A003161 -id:A003161 - cp's OEIS frontend

A053121 Catalan triangle (with 0's) read by rows.

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

Offset: 0

Wolfdieter Lang

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of n-step walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b-1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,-1). (This is basically a different formulation of the previous - walks with a wall - property.) - Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1-x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bell-matrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(-1)}(x))/x, with f(x) := x*G(x) and f^{(-1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231. - Emeric Deutsch, Oct 12 2006
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
Riordan array (c(x^2),xc(x^2)), where c(x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Nov 25 2007
A053121^2 = triangle A145973. Convolved with A001405 = triangle A153585. - Gary W. Adamson, Dec 28 2008
By columns without the zeros, n-th row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then n-th row = coefficients of A^(n+1). - Gary W. Adamson, May 13 2009
Triangle read by rows,product of A130595 and A064189 considered as infinite lower triangular arrays; A053121 = A130595*A064189 = B^(-1)*A097609*B where B = A007318. - Philippe Deléham, Dec 07 2009
From Mark Dols, Aug 17 2010: (Start)
As an upper right triangle, rows represent powers of 5-sqrt(24):
  5 - sqrt(24)^1 = 0.101020514...
  5 - sqrt(24)^2 = 0.010205144...
  5 - sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)-steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,-1), H=(1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH. - Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the N-th Chebyshev S-polynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x). - L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular N-gon:
  rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j-1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310. - Wolfdieter Lang, Sep 21 2013
The so-called A-sequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n-1, m-1) + a(n-1, m+1) for n>=1 and m>=1, with inputs. The Z-sequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n-1,1), n>=1, a(0,0) = 1. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. - Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2-dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the m-th ((m+1)-dimensional) irreducible representation in the n-th tensor power of the standard one. - Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n-1} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)*p(n-1-j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section. - Wolfdieter Lang, Aug 11 2017
From Roger Ford, Jan 22 2018: (Start)
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the x-axis with the following constraints:  The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches.  If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3.  c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
  top UuUdUddd c=3   top UdUuUddd c=1     top UdUdUudd c=1
       /\                    /\
      //\\                  /  \
     //  \\                / /\ \                    /\
    //    \\              / /  \ \                  /  \
   ///\  /\\\        /\  / / /\ \ \        /\  /\  / /\ \
   \\\ \/ ///        \ \ \ \/ / / /        \ \ \ \/ / / /
    \\\  ///          \ \ \  / / /          \ \ \  / / /
     \\\///            \ \ \/ / /            \ \ \/ / /
      \\//              \ \  / /              \ \  / /
       \/                \ \/ /                \ \/ /
                          \  /                  \  /
                           \/                    \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)

			Triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6  7  8  9 10 ...
  0:   1
  1:   0   1
  2:   1   0   1
  3:   0   2   0   1
  4:   2   0   3   0   1
  5:   0   5   0   4   0   1
  6:   5   0   9   0   5   0   1
  7:   0  14   0  14   0   6   0  1
  8:  14   0  28   0  20   0   7  0  1
  9:   0  42   0  48   0  27   0  8  0  1
  10: 42   0  90   0  75   0  35  0  9  0  1
  ... (Reformatted by _Wolfdieter Lang_, Sep 20 2013)
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
From _Paul Barry_, May 29 2009: (Start)
Production matrix is
  0, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
Boas-Buck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9. - _Wolfdieter Lang_, Aug 11 2017

J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
I. Bajunaid et al., Function series, Catalan numbers and random walks on trees, Amer. Math. Monthly 112 (2005), 765-785.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 3.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 3.
Xiang-Ke Chang, X.-B. Hu, H. Lei, and Y.-N. Yeh, Combinatorial proofs of addition formulas, The Electronic Journal of Combinatorics, 23(1) (2016), #P1.8.
J. Cigler, Some q-analogues of Fibonacci, Lucas and Chebyshev polynomials with nice moments, 2013.
J. Cigler, Some remarks about q-Chebyshev polynomials and q-Catalan numbers and related results, 2013.
J. Cigler, Some notes on q-Gould polynomials, 2013.
Emeric Deutsch, A. Robertson and D. Saracino, Refined restricted involutions, European Journal of Combinatorics 28 (2007), 481-498 (see pp. 486 and 498).
J. East and R. D. Gray, Idempotent generators in finite partition monoids and related semigroups, arXiv preprint arXiv:1404.2359, 2014
D. Gouyou-Beauchamps, Chemins sous-diagonaux et tableau de Young, pp. 112-125 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986 (see |F_{l,p}| on page 114). - _N. J. A. Sloane_, Jan 29 2011
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., 14 (1976), 395-405.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Note 4, pp. 414-415.
MathOverflow, Catalan numbers as sums of squares of numbers in the rows of the Catalan triangle - is there a combinatorial explanation?
A. Nkwanta and A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
Karim Ritter von Merkl, Computing colored Khovanov homology, arXiv:2505.03916 [math.QA], 2025. See p. 2.
Frank Ruskey and Mark Weston, Spherical Venn Diagrams with Involutory Isometries, Electronic Journal of Combinatorics, 18 (2011), #P191.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014).
Mark C. Wilson, Diagonal asymptotics for products of combinatorial classes.
W.-J. Woan, Area of Catalan Paths, Discrete Math., 226 (2001), 439-444.
Index entries for sequences related to Chebyshev polynomials.

Cf. A008315, A049310, A000108, A001405 (row sums), A145973, A153585, A108786, A037952. Another version: A008313. A039598 and A039599 without zeros, and odd and even numbered rows.

Variant without zero-diagonals: A033184 and with rows reversed: A009766.

a053121 n k = a053121_tabl !! n !! k
a053121_row n = a053121_tabl !! n
a053121_tabl = iterate
   (\row -> zipWith (+) ([0] ++ row) (tail row ++ [0,0])) [1]
-- Reinhard Zumkeller, Feb 24 2012

T:=proc(n,k): if n+k mod 2 = 0 then (k+1)*binomial(n+1,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
F:=proc(l,p) if ((l-p) mod 2) = 1 then 0 else (p+1)*l!/( ( (l-p)/2 )! * ( (l+p)/2 +1)! ); fi; end;
r:=n->[seq( F(n,p),p=0..n)]; [seq(r(n),n=0..15)]; # N. J. A. Sloane, Jan 29 2011
A053121 := proc(n,k) option remember; `if`(k>n or k<0,0,`if`(n=k,1,
procname(n-1,k-1)+procname(n-1,k+1))) end proc:
seq(print(seq(A053121(n,k), k=0..n)),n=0..12); # Peter Luschny, May 01 2011

a[n_, m_] /; n < m || OddQ[n-m] = 0; a[n_, m_] = (m+1) Binomial[n+1, (n-m)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* Jean-François Alcover, May 18 2011 *)
T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = T[n-1, k-1]+T[n-1, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Dec 31 2024 *)

T(n, m)=if(nCharles R Greathouse IV, Mar 09 2016

def A053121_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1] + M[n-1,k+1]
    return M
A053121_triangle(13) # Peter Luschny, Sep 19 2012

a(n, m) := 0 if n

a(n, m) = (4*(n-1)*a(n-2, m) + 2*(m+1)*a(n-1, m-1))/(n+m+2), a(n, m)=0 if n

G.f. for m-th column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.

G.f.: G(t,z) = c(z^2)/(1 - t*z*c(z^2)), where c(z) = (1 - sqrt(1-4*z))/(2*z) is the g.f. for the Catalan numbers (A000108). - Emeric Deutsch, Jun 16 2011

a(n, m) = a(n-1, m-1) + a(n-1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0. - Henry Bottomley, Jan 25 2001

Sum_{k>=0} T(m,k)^2 = A000108(m). - Paul D. Hanna, Apr 23 2005

Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even. - Philippe Deléham, May 26 2005

T(n,k)=sum{i=0..n, (-1)^(n-i)*C(n,i)*sum{j=0..i, C(i,j)*(C(i-j,j+k)-C(i-j,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)-BesselI(k+2,2x). - Paul Barry, Feb 16 2006

Sum_{k=0..n} T(n,k)*(k+1) = 2^n. - Philippe Deléham, Mar 22 2007

Sum_{j>=0} T(n,j)*binomial(j,k) = A054336(n,k). - Philippe Deléham, Mar 30 2007

T(2*n+1,2*k+1) = A039598(n,k), T(2*n,2*k) = A039599(n,k). - Philippe Deléham, Apr 16 2007

Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 22 2009

Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 28 2009

Sum_{k=0..n} T(n,k)*A000045(k+1) = A098615(n). - Philippe Deléham, Feb 03 2012

Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n-1-k, x), n >= 0, with C(-1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108. - Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015

The Boas-Buck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(n-m))*Sum_{j=0..n-1-m} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)* a(n-1-j, k), for n > m >= 0 and input a(m, m) = 1. - Wolfdieter Lang, Aug 11 2017

Sum_{m=1..n} a(n,m) = A037952(n). - R. J. Mathar, Sep 23 2021

Edited by N. J. A. Sloane, Jan 29 2011

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)

A129123 Number of 4-tuples of standard tableau with height less than or equal to 2.

Original entry on oeis.org

1, 1, 2, 17, 98, 882, 7812, 78129, 815474, 8955650, 101869508, 1194964498, 14374530436, 176681194276, 2212121332488, 28145258688369, 363177582488274, 4745064935840178, 62687665026816228, 836447728509168930, 11261240896657686660, 152847558411986548260

Offset: 0

Mike Zabrocki, Mar 29 2007

nonn

Number of pairs of Dyck paths of semilength n with equal midpoint. - Alois P. Heinz, Oct 07 2022

Alois P. Heinz, Table of n, a(n) for n = 0..839
F. Bergeron, L. Favreau and D. Krob, Conjectures on the enumeration of tableaux of bounded height, Discrete Math, vol. 139, no. 1-3 (1995), 463-468.

Cf. A000108, A001405, A003161, A120730, A355481, A357652.

Column k=4 of A357824.

[(&+[((n-2*j+1)/(n-j+1))^4*Binomial(n,j)^4: j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 08 2022

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:
a:= n-> add(b(n, n-2*j)^4, j=0..n/2):
seq(a(n), n=0..21);  # Alois P. Heinz, Mar 25 2025

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^4, {k,0, Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Dec 16 2017 *)

a(n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^4);

from math import comb
def A129123(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**4 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

def A129123(n): return sum(((n-2*j+1)/(n-j+1))^4*binomial(n,j)^4 for j in range((n//2)+1))
[A129123(n) for n in range(31)] # G. C. Greubel, Nov 08 2022

a(n) = Sum_{k=0..n} A120730(n,k)^4. - Philippe Deléham, Oct 18 2008
From Vaclav Kotesovec, Dec 16 2017: (Start)
Recurrence: n*(n+1)^3*(15*n^2 - 34*n + 7)*a(n) = 2*n*(90*n^5 - 309*n^4 + 147*n^3 + 124*n^2 - 135*n + 35)*a(n-1) + 4*(n-1)^2*(4*n - 5)*(4*n - 3)*(15*n^2 - 4*n - 12)*a(n-2).
a(n) ~ 3* 2^(4*n - 1/2) / (Pi^(3/2) * n^(7/2)). (End)
a(n) = A357652(n) - A355481(n). - Alois P. Heinz, Oct 13 2022
a(n) = Sum_{j=0..floor(n/2)} ((n-2*j+1)/(n-j+1))^4 * binomial(n,j)^4. - G. C. Greubel, Nov 08 2022
a(n) = Sum_{k=0..n} binomial(n,k) * ( binomial(n,k) - binomial(n,k-1) )^3. - Seiichi Manyama, Mar 25 2025

A003162 A binomial coefficient summation.

Original entry on oeis.org

1, 1, 1, 3, 6, 19, 49, 163, 472, 1626, 5034, 17769, 57474, 206487, 688881, 2508195, 8563020, 31504240, 109492960, 406214878, 1432030036, 5349255726, 19077934506, 71672186953, 258095737156, 974311431094, 3537275250214, 13408623649893

Offset: 0

N. J. A. Sloane

From Peter Bala, Mar 26 2023: (Start)
For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). For other cases see A361888 ({S(5,n)/S(1,n)}) and A361891 ({S(7,n)/ S(1,n)}).
Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^k) == b(n*p^(k-1)) (mod p^(3*k)) holds for positive integers n and k and all primes p >= 5. See A183069. (End)

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

Seiichi Manyama, Table of n, a(n) for n = 0..1000
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161, A183069, A361887, A361888, A361889, A361890, A361891, A361892.

H := hypergeom([1/2,1/2],[1],16*x^2);
ogf := (Int(6*H*(4*x^2+5)/(4-x^2)^(3/2),x)+H*(16*x^2-1)/(4-x^2)^(1/2))*((2-x)/(2+x))^(1/2)/(4*x)+1/(8*x);
series(ogf,x=0,20);  # Mark van Hoeij, May 06 2013

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^3/Binomial[n, Floor[n/2]],{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)

a(n)=if(n<0, 0, sum(k=0,n\2, (binomial(n,k)-binomial(n,k-1))^3)/binomial(n,n\2)) /* Michael Somos, Jun 02 2005 */

G.f.: hypergeometric expression with an antiderivative, see Maple program. - Mark van Hoeij, May 06 2013
Recurrence: 4*n*(n+1)^2*(196*n^3 - 819*n^2 + 530*n + 528)*a(n) = 2*n*(1372*n^4 - 3633*n^3 - 7455*n^2 + 21934*n - 8448)*a(n-1) + (12740*n^6 - 90867*n^5 + 195310*n^4 - 13277*n^3 - 452690*n^2 + 528384*n - 174960)*a(n-2) + 8*(n-2)*(686*n^4 - 3010*n^3 + 1176*n^2 + 6543*n - 4725)*a(n-3) - 16*(n-3)^2*(n-2)*(196*n^3 - 231*n^2 - 520*n + 435)*a(n-4). - Vaclav Kotesovec, Mar 06 2014
a(n) ~ 4^(n+2)/(9*Pi*n^2). - Vaclav Kotesovec, Mar 06 2014

A361887 a(n) = S(5,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 2, 33, 276, 4150, 65300, 1083425, 20965000, 399876876, 8461219032, 178642861782, 4010820554664, 90684123972156, 2130950905378152, 50560833176021025, 1231721051614138800, 30294218438009039800, 759645100717216142000, 19213764100954274616908, 493269287121905287769776

Offset: 0

Peter Bala, Mar 28 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(5,n)}. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

a(n) is the total number of 5-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Seiichi Manyama, Table of n, a(n) for n = 0..673
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).
Column k=5 of A357824.
Cf. A008315, A120730.

seq(add( ( binomial(n,k) - binomial(n,k-1) )^5, k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^5, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)

from math import comb
def A361887(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**5 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^5.
a(n) = Sum_{j=0..n} A120730(n,j)^5.
a(n) = A357824(n,5). (End)
a(n) ~ 2^(5*n + 19/2) / (125 * Pi^(5/2) * n^(9/2)). - Vaclav Kotesovec, Aug 27 2023

A361890 a(n) = S(7,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 2, 129, 2316, 94510, 4939220, 211106945, 14879165560, 828070125876, 61472962084968, 4223017425122958, 325536754765395096, 25399546083773839692, 2059386837863675003112, 173281152533121109073025, 14789443838781868027714800, 1307994690673355979749969800

Offset: 0

Peter Bala, Mar 30 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(7,n)}. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

a(n) is the total number of 7-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Seiichi Manyama, Table of n, a(n) for n = 0..483
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/ S(1,2*n-1) ).
Column k=7 of A357824.
Cf. A008315, A120730.

seq(add( ( binomial(n,k) - binomial(n,k-1) )^7, k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^7, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)

from math import comb
def A361890(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^7.
a(n) = Sum_{j=0..n} A120730(n,j)^7.
a(n) = A357824(n,7). (End)
a(n) ~ 3 * 2^(7*n + 27/2) / (2401 * Pi^(7/2) * n^(13/2)). - Vaclav Kotesovec, Aug 27 2023

A357824 Total number A(n,k) of k-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 5, 6, 3, 1, 1, 2, 9, 14, 10, 4, 1, 1, 2, 17, 36, 42, 20, 4, 1, 1, 2, 33, 98, 190, 132, 35, 5, 1, 1, 2, 65, 276, 882, 980, 429, 70, 5, 1, 1, 2, 129, 794, 4150, 7812, 5705, 1430, 126, 6, 1, 1, 2, 257, 2316, 19722, 65300, 78129, 33040, 4862, 252, 6

Offset: 0

Alois P. Heinz, Oct 14 2022

			Square array A(n,k) begins:
  1,  1,   1,    1,     1,       1,        1,         1, ...
  1,  1,   1,    1,     1,       1,        1,         1, ...
  2,  2,   2,    2,     2,       2,        2,         2, ...
  2,  3,   5,    9,    17,      33,       65,       129, ...
  3,  6,  14,   36,    98,     276,      794,      2316, ...
  3, 10,  42,  190,   882,    4150,    19722,     94510, ...
  4, 20, 132,  980,  7812,   65300,   562692,   4939220, ...
  4, 35, 429, 5705, 78129, 1083425, 15105729, 211106945, ...

Alois P. Heinz, Antidiagonals n = 0..121, flattened
Wikipedia, Counting lattice paths

Columns k=0-7 give A008619, A001405, A000108, A003161, A129123, A361887, A382433, A361890.
Rows n=1-5 give: A000012, A007395, A000051, A001550, A074511.
Main diagonal gives A357825.
Cf. A008315, A120730.

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:
A:= (n, k)-> add(b(n, n-2*j)^k, j=0..n/2):
seq(seq(A(n, d-n), n=0..d), d=0..12);

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}}]]];
A[n_, k_] := Sum[b[n, n - 2*j]^k, { j, 0, n/2}];
Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Oct 18 2022, after Alois P. Heinz *)

A(n,k) = Sum_{j=0..floor(n/2)} A008315(n,j)^k.

A(n,k) = Sum_{j=0..n} A120730(n,j)^k for k>=1, A(n,0) = A008619(n).

A361888 a(n) = S(5,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 1, 11, 46, 415, 3265, 30955, 299500, 3173626, 33576266, 386672861, 4340714886, 52846226091, 620906440961, 7857161332715, 95704821415240, 1246162831674580, 15624127945644100, 207990691516965886, 2669841775757784796, 36176886727828945286, 473508685502539872586

Offset: 0

Peter Bala, Mar 29 2023

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). The present sequence is {S(5,n)/S(1,n)}.

Seiichi Manyama, Table of n, a(n) for n = 0..840
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(n,k) - binomial(n,k-1) )^5/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^5/Binomial[n, Floor[n/2]], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
a(n) = s(5, n)/s(1, n); \\ Seiichi Manyama, Mar 24 2025

a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.

a(n) ~ 2^(4*n + 9) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025

A361889 a(n) = S(5,2n-1)/S(1,2n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 11, 415, 30955, 3173626, 386672861, 52846226091, 7857161332715, 1246162831674580, 207990691516965886, 36176886727828945286, 6510211391453319830461, 1205449991704260042021490, 228686327051301858363357905, 44299708036441260810228742915, 8738765548899621077157770551275

Offset: 1

Peter Bala, Mar 29 2023

Odd bisection of A361888.

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

			Examples of supercongruences:
a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).
a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)

Seiichi Manyama, Table of n, a(n) for n = 1..420
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^5/binomial(2*n-1,n-1), k = 0..n-1), n = 1..20);

Table[Sum[(Binomial[2*n-1, k]-Binomial[2*n-1, k-1])^5 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

from math import comb
def A361889(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**5 for j in range(n))//comb((n<<1)-1,n-1) # Chai Wah Wu, Mar 25 2025

a(n) = 1/binomial(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n - k) * binomial(2*n-1,k) )^5 for n >= 1.

a(n) ~ 2^(8*n + 1) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025

A361891 a(n) = S(7,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 1, 43, 386, 9451, 246961, 6031627, 212559508, 6571985126, 243940325734, 9140730357409, 352312505157354, 14801600281919487, 600054439936968241, 26927918031565051915, 1149140935414286560040, 53804800109969394477580, 2401141625752684697505820

Offset: 0

Peter Bala, Mar 30 2023

Seiichi Manyama, Table of n, a(n) for n = 0..563
H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

seq(add( ( binomial(n,k) - binomial(n,k-1) )^7/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^7/Binomial[n, Floor[n/2]], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
a(n) = s(7, n)/s(1, n); \\ Seiichi Manyama, Mar 24 2025

from math import comb
def A361891(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1))//comb(n,n>>1) # Chai Wah Wu, Mar 25 2025

a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.

a(n) ~ 3 * 2^(6*n+13) / (2401 * Pi^3 * n^6). - Vaclav Kotesovec, Mar 24 2025

Showing 1-10 of 13 results. Next

cp's OEIS Frontend

A053121 Catalan triangle (with 0's) read by rows.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Sage

Formula

Extensions

A120730 Another version of Catalan triangle A009766.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A129123 Number of 4-tuples of standard tableau with height less than or equal to 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

SageMath

Formula

A003162 A binomial coefficient summation.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A361887 a(n) = S(5,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A361890 a(n) = S(7,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

A361889 a(n) = S(5,2n-1)/S(1,2n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.