A000096 -id:A000096 - cp's OEIS frontend

A014132 Complement of triangular numbers (A000217); also array T(n,k) = ((n+k)^2 + n-k)/2, n, k > 0, read by antidiagonals.

2, 4, 5, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79

Offset: 1

N. J. A. Sloane

Numbers that are not triangular (nontriangular numbers).
Also definable as follows: a(1)=2; for n>1, a(n) is smallest integer greater than a(n-1) such that the condition "n and a(a(n)) have opposite parities" can always be satisfied. - Benoit Cloitre and Matthew Vandermast, Mar 10 2003
Record values in A256188 that are greater than 1. - Reinhard Zumkeller, Mar 26 2015
From Daniel Forgues, Apr 10 2015: (Start)
With n >= 1, k >= 1:
  t(n+k) - k, 1 <= k <= n+k-1, n >= 1;
  t(n+k-1) + n, 1 <= n <= n+k-1, k >= 1;
where t(n+k) = t(n+k-1) + (n+k) is the (n+k)-th triangular number, while the number of compositions of n+k into 2 parts is C(n+k-1, 2-1) = n+k-1, the number of nontriangular numbers between t(n+k-1) and t(n+k), just right!
Related to Hilbert's Infinite Hotel:
0) All rooms, numbered through the positive integers, are full;
1) An infinite number of trains, each containing an infinite number of passengers, arrives: i.e., a 2-D lattice of pairs of positive integers;
2) Move occupant of room m, m >= 1, to room t(m) = m*(m+1)/2, where t(m) is the m-th triangular number;
3) Assign n-th passenger from k-th train to room t(n+k-1) + n, 1 <= n <= n+k-1, k >= 1;
4) Everybody has his or her own room, no room is empty, for m >= 1.
If situation 1 happens again, repeat steps 2 and 3, you're back to 4.
(End)
1711 + 2*a(n)*(58 + a(n)) is prime for n<=21. The terms that do not have this property start 29,32,34,43,47,58,59,60,62,63,65,68,70,73,...  - Benedict W. J. Irwin, Nov 22 2016
Also numbers k with the property that in the symmetric representation of sigma(k) both Dyck paths have a central peak or both Dyck paths have a central valley. (Cf. A237593.) - Omar E. Pol, Aug 28 2018

			From _Boris Putievskiy_, Jan 14 2013: (Start)
Start of the sequence as a table (read by antidiagonals, right to left), where the k-th row corresponds to the k-th column of the triangle (shown thereafter):
   2,  4,  7, 11, 16, 22, 29, ...
   5,  8, 12, 17, 23, 30, 38, ...
   9, 13, 18, 24, 31, 39, 48, ...
  14, 19, 25, 32, 40, 49, 59, ...
  20, 26, 33, 41, 50, 60, 71, ...
  27, 34, 42, 51, 61, 72, 84, ...
  35, 43, 52, 62, 73, 85, 98, ...
  (...)
Start of the sequence as a triangle (read by rows), where the i elements of the i-th row are t(i) + 1 up to t(i+1) - 1, i >= 1:
   2;
   4,  5;
   7,  8,  9;
  11, 12, 13, 14;
  16, 17, 18, 19, 20;
  22, 23, 24, 25, 26, 27;
  29, 30, 31, 32, 33, 34, 35;
  (...)
Row number i contains i numbers, where t(i) = i*(i+1)/2:
  t(i) + 1, t(i) + 2, ..., t(i) + i = t(i+1) - 1
(End) [Edited by _Daniel Forgues_, Apr 11 2015]

T. D. Noe, Table of n, a(n) for n = 1..1000
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence arXiv:math/0305308 [math.NT], 2003.
Bakir Farhi, An explicit formula generating the non-Fibonacci numbers, arXiv:1105.1127 [math.NT], May 05 2011. See Example 5 p. 456.
J. Lambek and L. Moser, Inverse and complementary sequences of natural numbers, Amer. Math. Monthly, 61 (1954), 454-458.
Cristinel Mortici, Remarks on Complementary Sequences, Fibonacci Quart. 48 (2010), no. 4, 343-347.
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A000217, A006002, A035214, A080036, A002024, A007401, A003057, A114327, A002260, A004736, A118011, A237593.
Cf. A000124 (left edge: quasi-triangular numbers), A000096 (right edge: almost-triangular numbers), A006002 (row sums), A001105 (central terms).
Cf. A242401 (subsequence).
Cf. A004202, A256188.
Cf. A145397 (the non-tetrahedral numbers).

a014132 n = n + round (sqrt $ 2 * fromInteger n)
a014132_list = filter ((== 0) . a010054) [0..]
-- Reinhard Zumkeller, Dec 12 2012

IsTriangular:=func< n | exists{ k: k in [1..Isqrt(2*n)] | n eq (k*(k+1) div 2)} >; [ n: n in [1..90] | not IsTriangular(n) ]; // Klaus Brockhaus, Jan 04 2011

f[n_] := n + Round[Sqrt[2n]]; Array[f, 71] (* or *)
Complement[ Range[83], Array[ #(# + 1)/2 &, 13]] (* Robert G. Wilson v, Oct 21 2005 *)
DeleteCases[Range[80],?(OddQ[Sqrt[8#+1]]&)] (* _Harvey P. Dale, Jul 24 2021 *)

PARI
```
a(n)=if(n<1,0,n+(sqrtint(8*n-7)+1)\2)
```

isok(n) = !ispolygonal(n,3); \\ Michel Marcus, Mar 01 2016

from math import isqrt
def A014132(n): return n+(isqrt((n<<3)-7)+1>>1) # Chai Wah Wu, Jun 17 2024

a(n) = n + round(sqrt(2*n)).
a(a(n)) = n + 2*floor(1/2 + sqrt(2n)) + 1.
a(n) = a(n-1) + A035214(n), a(1)=2.
a(n) = A080036(n) - 1.
a(n) = n + A002024(n). - Vincenzo Librandi, Jul 08 2010
A010054(a(n)) = 0. - Reinhard Zumkeller, Dec 10 2012
From Boris Putievskiy, Jan 14 2013: (Start)
a(n) = A007401(n)+1.
a(n) = A003057(n)^2 - A114327(n).
a(n) = ((t+2)^2 + i - j)/2, where
i = n-t*(t+1)/2,
j = (t*t+3*t+4)/2-n,
t = floor((-1+sqrt(8*n-7))/2). (End)
A248952(a(n)) < 0. - Reinhard Zumkeller, Oct 20 2014
a(n) = A256188(A004202(n)). - Reinhard Zumkeller, Mar 26 2015
From Robert Israel, Apr 20 2015 (Start):
a(n) = A118011(n) - n.
G.f.: x/(1-x)^2 + x/(1-x) * Sum(j>=0, x^(j*(j+1)/2)) = x/(1-x)^2 + x^(7/8)/(2-2*x) * Theta2(0,sqrt(x)), where Theta2 is a Jacobi theta function. (End)
G.f. as array: x*y*(2 - 2*y + x^2*y + y^2 - x*(1 + y))/((1 - x)^3*(1 - y)^3). - Stefano Spezia, Apr 22 2024

Following Alford Arnold's comment: keyword tabl and correspondent crossrefs added by Reinhard Zumkeller, Dec 12 2012

I restored the original definition. - N. J. A. Sloane, Jan 27 2019

A144064 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is Euler transform of (j->k).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 9, 10, 5, 0, 1, 5, 14, 22, 20, 7, 0, 1, 6, 20, 40, 51, 36, 11, 0, 1, 7, 27, 65, 105, 108, 65, 15, 0, 1, 8, 35, 98, 190, 252, 221, 110, 22, 0, 1, 9, 44, 140, 315, 506, 574, 429, 185, 30, 0, 1, 10, 54, 192, 490, 918, 1265, 1240, 810, 300, 42, 0

Offset: 0

Alois P. Heinz, Sep 09 2008

A(n,k) is also the number of partitions of n into parts of k kinds.
In general, column k > 0 is asymptotic to k^((k+1)/4) * exp(Pi*sqrt(2*k*n/3)) / (2^((3*k+5)/4) * 3^((k+1)/4) * n^((k+3)/4)) * (1 - (Pi*k^(3/2)/(24*sqrt(6)) + sqrt(3)*(k+1)*(k+3)/(8*Pi*sqrt(2*k))) / sqrt(n)). - Vaclav Kotesovec, Feb 28 2015, extended Jan 16 2017
When k is a prime power greater than 1, A(n,k) is the number of conjugacy classes of n X n matrices over a field with k elements that contain an upper-triangular matrix. - Geoffrey Critzer, Nov 11 2022

			Square array begins:
  1,   1,   1,   1,   1,   1, ...
  0,   1,   2,   3,   4,   5, ...
  0,   2,   5,   9,  14,  20, ...
  0,   3,  10,  22,  40,  65, ...
  0,   5,  20,  51, 105, 190, ...
  0,   7,  36, 108, 252, 506, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
G. E. Bergum and V. E. Hoggatt, Jr., Numerator polynomial coefficient array for the convolved Fibonacci sequence, Fib. Quart., 14 (1976), 43-44. (Annotated scanned copy)
G. E. Bergum and V. E. Hoggatt, Jr., Numerator polynomial coefficient array for the convolved Fibonacci sequence, Fib. Quart., 14 (1976), 43-48. See Table 1.
Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 8.
N. J. A. Sloane, Transforms
Index entries for expansions of Product_{k >= 1} (1-x^k)^m

Columns k=0-24 give: A000007, A000041, A000712, A000716, A023003, A023004, A023005, A023006, A023007, A023008, A023009, A023010, A005758, A023011, A023012, A023013, A023014, A023015, A023016, A023017, A023018, A023019, A023020, A023021, A006922.
Cf. A082556 (k=30), A082557 (k=32), A082558 (k=48), A082559 (k=64).
Rows n=0-4 give: A000012, A001477, A000096, A006503, A006504.
Main diagonal gives A008485.
Antidiagonal sums give A067687.
Cf. A060642, A122768, A246935, A255961, A261718.

# DedekindEta is defined in A000594.
A144064Column(k, len) = DedekindEta(len, -k)
for n in 0:8 A144064Column(n, 6) |> println end # Peter Luschny, Mar 10 2018

with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; `if`(n=0, 1, add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n) end end: A:= (n,k)-> etr(j->k)(n): seq(seq(A(n, d-n), n=0..d), d=0..14);

a[0, ] = 1; a[, 0] = 0; a[n_, k_] := SeriesCoefficient[ Product[1/(1 - x^j)^k, {j, 1, n}], {x, 0, n}]; Table[a[n - k, k], {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 06 2013 *)
etr[p_] := Module[{b}, b[n_] := b[n] = If[n==0, 1, Sum[Sum[d*p[d], {d, Divisors[j]} ]*b[n-j], {j, 1, n}]/n]; b]; A[n_, k_] := etr[k&][n]; Table[A[n, d-n], {d, 0, 14}, {n, 0, d}] // Flatten (* Jean-François Alcover, Mar 30 2015, after Alois P. Heinz *)

Mat(apply( {A144064_col(k,nMax=9)=Col(1/eta('x+O('x^nMax))^k,nMax)}, [0..9])) \\ M. F. Hasler, Aug 04 2024

G.f. of column k: Product_{j>=1} 1/(1-x^j)^k.

A(n,k) = Sum_{i=0..k} binomial(k,i) * A060642(n,k-i):

A014107 a(n) = n(2n-3).

Original entry on oeis.org

0, -1, 2, 9, 20, 35, 54, 77, 104, 135, 170, 209, 252, 299, 350, 405, 464, 527, 594, 665, 740, 819, 902, 989, 1080, 1175, 1274, 1377, 1484, 1595, 1710, 1829, 1952, 2079, 2210, 2345, 2484, 2627, 2774, 2925, 3080, 3239, 3402, 3569, 3740, 3915, 4094, 4277

Offset: 0

N. J. A. Sloane

Positive terms give a bisection of A000096. - Omar E. Pol, Dec 16 2016

G. C. Greubel, Table of n, a(n) for n = 0..5000
Emily Barnard and Nathan Reading, Coxeter-biCatalan combinatorics, arXiv:1605.03524 [math.CO], 2016. See p. 51.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A000108, A014105, A033537, A097070, A100035, A100036, A100037, A100038, A100039, A100345, A002378, A000384.

A014107:=n->n*(2*n-3); seq(A014107(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013

Table[2n^2 - 3n, {n, 0, 49}] (* Alonso del Arte, Dec 15 2012 *)
LinearRecurrence[{3,-3,1},{0,-1,2},50] (* Harvey P. Dale, Sep 18 2018 *)

PARI
```
a(n)=n*(2*n-3)
```

a(n) = A100345(n, n - 3) for n > 2.
a(n) = A033537(n) - 8*n^2; A100035(a(n)) = 2 for n > 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A014106(-n) for all n in Z. - Michael Somos, Nov 06 2005
From Michael Somos, Nov 06 2005: (Start)
G.f.: x*(-1 + 5*x)/(1 - x)^3.
E.g.f: x*(-1 + 2*x)*exp(x). (End)
a(n) = A097070(n)/A000108(n - 2), n >= 2. - Philippe Deléham, Apr 12 2007
a(n) = 2*a(n-1) - a(n-2) + 4, n > 1; a(0) = 0, a(1) = -1, a(2) = 2. - Zerinvary Lajos, Feb 18 2008
a(n) = a(n-1) + 4*n - 5 with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = (2*n-1)*(n-1) - 1. Also, with an initial offset of -1, a(n) = (2*n-1)*(n+1) = 2*n^2 + n - 1. - Alonso del Arte, Dec 15 2012
(a(n) + 1)^2 + (a(n) + 2)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2n - 1)^2 starting with a(1) = -1. - Jeffreylee R. Snow, Sep 17 2013
a(n) = A014105(n-1) - 1 for all n in Z. - Michael Somos, Nov 23 2021
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = -2*(1 - log(2))/3.
Sum_{n>=1} (-1)^n/a(n) = Pi/6 + log(2)/3 + 2/3. (End)
For n > 0, A002378(a(n)) = A000384(n-1)*A000384(n). - Charlie Marion, May 21 2023

A033282 Triangle read by rows: T(n, k) is the number of diagonal dissections of a convex n-gon into k+1 regions.

Original entry on oeis.org

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

Offset: 3

N. J. A. Sloane

T(n+3, k) is also the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's cluster algebra of finite type A_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of the triangle of Narayana numbers A001263. For example, x^2 + 5*x + 5 = y^2 + 3*y + 1. - Paul Boddington, Mar 07 2003
Number of standard Young tableaux of shape (k+1,k+1,1^(n-k-3)), where 1^(n-k-3) denotes a sequence of n-k-3 1's (see the Stanley reference).
Number of k-dimensional 'faces' of the n-dimensional associahedron (see Simion, p. 168). - Mitch Harris, Jan 16 2007
Mirror image of triangle A126216. - Philippe Deléham, Oct 19 2007
For relation to Lagrange inversion or series reversion and the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see A133437. - Tom Copeland, Sep 29 2008
Row generating polynomials 1/(n+1)*Jacobi_P(n,1,1,2*x+1). Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p. 60]. See A001263 for the corresponding array of h-vectors for associahedra of type A_n. See A063007 and A080721 for the f-vectors for associahedra of type B and type D respectively. - Peter Bala, Oct 28 2008
f-vectors of secondary polytopes for Grobner bases for optimization and integer programming (see De Loera et al. and Thomas). - Tom Copeland, Oct 11 2011
From Devadoss and O'Rourke's book: The Fulton-MacPherson compactification of the configuration space of n free particles on a line segment with a fixed particle at each end is the n-Dim Stasheff associahedron whose refined f-vector is given in A133437 which reduces to A033282. - Tom Copeland, Nov 29 2011
Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012
The general results on the convolution of the refined partition polynomials of A133437, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016
The signed triangle t(n, k) =(-1)^k* T(n+2, k-1), n >= 1, k = 1..n, seems to be obtainable from the partition array A111785 (in Abramowitz-Stegun order) by adding the entries corresponding to the partitions of n with the number of parts k. E.g., triangle t, row n=4: -1, (6+3) = 9, -21, 14. - Wolfdieter Lang, Mar 17 2017
The preceding conjecture by Lang is true. It is implicit in Copeland's 2011 comments in A086810 on the relations among a gf and its compositional inverse for that entry and inversion through A133437 (a differently normalized version of A111785), whose integer partitions are the same as those for A134685. (An inversion pair in Copeland's 2008 formulas below can also be used to prove the conjecture.) In addition, it follows from the relation between the inversion formula of A111785/A133437 and the enumeration of distinct faces of associahedra. See the MathOverflow link concernimg Loday and the Aguiar and Ardila reference in A133437 for proofs of the relations between the partition polynomials for inversion and enumeration of the distinct faces of the A_n associahedra, or Stasheff polytopes. - Tom Copeland, Dec 21 2017
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/(n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 07 2022
Chapoton's observation above is correct: the precise expansion is (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/ (n!*(n+1)!) = Sum_{k = 0..n-1} (-1)^k*T(n+2,n-k-1)*binomial(x+2*n-k,2*n-k), as can be verified using the WZ algorithm. For example, n = 4 gives (x+1)*(x+2)^2*(x+3)^2*(x+4)^2*(x+5)/(4!*5!) = 14*binomial(x+8,8) - 21*binomial(x+7,7) + 9*binomial(x+6,6) - binomial(x+5,5). - Peter Bala, Jun 24 2023

			The triangle T(n, k) begins:
n\k  0  1   2    3     4     5      6      7     8     9
3:   1
4:   1  2
5:   1  5   5
6:   1  9  21   14
7:   1 14  56   84    42
8:   1 20 120  300   330   132
9:   1 27 225  825  1485  1287    429
10:  1 35 385 1925  5005  7007   5005   1430
11:  1 44 616 4004 14014 28028  32032  19448  4862
12:  1 54 936 7644 34398 91728 148512 143208 75582 16796
... reformatted. - _Wolfdieter Lang_, Mar 17 2017

S. Devadoss and J. O'Rourke, Discrete and Computational Geometry, Princeton Univ. Press, 2011 (See p. 241.)
Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Exercise 7.50, pages 379, 573.
T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.8.

Vincenzo Librandi, Table of n, a(n) for n = 3..2000
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
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Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Karin Baur and P. P. Martin, The fibres of the Scott map on polygon tilings are the flip equivalence classes, arXiv:1601.05080 [math.CO], 2016.
David Beckwith, Legendre polynomials and polygon dissections?, Amer. Math. Monthly, 105 (1998), 256-257.
William Butler, Andrew Kalotay and N. J. A. Sloane, Correspondence, 1974
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Adrian Celestino and Yannic Vargas, Schröder trees, antipode formulas and non-commutative probability, arXiv:2311.07824 [math.CO], 2023.
Frédéric Chapoton, Enumerative properties of generalized associahedra, Séminaire Lotharingien de Combinatoire, B51b (2004), 16 pp.
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015.
Manosij Ghosh Dastidar and Michael Wallner, Bijections and congruences involving lattice paths and integer compositions, arXiv:2402.17849 [math.CO], 2024. See p. 16.
Jesús A. De Loera, Jörg Rambau, and Francisco Santos Leal, Triangulations of Point Sets [From Tom Copeland Oct 11 2011]
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Cassandra Durell and Stefan Forcey, Level-1 Phylogenetic Networks and their Balanced Minimum Evolution Polytopes, arXiv:1905.09160 [math.CO], 2019.
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Germain Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30.
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30. (Annotated scanned copy)
Thibault Manneville and Vincent Pilaud, Compatibility fans for graphical nested complexes, arXiv:1501.07152 [math.CO], 2015.
Sebastian Mizera, Combinatorics and topology of Kawai-Lewellen-Tye relations J. High Energy Phys. 2017, No. 8, Paper No. 97, 54 p. (2017).
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.
Vincent Pilaud, Brick polytopes, lattice quotients, and Hopf algebras, arXiv:1505.07665 [math.CO], 2015.
Vincent Pilaud and V. Pons, Permutrees, arXiv:1606.09643 [math.CO], 2016-2017.
Ronald C. Read, On general dissections of a polygon, Aequat. Math. 18 (1978), 370-388.
Dean Rubine, Exercises for A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, arXiv:2507.13045 [math.CO], 2025. See p. 9.
Rodica Simion, Convex Polytopes and Enumeration, Adv. in Appl. Math. 18 (1997) pp. 149-180.
Richard P. Stanley, Polygon dissections and standard Young tableaux, J. Comb. Theory, Ser. A, 76, 175-177, 1996.
Rekha R. Thomas, Lectures in Geometric Combinatorics [_Tom Copeland_, Oct 11 2011]

Cf. diagonals: A000012, A000096, A033275, A033276, A033277, A033278, A033279; A000108, A002054, A002055, A002056, A007160, A033280, A033281; row sums: A001003 (Schroeder numbers, first term omitted). See A086810 for another version.
A007160 is a diagonal. Cf. A001263.
With leading zero: A086810.
Cf. A019538 'faces' of the permutohedron.
Cf. A063007 (f-vectors type B associahedra), A080721 (f-vectors type D associahedra), A126216 (mirror image).
Cf. A248727 for a relation to f-polynomials of simplices.
Cf. A111785 (contracted partition array, unsigned; see a comment above).
Antidiagonal sums give A005043. - Jordan Tirrell, Jun 01 2017

[[Binomial(n-3, k)*Binomial(n+k-1, k)/(k+1): k in [0..(n-3)]]: n in [3..12]];  // G. C. Greubel, Nov 19 2018

T:=(n,k)->binomial(n-3,k)*binomial(n+k-1,k)/(k+1): seq(seq(T(n,k),k=0..n-3),n=3..12); # Muniru A Asiru, Nov 24 2018

t[n_, k_] = Binomial[n-3, k]*Binomial[n+k-1, k]/(k+1);
Flatten[Table[t[n, k], {n, 3, 12}, {k, 0, n-3}]][[1 ;; 52]] (* Jean-François Alcover, Jun 16 2011 *)

Q=(1+z-(1-(4*w+2+O(w^20))*z+z^2+O(z^20))^(1/2))/(2*(1+w)*z);for(n=3,12,for(m=1,n-2,print1(polcoef(polcoef(Q,n-2,z),m,w),", "))) \\ Hugo Pfoertner, Nov 19 2018

for(n=3,12, for(k=0,n-3, print1(binomial(n-3,k)*binomial(n+k-1,k)/(k+1), ", "))) \\ G. C. Greubel, Nov 19 2018

[[ binomial(n-3,k)*binomial(n+k-1,k)/(k+1) for k in (0..(n-3))] for n in (3..12)] # G. C. Greubel, Nov 19 2018

G.f. G = G(t, z) satisfies (1+t)*G^2 - z*(1-z-2*t*z)*G + t*z^4 = 0.
T(n, k) = binomial(n-3, k)*binomial(n+k-1, k)/(k+1) for n >= 3, 0 <= k <= n-3.
From Tom Copeland, Nov 03 2008: (Start)
Two g.f.s (f1 and f2) for A033282 and their inverses (x1 and x2) can be derived from the Drake and Barry references.
1. a: f1(x,t) = y = {1 - (2t+1) x - sqrt[1 - (2t+1) 2x + x^2]}/[2x (t+1)] = t x + (t + 2 t^2) x^2 + (t + 5 t^2 + 5 t^3) x^3 + ...
b: x1 = y/[t + (2t+1)y + (t+1)y^2] = y {1/[t/(t+1) + y] - 1/(1+y)} = (y/t) - (1+2t)(y/t)^2 + (1+ 3t + 3t^2)(y/t)^3 +...
2. a: f2(x,t) = y = {1 - x - sqrt[(1-x)^2 - 4xt]}/[2(t+1)] = (t/(t+1)) x + t x^2 + (t + 2 t^2) x^3 + (t + 5 t^2 + 5 t^3) x^4 + ...
b: x2 = y(t+1) [1- y(t+1)]/[t + y(t+1)] = (t+1) (y/t) - (t+1)^3 (y/t)^2 + (t+1)^4 (y/t)^3 + ...
c: y/x2(y,t) = [t/(t+1) + y] / [1- y(t+1)] = t/(t+1) + (1+t) y + (1+t)^2 y^2 + (1+t)^3 y^3 + ...
x2(y,t) can be used along with the Lagrange inversion for an o.g.f. (A133437) to generate A033282 and show that A133437 is a refinement of A033282, i.e., a refinement of the f-polynomials of the associahedra, the Stasheff polytopes.
y/x2(y,t) can be used along with the indirect Lagrange inversion (A134264) to generate A033282 and show that A134264 is a refinement of A001263, i.e., a refinement of the h-polynomials of the associahedra.
f1[x,t](t+1) gives a generator for A088617.
f1[xt,1/t](t+1) gives a generator for A060693, with inverse y/[1 + t + (2+t) y + y^2].
f1[x(t-1),1/(t-1)]t gives a generator for A001263, with inverse y/[t + (1+t) y + y^2].
The unsigned coefficients of x1(y t,t) are A074909, reverse rows of A135278. (End)
G.f.: 1/(1-x*y-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A033282 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Sep 06 2011
With a different offset, the row polynomials equal 1/(1 + x)*Integral_{0..x} R(n,t) dt, where R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*t^k are the row polynomials of A063007. - Peter Bala, Jun 23 2016
n-th row polynomial = ( LegendreP(n-1,2*x + 1) - LegendreP(n-3,2*x + 1) )/((4*n - 6)*x*(x + 1)), n >= 3. - Peter Bala, Feb 22 2017
n*T(n+1, k) = (4n-6)*T(n, k-1) + (2n-3)*T(n, k) - (n-3)*T(n-1, k) for n >= 4. - Fang Lixing, May 07 2019

Missing factor of 2 for expansions of f1 and f2 added by Tom Copeland, Apr 12 2009

A080956 a(n) = (n+1)*(2-n)/2.

Original entry on oeis.org

1, 1, 0, -2, -5, -9, -14, -20, -27, -35, -44, -54, -65, -77, -90, -104, -119, -135, -152, -170, -189, -209, -230, -252, -275, -299, -324, -350, -377, -405, -434, -464, -495, -527, -560, -594, -629, -665, -702, -740, -779, -819, -860, -902, -945, -989, -1034, -1080, -1127, -1175, -1224, -1274, -1325, -1377

Offset: 0

Paul Barry, Mar 01 2003

Coefficient of x in the polynomial C(n,0)+C(n+1,1)x+C(n+2,2)x(x-1)/2.
Equals A154990 * [1,2,3,...]. - Gary W. Adamson & Mats Granvik, Jan 19 2009
a(n) is essentially the case 1 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_1(n) = n*(3-n)/2 and a(n) = P_1(n+1). See A005563 for the case k=0. - Peter Luschny, Jul 08 2011
This is the case k=-1 of the formula (k*m*(m+1)-(-1)^k+1)/2. See similar sequences listed in A262221. - Bruno Berselli, Sep 17 2015

			a(5) = 6-(1+2+3+4+5). - _Stanislav Sykora_, Feb 19 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000027, A000096, A000217, A154990, A214292, A262221.

[(n+1)*(2-n)/2: n in [0..80]]; // Vincenzo Librandi, Jul 08 2011

G(x):=exp(x)*(x-x^2/2): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=1..54 ); # Zerinvary Lajos, Apr 05 2009

FoldList[#1 - #2 &, 1, Range[0, 44]] (* Arkadiusz Wesolowski, May 26 2013 *)
LinearRecurrence[{3,-3,1},{1,1,0},60] (* Harvey P. Dale, Nov 29 2019 *)

PARI
```
a(n)=(n+1)*(2-n)/2;
```

def A080956(n): return (2-n)*(n+1)//2 # G. C. Greubel, May 08 2025

a(n) = 2*(C(n+1, 1)-C(n+2, 2)) = (n+1)*(2-n)/2.
G.f.: (1-2*x)/(1-x)^3. - R. J. Mathar, Jun 11 2009
If we define f(n,i,a) = Sum_{k=0..n-i} (binomial(n,k)*stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = f(n,n-1,2), for n>=3. - Milan Janjic, Dec 20 2008
E.g.f.: exp(x)*(1-x^2/2). - Zerinvary Lajos, Apr 05 2009, R. J. Mathar, Jun 11 2009
a(n) = - A214292(n,1) for n > 0. - Reinhard Zumkeller, Jul 12 2012
Recurrence: a(0)=1, a(n+1) = a(n) - n. Also a(n)=(n+1)-Sum[k=1..n](k). Also a(n) = A000027(n+1) - A000217(n). Also, for n>1, a(n) = - A000096(n-2). - Stanislav Sykora, Feb 19 2014
Sum_{n>=3} 1/a(n) = -11/9. - Amiram Eldar, Sep 26 2022

Lajos e.g.f. adapted to offset zero by R. J. Mathar, Jun 11 2009

A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.

Original entry on oeis.org

1, 2, 2, 3, 6, 9, 15, 30, 54, 97, 189, 360, 675, 1304, 2522, 4835, 9358, 18193, 35269, 68568, 133737, 260802, 509132, 995801, 1948931, 3816904, 7483636, 14683721, 28827798, 56637969, 111347879, 219019294, 431043814, 848764585, 1672056525, 3295390800, 6497536449

Offset: 0

Christopher Carl Heckman, Jul 29 2009

nonn

A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
Proposed Problem 11610 in the Dec 2011 A.M.M.
From Gus Wiseman, Jul 27 2021: (Start)
Also the antidiagonal sums of the matrices counting integer compositions by length and alternating sum (A345197). So a(n) is the number of integer compositions of n + 1 of length (n - s + 3)/2, where s is the alternating sum of the composition. For example, the a(0) = 1 through a(6) = 7 compositions are:
  (1)  (2)   (3)   (4)    (5)     (6)     (7)
       (11)  (21)  (31)   (41)    (51)    (61)
                   (121)  (122)   (123)   (124)
                          (221)   (222)   (223)
                          (1112)  (321)   (322)
                          (1211)  (1122)  (421)
                                  (1221)  (1132)
                                  (2112)  (1231)
                                  (2211)  (2122)
                                          (2221)
                                          (3112)
                                          (3211)
                                          (11131)
                                          (12121)
                                          (13111)
For a bijection with the main (binary string) interpretation, take the run-lengths of each binary string of length n + 1 that satisfies the condition and starts with 1.
(End)

			1 + 2*x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 15*x^6 + 30*x^7 + 54*x^8 + 97*x^9 + ...
From _Gus Wiseman_, Jul 27 2021: (Start)
The a(0) = 1 though a(6) = 15 binary strings:
  ()  (0)  (1,0)  (0,0,1)  (0,0,1,0)  (0,0,1,1,0)  (0,0,0,1,0,1)
      (1)  (1,1)  (1,1,0)  (0,0,1,1)  (0,0,1,1,1)  (0,0,1,0,0,1)
                  (1,1,1)  (0,1,0,0)  (0,1,1,0,0)  (0,0,1,1,1,0)
                           (1,0,0,1)  (1,0,0,1,0)  (0,0,1,1,1,1)
                           (1,1,1,0)  (1,0,0,1,1)  (0,1,0,0,0,1)
                           (1,1,1,1)  (1,0,1,0,0)  (0,1,1,1,0,0)
                                      (1,1,0,0,1)  (1,0,0,1,1,0)
                                      (1,1,1,1,0)  (1,0,0,1,1,1)
                                      (1,1,1,1,1)  (1,0,1,1,0,0)
                                                   (1,1,0,0,1,0)
                                                   (1,1,0,0,1,1)
                                                   (1,1,0,1,0,0)
                                                   (1,1,1,0,0,1)
                                                   (1,1,1,1,1,0)
                                                   (1,1,1,1,1,1)
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..3328 (first 501 terms from R. H. Hardin)
Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv preprint arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of g.f., recurrence, asymptotics for this sequence. [From _N. J. A. Sloane_, Apr 07 2012]
R. Stanley, Problem 11610, Amer. Math. Monthly, 118 (2011), 937; 120 (2013), 943-944.

Antidiagonal sums of the matrices A345197.
Row sums of A345907.
Taking diagonal instead of antidiagonal sums gives A345908.
A011782 counts compositions (or binary strings).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000041, A000070, A000096, A000097, A000124, A000346, A007318, A008549, A025047, A131577, A238279.

with(combinat): count := proc(n) local S, matches, A, k, i; S := subsets(\{seq(i, i=1..n)\}): matches := 0: while not S[finished] do A := S[nextvalue](): k := 0: for i from 1 to n-1 do: if not (i in A) and not (i+1 in A) then k := k + 1: fi: if not (i in A) and (i+1 in A) then k := k - 1: fi: od: if (k = 0) then matches := matches + 1: fi: end do; return(matches); end proc:
# second Maple program:
b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 1, 0):
seq(a(n), n=0..36);  # Alois P. Heinz, Mar 20 2024

a[0] = 1; a[n_] := Sum[Binomial[2*k - 1, k]*Binomial[n - 2*k, k] + Binomial[2*k, k]*Binomial[n - 2*k - 1, k], {k, 0, n/3}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 28 2017, after Joel B. Lewis *)
Table[Length[Select[Tuples[{0,1},n],Count[Partition[#,2,1],{0,0}]==Count[Partition[#,2,1],{0,1}]&]],{n,0,10}] (* Gus Wiseman, Jul 27 2021 *)
a[0]:=1; a[n_]:=(1 + 3*HypergeometricPFQ[{1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3},{1, (1-n)/2, 1-n/2, -3*n/8}, -27])/2; Array[a,37,0] (* Stefano Spezia, Apr 26 2024 *)

from math import comb
def A163493(n): return 2+sum((x:=comb((k:=m<<1)-1,m)*comb(n-k,m))+(x*(n-3*m)<<1)//(n-k) for m in range(1,n//3+1)) if n else 1 # Chai Wah Wu, May 01 2024

G.f.: 1/2/(1-x) + (1+2*x)/2/sqrt((1-x)*(1-2*x)*(1+x+2*x^2)). - Richard Stanley, corrected Apr 29 2011
G.f.: (1 + sqrt( 1 + 4*x / ((1 - x) * (1 - 2*x) * (1 + x + 2*x^2)))) / (2*(1 - x)). - Michael Somos, Jan 30 2012
a(n) = sum( binomial(2*k-1, k)*binomial(n-2*k,k) + binomial(2*k, k)*binomial(n-2*k-1, k), k=0..floor(n/3)). - Joel B. Lewis, May 21 2011
Conjecture: -n*a(n) +(2+n)*a(n-1) +(3n-12)*a(n-2) +(12-n)*a(n-3) +(2n-18)*a(n-4)+(56-12n)*a(n-5) +(8n-40)*a(n-6)=0. - R. J. Mathar, Nov 28 2011
G.f. y = A(x) satisfies x = (1 - x) * (1 - 2*x) * (1 + x + 2*x^2) * y * (y * (1 - x) - 1). - Michael Somos, Jan 30 2012
Sequence a(n) satisfies 0 = a(n) * (n^2-2*n) + a(n-1) * (-3*n^2+8*n-2) + a(n-2) * (3*n^2-10*n+2) + a(n-3) * (-5*n^2+18*n-6) + a(n-4) * (8*n^2-34*n+22) + a(n-5) * (-4*n^2+20*n-16) except if n=1 or n=2. - Michael Somos, Jan 30 2012
a(n) = (1 + 3*hypergeom([1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3],[1, (1-n)/2, 1-n/2, -3*n/8],-27))/2 for n > 0. - Stefano Spezia, Apr 26 2024
a(n) ~ 2^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 26 2024

A147875 Second heptagonal numbers: a(n) = n(5n+3)/2.

Original entry on oeis.org

0, 4, 13, 27, 46, 70, 99, 133, 172, 216, 265, 319, 378, 442, 511, 585, 664, 748, 837, 931, 1030, 1134, 1243, 1357, 1476, 1600, 1729, 1863, 2002, 2146, 2295, 2449, 2608, 2772, 2941, 3115, 3294, 3478, 3667, 3861, 4060, 4264, 4473, 4687, 4906, 5130, 5359, 5593

Offset: 0

Vladimir Joseph Stephan Orlovsky, Nov 16 2008

Zero followed by partial sums of A016897.
Apparently = every 2nd term of A111710 and A085787.
Bisection of A085787. Sequence found by reading the line from 0, in the direction 0, 13, ... and the line from 4, in the direction 4, 27, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Numbers of the form m^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

			G.f. = 4*x + 13*x^2 + 27*x^3 + 46*x^4 + 70*x^5 + 99*x^6 + 133*x^7 + ... - _Michael Somos_, Jan 25 2019

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Sliced Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016897, A111710, A000217, A085787, A224419 (positions of squares).
Second n-gonal numbers: A005449, A014105, A045944, A179986, A033954, A062728, A135705.
Cf. A000566.
Cf. A003215, A000096, A000290.

List([0..50], n-> n*(5*n+3)/2); # G. C. Greubel, Jul 04 2019

[n*(5*n+3)/2: n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[(n(5n+3))/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 4, 13}, 50] (* Harvey P. Dale, May 15 2013 *)

a(n)=n*(5*n+3)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(5*n+3)/2 for n in (0..50)] # G. C. Greubel, Jul 04 2019

G.f.: x*(4+x)/(1-x)^3.
a(n) = Sum_{k=0..n-1} A016897(k).
a(n) - a(n-1) = 5*n -1. - Vincenzo Librandi, Nov 26 2010
G.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3) + (2*k+2)*(2*k+3)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
E.g.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - 2*x*(k+2)^2*(k+3)/(2*x*(k+2)*(k+3) + (2*k+2)^2*(2*k+3)/U(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
a(n) = A130520(5n+3). - Philippe Deléham, Mar 26 2013
a(n) = A131242(10n+7)/2. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=13. - Harvey P. Dale, May 15 2013
Sum_{n>=1} 1/a(n) = 10/9 + sqrt(1 - 2/sqrt(5))*Pi/3 - 5*log(5)/6 + sqrt(5)*log((1 + sqrt(5))/2)/3 = 0.4688420784500060750083432... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n) + A000217(2*n). - Bruno Berselli, Jul 01 2016
From Ilya Gutkovskiy, Jul 01 2016: (Start)
E.g.f.: x*(8 + 5*x)*exp(x)/2.
Dirichlet g.f.: (5*zeta(s-2) + 3*zeta(s-1))/2. (End)
a(n) = A000566(-n) for all n in Z. - Michael Somos, Jan 25 2019
From Leo Tavares, Feb 14 2022: (Start)
a(n) = A003215(n) - A000217(n+1). See Sliced Hexagons illustration in links.
a(n) = A000096(n) + 2*A000290(n). (End)

Edited by Klaus Brockhaus and R. J. Mathar, Nov 20 2008

New name from Bruno Berselli, Jan 13 2011

A214292 Triangle read by rows: T(n,k) = T(n-1,k-1) + T(n-1,k), 0 < k < n with T(n,0) = n and T(n,n) = -n.

Original entry on oeis.org

0, 1, -1, 2, 0, -2, 3, 2, -2, -3, 4, 5, 0, -5, -4, 5, 9, 5, -5, -9, -5, 6, 14, 14, 0, -14, -14, -6, 7, 20, 28, 14, -14, -28, -20, -7, 8, 27, 48, 42, 0, -42, -48, -27, -8, 9, 35, 75, 90, 42, -42, -90, -75, -35, -9, 10, 44, 110, 165, 132, 0, -132, -165, -110, -44, -10

Offset: 0

Reinhard Zumkeller, Jul 12 2012

			The triangle begins:
    0:                              0
    1:                            1   -1
    2:                          2   0   -2
    3:                       3    2   -2   -3
    4:                     4    5   0   -5   -4
    5:                  5    9    5   -5   -9   -5
    6:                6   14   14   0  -14  -14   -6
    7:             7   20   28   14  -14  -28  -20   -7
    8:           8   27   48   42   0  -42  -48  -27   -8
    9:        9   35   75   90   42  -42  -90  -75  -35   -9
   10:     10   44  110  165  132   0 -132 -165 -110  -44  -10
   11:  11   54  154  275  297  132 -132 -297 -275 -154  -54  -11  .

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A000004, A000096, A000108, A000245, A000344, A000588, A000589, A000590, A001392, A002057, A003517, A003518, A003519, A005557, A005586, A005587, A008313, A014495, A064059, A064061, A080956, A090749, A097808, A112467, A124087, A124088, A129936, A259525.

a214292 n k = a214292_tabl !! n !! k
a214292_row n = a214292_tabl !! n
a214292_tabl = map diff $ tail a007318_tabl
   where diff row = zipWith (-) (tail row) row

row[n_] := Table[Binomial[n, k], {k, 0, n}] // Differences;
T[n_, k_] := row[n + 1][[k + 1]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 31 2018 *)

T(n,k) = A007318(n+1,k+1) - A007318(n+1,k), 0<=k<=n, i.e. first differences of rows in Pascal's triangle;
T(n,k) = -T(n,k);
row sums and central terms equal 0, cf. A000004;
sum of positive elements of n-th row = A014495(n+1);
T(n,0) = n;
T(n,1) = A000096(n-2) for n > 1; T(n,1) = - A080956(n) for n > 0;
T(n,2) = A005586(n-4) for n > 3; T(n,2) = A129936(n-2);
T(n,3) = A005587(n-6) for n > 5;
T(n,4) = A005557(n-9) for n > 8;
T(n,5) = A064059(n-11) for n > 10;
T(n,6) = A064061(n-13) for n > 12;
T(n,7) = A124087(n) for n > 14;
T(n,8) = A124088(n) for n > 16;
T(2*n+1,n) = T(2*n+2,n) = A000108(n+1), Catalan numbers;
T(2*n+3,n) = A000245(n+2);
T(2*n+4,n) = A002057(n+1);
T(2*n+5,n) = A000344(n+3);
T(2*n+6,n) = A003517(n+3);
T(2*n+7,n) = A000588(n+4);
T(2*n+8,n) = A003518(n+4);
T(2*n+9,n) = A001392(n+5);
T(2*n+10,n) = A003519(n+5);
T(2*n+11,n) = A000589(n+6);
T(2*n+12,n) = A090749(n+6);
T(2*n+13,n) = A000590(n+7).

A120730 Another version of Catalan triangle A009766.

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Aug 17 2006, corrected Sep 15 2006

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, ...] DELTA [1, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, -1, 1, ...] where DELTA is the operator defined in A084938.
Aerated version gives A165408. - Philippe Deléham, Sep 22 2009
T(n,k) is the number of length n left factors of Dyck paths having k up steps. Example: T(5,4)=4 because we have UDUUU, UUDUU, UUUDU, and UUUUD, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Jun 19 2011
With zeros omitted: 1,1,1,1,2,1,2,3,1,5,4,1,... = A008313. - Philippe Deléham, Nov 02 2011

			As a triangle, this begins:
  1;
  0,  1;
  0,  1,  1;
  0,  0,  2,  1;
  0,  0,  2,  3,  1;
  0,  0,  0,  5,  4,  1;
  0,  0,  0,  5,  9,  5,  1;
  0,  0,  0,  0, 14, 14,  6,  1;
  ...

Alois P. Heinz, Rows n = 0..200, flattened

Cf. A000007, A000096, A000108, A001405, A001477, A003161, A005586, A005587.
Cf. A008313, A009766, A039598, A039599, A084938, A099376, A105523, A121725.
Cf. A126087, A128386, A121724, A128387, A129123, A132373, A132374, A132375.
Cf. A132889, A151162, A151254, A151281, A156195, A156361, A156362, A156566.
Cf. A156577, A165408, A357654.

A120730:= func< n,k | n gt 2*k select 0 else Binomial(n, k)*(2*k-n+1)/(k+1) >;
[A120730(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Nov 07 2022

G := 4*z/((2*z-1+sqrt(1-4*z^2*t))*(1+sqrt(1-4*z^2*t))): Gser := simplify(series(G, z = 0, 13)): for n from 0 to 12 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 12 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form  # Emeric Deutsch, Jun 19 2011
# second Maple program:
b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
    end:
T:= (n, k)-> b(n, 2*k-n):
seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Oct 13 2022

b[x_, y_]:= b[x, y]= If[y<0 || y>x, 0, If[x==0, 1, Sum[b[x-1, y+j], {j, {-1, 1}}] ]];
T[n_, k_] := b[n, 2 k - n];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Oct 21 2022, after Alois P. Heinz *)
T[n_, k_]:= If[n>2*k, 0, Binomial[n, k]*(2*k-n+1)/(k+1)];
Table[T[n, k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 07 2022 *)

def A120730(n,k): return 0 if (n>2*k) else binomial(n, k)*(2*k-n+1)/(k+1)
flatten([[A120730(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Nov 07 2022

G.f.: G(t,z) = 4*z/((2*z-1+sqrt(1-4*t*z^2))*(1+sqrt(1-4*t*z^2))). - Emeric Deutsch, Jun 19 2011
Sum_{k=0..n} x^k*T(n,n-k) = A001405(n), A126087(n), A128386(n), A121724(n), A128387(n), A132373(n), A132374(n), A132375(n), A121725(n) for x=1,2,3,4,5,6,7,8,9 respectively. [corrected by Philippe Deléham, Oct 16 2008]
T(2*n,n) = A000108(n); A000108: Catalan numbers.
From Philippe Deléham, Oct 18 2008: (Start)
Sum_{k=0..n} T(n,k)^2 = A000108(n) and Sum_{n>=k} T(n,k) = A000108(k+1).
Sum_{k=0..n} T(n,k)^3 = A003161(n).
Sum_{k=0..n} T(n,k)^4 = A129123(n). (End)
Sum_{k=0..n}, T(n,k)*x^k = A000007(n), A001405(n), A151281(n), A151162(n), A151254(n), A156195(n), A156361(n), A156362(n), A156566(n), A156577(n) for x=0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Feb 10 2009
From G. C. Greubel, Nov 07 2022: (Start)
T(n, k) = 0 if n > 2*k, otherwise binomial(n, k)*(2*k-n+1)/(k+1).
Sum_{k=0..n} (-1)^k*T(n,k) = A105523(n).
Sum_{k=0..n} (-1)^k*T(n,k)^2 = -A132889(n), n >= 1.
Sum_{k=0..floor(n/2)} T(n-k, k) = A357654(n).
T(n, n-1) = A001477(n).
T(n, n-2) = [n=2] + A000096(n-3), n >= 2.
T(n, n-3) = 2*[n<5] + A005586(n-5), n >= 3.
T(n, n-4) = 5*[n<7] - 2*[n=4] + A005587(n-7), n >= 4.
T(2*n+1, n+1) = A000108(n+1), n >= 0.
T(2*n-1, n+1) = A099376(n-1), n >= 1. (End)

cp's OEIS Frontend

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A014107 a(n) = n*(2*n-3).

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A033282 Triangle read by rows: T(n, k) is the number of diagonal dissections of a convex n-gon into k+1 regions.

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A080956 a(n) = (n+1)*(2-n)/2.

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