15425, (1565), 105, 6, 1;
7444172, 2725402, 490806, (48950), 3080, 141, 6, 1; ...
where column 3 of T^6 equals left-shifted column 2 of T.

Columns: A126461, A126462, A126463, A126464; A126465 (dual); A107876 (variant); subpartitions defined: A115728.

{T(n,k)=abs((matrix(n+1,n+1,r,c, binomial((c-1)*c*(c+1)/3!-k*(k+1)*(k+2)/3!+r-c,r-c))^-1)[n+1,k+1])}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

/* As Defined by Matrix Product A126460 = A126445^-1*A126450: */
{T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((r-1)*r*(r+1)/3!-(c-1)*c*(c+1)/3!,r-c))), N=matrix(n+1,n+1,r,c,if(r>=c,binomial((r-1)*r*(r+1)/3!-(c-1)*c*(c+1)/3!+1,r-c)))); (M^-1*N)[n+1,k+1]}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

G.f. of column k: 1/(1-x) = Sum_{n>=0} T(n+k,k)*x^n*(1-x)^p_k(n), so that column k equals the number of subpartitions of the partition p_k defined by: p_k(n) = (n^2 + (3*k+3)*n + (3*k^2+6*k-4))*n/6 for n>=0.

cp's OEIS Frontend

A006013 a(n) = binomial(3*n+1,n)/(n+1).

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A121412 Triangular matrix T, read by rows, where row n of T equals row (n-1) of T^(n+1) with an appended '1'.

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A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

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A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)*prime_m(2)*...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).

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A001192 Number of full sets of size n.

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A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

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Haskell

A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)prime_m(2)...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).