A002054 -id:A002054 - cp's OEIS frontend

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

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.
Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6.
Paul Barry, On the Inverses of a Family of Pascal-Like Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.6.
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
Arthur Cayley, On the partitions of a polygon, Proc. London Math. Soc., 22 (1891), 237-262 = Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 93ff. (See p. 239.)
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]
Satyan L. Devadoss, Combinatorial Equivalence of Real Moduli Spaces, Notices Amer. Math. Soc. 51 (2004), no. 6, 620-628.
Satyan L. Devadoss and Ronald C. Read, Cellular structures determined by polygons and trees, arXiv/0008145 [math.CO], 2000. [From Tom Copeland Nov 21 2017]
Anton Dochtermann, Face rings of cycles, associahedra, and standard Young tableaux, arXiv preprint arXiv:1503.06243 [math.CO], 2015.
Brian Drake, Ira M. Gessel, and Guoce Xin, Three Proofs and a Generalization of the Goulden-Litsyn-Shevelev Conjecture on a Sequence Arising in Algebraic Geometry, J. of Integer Sequences, Vol. 10 (2007), Article 07.3.7.
Cassandra Durell and Stefan Forcey, Level-1 Phylogenetic Networks and their Balanced Minimum Evolution Polytopes, arXiv:1905.09160 [math.CO], 2019.
P. Flajolet and M. Noy, Analytic Combinatorics of Non-crossing Configurations, Discrete Math., 204, 1999, 203-229.
Sergey Fomin and Nathan Reading, Root systems and generalized associahedra, Lecture notes for IAS/Park-City 2004, arXiv:math/0505518 [math.CO], 2005-2008. [From _Peter Bala_, Oct 28 2008]
Sergey Fomin and Andrei Zelevinsky, Cluster algebras I: Foundations, arXiv:math/0104151 [math.RT], 2001.
Sergey Fomin and Andrei Zelevinsky, Cluster algebras I: Foundations, J. Amer. Math. Soc. 15 (2002) no.2, 497-529.
Sergey Fomin and Andrei Zelevinsky, Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3, 977-1018.
Ivan Geffner and Marc Noy, Counting Outerplanar Maps, Electronic Journal of Combinatorics 24(2) (2017), #P2.3.
Rijun Huang, Fei Teng, and Bo Feng, Permutation in the CHY-Formulation, arXiv:1801.08965 [hep-th], 2018.
Germain Kreweras, Sur les partitions non croisées d'un cycle, (French) Discrete Math. 1 (1972), no. 4, 333--350. MR0309747 (46 #8852).
Germain Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
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

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

A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

Original entry on oeis.org

1, 3, 1, 10, 5, 1, 35, 21, 7, 1, 126, 84, 36, 9, 1, 462, 330, 165, 55, 11, 1, 1716, 1287, 715, 286, 78, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1, 92378, 75582, 50388

Offset: 0

Philippe Deléham, Nov 13 2005

Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) where c(x) is g.f. of A000108. Unsigned version of A113187. Diagonal sums are A014301(n+1).
Triangle T(n,k),0<=k<=n, read by rows defined by :T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=3*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+2*T(n-1,k)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 22 2007
Reversal of A122366. - Philippe Deléham, Mar 22 2007
Column k has e.g.f. exp(2x)(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Jun 06 2007
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Diagonal sums are A014301(n+1). - Paul Barry, Mar 08 2011
This triangle T(n,k) appears in the expansion of odd powers of Fibonacci numbers F=A000045 in terms of F-numbers with multiples of odd numbers as indices. See the Ozeki reference, p. 108, Lemma 2. The formula is: F_l^(2*n+1) = sum(T(n,k)*(-1)^((n-k)*(l+1))* F_{(2*k+1)*l}, k=0..n)/5^n, n >= 0, l >= 0. - Wolfdieter Lang, Aug 24 2012
Central terms give A052203. - Reinhard Zumkeller, Mar 14 2014
This triangle appears in the expansion of (4*x)^n in terms of the polynomials Todd(n, x):= T(2*n+1, sqrt(x))/sqrt(x) = sum(A084930(n,m)*x^m), n >= 0. This follows from the inversion of the lower triangular Riordan matrix built from A084930 and comparing the g.f. of the row polynomials. - Wolfdieter Lang, Aug 05 2014
From Wolfdieter Lang, Aug 15 2014: (Start)
This triangle is the inverse of the signed Riordan triangle (-1)^(n-m)*A111125(n,m).
This triangle T(n,k) appears in the expansion of x^n in terms of the polynomials todd(k, x):= T(2*k+1, sqrt(x)/2)/(sqrt(x)/2) = S(k, x-2) - S(k-1, x-2) with the row polynomials T and S for the triangles A053120 and A049310, respectively: x^n = sum(T(n,k)*todd(k, x), k=0..n). Compare this with the preceding comment.
The A- and Z-sequences for this Riordan triangle are [1, 2, 1, repeated 0] and [3, 1, repeated 0]. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. This corresponds to the recurrences given in the Philippe Deléham, Mar 22 2007 comment above. (End)

			From _Wolfdieter Lang_, Aug 05 2014: (Start)
The triangle T(n,k) begins:
n\k      0      1      2      3     4     5    6    7   8  9  10 ...
0:       1
1:       3      1
2:      10      5      1
3:      35     21      7      1
4:     126     84     36      9     1
5:     462    330    165     55    11     1
6:    1716   1287    715    286    78    13    1
7:    6435   5005   3003   1365   455   105   15    1
8:   24310  19448  12376   6188  2380   680  136   17   1
9:   92378  75582  50388  27132 11628  3876  969  171  19  1
10: 352716 293930 203490 116280 54264 20349 5985 1330 210 21   1
...
Expansion examples (for the Todd polynomials see A084930 and a comment above):
(4*x)^2 = 10*Todd(n,  0) + 5*Todd(n, 1) + 1*Todd(n, 2) = 10*1 + 5*(-3 + 4*x) + 1*(5 - 20*x + 16*x^2).
(4*x)^3 =  35*1 + 21*(-3 + 4*x) + 7*(5 - 20*x + 16*x^2) + (-7 + 56*x - 112*x^2 +64*x^3)*1. (End)
---------------------------------------------------------------------
Production matrix is
3, 1,
1, 2, 1,
0, 1, 2, 1,
0, 0, 1, 2, 1,
0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 0, 1, 2, 1
- _Paul Barry_, Mar 08 2011
Application to odd powers of Fibonacci numbers F, row n=2:
F_l^5 = (10*(-1)^(2*(l+1))*F_l + 5*(-1)^(1*(l+1))*F_{3*l} + 1*F_{5*l})/5^2, l >= 0. - _Wolfdieter Lang_, Aug 24 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012.
A. Nkwanta, A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
Sun, Yidong; Ma, Luping Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.

Cf. A000108, A113187.

Columns are : A001700, A002054, A003516, A030053, A030054, A030055, A030056.

a111418 n k = a111418_tabl !! n !! k
a111418_row n = a111418_tabl !! n
a111418_tabl = map reverse a122366_tabl
-- Reinhard Zumkeller, Mar 14 2014

Table[Binomial[2*n+1, n-k], {n,0,10}, {k,0,n}] (* G. C. Greubel, May 22 2017 *)
T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
Table[T[n, k, 3, 2], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

T(n, k) = C(2*n+1, n-k).
Sum_{k=0..n} T(n, k) = 4^n.
Sum_{k, 0<=k<=n}(-1)^k *T(n,k) = binomial(2*n,n) = A000984(n). - Philippe Deléham, Mar 22 2007
T(n,k) = sum{j=k..n, C(n,j)*2^(n-j)*C(j,floor((j-k)/2))}. - Paul Barry, Jun 06 2007
Sum_{k, k>=0} T(m,k)*T(n,k) = T(m+n,0)= A001700(m+n). - Philippe Deléham, Nov 22 2009
G.f. row polynomials: ((1+x) - (1-x)/sqrt(1-4*z))/(2*(x - (1+x)^2*z))
(see the Riordan property mentioned in a comment above). - Wolfdieter Lang, Aug 05 2014

A345910 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum -1.

Original entry on oeis.org

6, 20, 25, 27, 30, 72, 81, 83, 86, 92, 98, 101, 103, 106, 109, 111, 116, 121, 123, 126, 272, 289, 291, 294, 300, 312, 322, 325, 327, 330, 333, 335, 340, 345, 347, 350, 360, 369, 371, 374, 380, 388, 393, 395, 398, 402, 405, 407, 410, 413, 415, 420, 425, 427

Offset: 1

Gus Wiseman, Jul 01 2021

nonn

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
      6: (1,2)
     20: (2,3)
     25: (1,3,1)
     27: (1,2,1,1)
     30: (1,1,1,2)
     72: (3,4)
     81: (2,4,1)
     83: (2,3,1,1)
     86: (2,2,1,2)
     92: (2,1,1,3)
     98: (1,4,2)
    101: (1,3,2,1)
    103: (1,3,1,1,1)
    106: (1,2,2,2)
    109: (1,2,1,2,1)

These compositions are counted by A001791.
A version using runs of binary digits is A031444.
These are the positions of -1's in A124754.
The opposite (positive 1) version is A345909.
The reverse version is A345912.
The version for alternating sum of prime indices is A345959.
Standard compositions: A000120, A066099, A070939, A124754, A228351, A344618.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A000070 counts partitions of 2n+1 with alternating sum 1, ranked by A001105.
A011782 counts compositions.
A097805 counts compositions by sum and alternating sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A345197 counts compositions by sum, length, and alternating sum.
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. A000097, A000346, A008549, A025047, A027187, A031443, A031448, A114121, A119899, A126869, A238279, A344617.

stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]==-1&]

A345912 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum -1.

Original entry on oeis.org

5, 18, 23, 25, 29, 68, 75, 78, 81, 85, 90, 95, 98, 103, 105, 109, 114, 119, 121, 125, 264, 275, 278, 284, 289, 293, 298, 303, 308, 315, 318, 322, 327, 329, 333, 338, 343, 345, 349, 356, 363, 366, 369, 373, 378, 383, 388, 395, 398, 401, 405, 410, 415, 418, 423

Offset: 1

Gus Wiseman, Jul 01 2021

nonn

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

			The sequence of terms together with the corresponding compositions begins:
      5: (2,1)
     18: (3,2)
     23: (2,1,1,1)
     25: (1,3,1)
     29: (1,1,2,1)
     68: (4,3)
     75: (3,2,1,1)
     78: (3,1,1,2)
     81: (2,4,1)
     85: (2,2,2,1)
     90: (2,1,2,2)
     95: (2,1,1,1,1,1)
     98: (1,4,2)
    103: (1,3,1,1,1)
    105: (1,2,3,1)

These compositions are counted by A001791.
These are the positions of -1's in A344618.
The non-reverse version is A345910.
The opposite (positive 1) version is A345911.
The version for Heinz numbers of partitions is A345959.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
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).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
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. A000070, A000346, A001105, A008549, A025047, A031444, A034871, A114121, A126869, A344608, A345958, A345959.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]==-1&]

A345917 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum > 0.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 9, 11, 14, 16, 17, 18, 19, 21, 22, 23, 26, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 42, 44, 45, 47, 52, 56, 57, 59, 62, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 82, 84, 85, 87, 88, 89, 90, 91, 93, 94, 95, 100, 104, 105, 107

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The initial terms and the corresponding compositions:
     1: (1)
     2: (2)
     4: (3)
     5: (2,1)
     7: (1,1,1)
     8: (4)
     9: (3,1)
    11: (2,1,1)
    14: (1,1,2)
    16: (5)
    17: (4,1)
    18: (3,2)
    19: (3,1,1)
    21: (2,2,1)
    22: (2,1,2)

The version for Heinz numbers of partitions is A026424.
These compositions are counted by A027306.
These are the positions of terms > 0 in A124754.
The weak (k >= 0) version is A345913.
The reverse-alternating version is A345918.
The opposite (k < 0) version is A345919.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
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).
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
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. A000070, A000346, A008549, A025047, A027187, A027193, A032443, A114121, A163493, A344609, A345908.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]>0&]

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 20, 10, 4, 1, 70, 35, 15, 5, 1, 252, 126, 56, 21, 6, 1, 924, 462, 210, 84, 28, 7, 1, 3432, 1716, 792, 330, 120, 36, 8, 1, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 48620, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 184756, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1

Offset: 0

Ralf Stephan, Mar 21 2004

First column is C(2*n,n) or A000984. Central coefficients are C(3*n,n) or A005809. - Paul Barry, Oct 14 2009
T(n,k) = A046899(n,n-k), k = 0..n-1. - Reinhard Zumkeller, Jul 27 2012
From  Peter Bala, Nov 03 2015: (Start)
Viewed as the square array [binomial (2*n + k, n + k)]n,k>=0  this is the generalized Riordan array ( 1/sqrt(1 - 4*x),c(x) ) in the sense of the Bala link, where c(x) is the o.g.f. for A000108.
The square array factorizes as ( 1/(2 - c(x)),x*c(x) ) * ( 1/(1 - x),1/(1 - x) ), which equals the matrix product of A100100 with the square Pascal matrix [binomial (n + k,k)]n,k>=0. See the example below. (End)

			From _Paul Barry_, Oct 14 2009: (Start)
Triangle begins
  1,
  2, 1,
  6, 3, 1,
  20, 10, 4, 1,
  70, 35, 15, 5, 1,
  252, 126, 56, 21, 6, 1,
  924, 462, 210, 84, 28, 7, 1,
  3432, 1716, 792, 330, 120, 36, 8, 1
Production array is
  2, 1,
  2, 1, 1,
  2, 1, 1, 1,
  2, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
As a square array = A100100 * square Pascal matrix:
  /1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
  |2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
  |6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
  |20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
  |70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n=0..149 of triangle, flattened
P. Bala, Notes on generalized Riordan arrays
P. Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Ik-Pyo Kim, Michael J. Tsatsomeros, Inverse Relations in Shapiro's Open Questions, arXiv:1707.06590 [math.CO], 2017. See p. 7.

Rows 0-14 are A000984, A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.
Columns are A000217, A000292, A000332, A000389, A000579.
Diagonals are A005809, A045721, A025174, A004319, A013698, A003408.
Cf. A100100.

a092392 n k = a092392_tabl !! (n-1) !! (k-1)
a092392_row n = a092392_tabl !! (n-1)
a092392_tabl = map reverse a046899_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As a triangle */ [[Binomial(2*n-k, n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 22 2017

A092392 := proc(n,k)
    binomial(2*n-k,n-k) ;
end proc:
seq(seq(A092392(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Table[Binomial[2 n - k, n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Mar 19 2016 *)

C(x):=(1-sqrt(1-4*x))/2;
A(x,y):=(1/sqrt(1-4*x))/(1-y*C(x));
taylor(A(x,y),y,0,10,x,0,10); /* Vladimir Kruchinin, Mar 19 2016 */

for(n=0,10, for(k=0,n, print1(binomial(2*n - k,n), ", "))) \\ G. C. Greubel, Nov 22 2017

As a number triangle, this is T(n, k) = if(k <= n, C(2*n - k, n), 0). Its row sums are C(2*n + 1, n + 1) = A001700. Its diagonal sums are A176287. - Paul Barry, Apr 23 2005
G.f. of column k: 2^k/[sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))^k].
As a number triangle, this is the Riordan array (1/sqrt(1 - 4*x), x*c(x)), c(x) the g.f. of A000108. - Paul Barry, Jun 24 2005
G.f.: A(x,y)=1/sqrt(1 - 4*x)/(1-y*x*C(x)), where C(x) is g.f. of Catalan numbers. - Vladimir Kruchinin, Mar 19 2016

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

Offset corrected by R. J. Mathar, Feb 08 2013

A345911 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum 1.

Original entry on oeis.org

1, 6, 7, 20, 21, 26, 27, 30, 31, 72, 73, 82, 83, 86, 87, 92, 93, 100, 101, 106, 107, 110, 111, 116, 117, 122, 123, 126, 127, 272, 273, 290, 291, 294, 295, 300, 301, 312, 313, 324, 325, 330, 331, 334, 335, 340, 341, 346, 347, 350, 351, 360, 361, 370, 371, 374

Offset: 1

Gus Wiseman, Jul 01 2021

nonn

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

			The sequence of terms together with the corresponding compositions begins:
     1: (1)
     6: (1,2)
     7: (1,1,1)
    20: (2,3)
    21: (2,2,1)
    26: (1,2,2)
    27: (1,2,1,1)
    30: (1,1,1,2)
    31: (1,1,1,1,1)
    72: (3,4)
    73: (3,3,1)
    82: (2,3,2)
    83: (2,3,1,1)
    86: (2,2,1,2)
    87: (2,2,1,1,1)

These compositions are counted by A000984 (bisection of A126869).
The version for Heinz numbers of partitions is A001105.
A version using runs of binary digits is A066879.
These are positions of 1's in A344618.
The non-reverse version is A345909.
The opposite (negative 1) version is A345912.
The version for prime indices is A345958.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
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).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
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. A000070, A000097, A000346, A008549, A025047, A027193, A031448, A034871, A114121, A120452, A344607.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]==1&]

A345913 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum >= 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 26, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 50, 52, 53, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 82

Offset: 1

Gus Wiseman, Jul 04 2021

nonn

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The sequence of terms together with the corresponding compositions begins:
     0: ()           17: (4,1)          37: (3,2,1)
     1: (1)          18: (3,2)          38: (3,1,2)
     2: (2)          19: (3,1,1)        39: (3,1,1,1)
     3: (1,1)        21: (2,2,1)        41: (2,3,1)
     4: (3)          22: (2,1,2)        42: (2,2,2)
     5: (2,1)        23: (2,1,1,1)      43: (2,2,1,1)
     7: (1,1,1)      26: (1,2,2)        44: (2,1,3)
     8: (4)          28: (1,1,3)        45: (2,1,2,1)
     9: (3,1)        29: (1,1,2,1)      46: (2,1,1,2)
    10: (2,2)        31: (1,1,1,1,1)    47: (2,1,1,1,1)
    11: (2,1,1)      32: (6)            50: (1,3,2)
    13: (1,2,1)      33: (5,1)          52: (1,2,3)
    14: (1,1,2)      34: (4,2)          53: (1,2,2,1)
    15: (1,1,1,1)    35: (4,1,1)        55: (1,2,1,1,1)
    16: (5)          36: (3,3)          56: (1,1,4)

These compositions are counted by A116406.
These are the positions of terms >= 0 in A124754.
The version for prime indices is A344609.
The reverse-alternating sum version is A345914.
The opposite (k <= 0) version is A345915.
The strict (k > 0) version is A345917.
The complement is A345919.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
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).
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
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. A000070, A000346, A008549, A025047, A027187, A032443, A034871, A114121, A163493, A236913, A238279, A344607, A344608, A344610, A344611.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]>=0&]

A345909 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum 1.

Original entry on oeis.org

1, 5, 7, 18, 21, 23, 26, 29, 31, 68, 73, 75, 78, 82, 85, 87, 90, 93, 95, 100, 105, 107, 110, 114, 117, 119, 122, 125, 127, 264, 273, 275, 278, 284, 290, 293, 295, 298, 301, 303, 308, 313, 315, 318, 324, 329, 331, 334, 338, 341, 343, 346, 349, 351, 356, 361

Offset: 1

Gus Wiseman, Jun 30 2021

nonn

The alternating sum of a composition (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The sequence of terms together with the corresponding compositions begins:
      1: (1)             87: (2,2,1,1,1)
      5: (2,1)           90: (2,1,2,2)
      7: (1,1,1)         93: (2,1,1,2,1)
     18: (3,2)           95: (2,1,1,1,1,1)
     21: (2,2,1)        100: (1,3,3)
     23: (2,1,1,1)      105: (1,2,3,1)
     26: (1,2,2)        107: (1,2,2,1,1)
     29: (1,1,2,1)      110: (1,2,1,1,2)
     31: (1,1,1,1,1)    114: (1,1,3,2)
     68: (4,3)          117: (1,1,2,2,1)
     73: (3,3,1)        119: (1,1,2,1,1,1)
     75: (3,2,1,1)      122: (1,1,1,2,2)
     78: (3,1,1,2)      125: (1,1,1,1,2,1)
     82: (2,3,2)        127: (1,1,1,1,1,1,1)
     85: (2,2,2,1)      264: (5,4)

These compositions are counted by A000984 (bisection of A126869).
The version for prime indices is A001105.
A version using runs of binary digits is A031448.
These are the positions of 1's in A124754.
The opposite (negative 1) version is A345910.
The reverse version is A345911.
The version for Heinz numbers of partitions is A345958.
Standard compositions: A000120, A066099, A070939, A124754, A228351, A344618.
A000070 counts partitions with alternating sum 1 (ranked by A345957).
A000097 counts partitions with alternating sum 2 (ranked by A345960).
A011782 counts compositions.
A097805 counts compositions by sum and alternating sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
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 (this sequence)/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. A000346, A008549, A025047, A031443, A031444, A034871, A114121, A238279, A304620, A344651.

stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]==1&]

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Sage

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Formula

A345910 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum -1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A345912 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum -1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A345917 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum > 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs