cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A026641 Number of nodes of even outdegree (including leaves) in all ordered trees with n edges.

Original entry on oeis.org

1, 1, 4, 13, 46, 166, 610, 2269, 8518, 32206, 122464, 467842, 1794196, 6903352, 26635774, 103020253, 399300166, 1550554582, 6031074184, 23493410758, 91638191236, 357874310212, 1399137067684, 5475504511858, 21447950506396
Offset: 0

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Author

Keywords

Comments

Number of lattice paths from (0,0) to (n,n) using steps (1,0),(0,2),(1,1). - Joerg Arndt, Jun 30 2011
From Emeric Deutsch, Jan 25 2004: (Start)
Let B = 1/sqrt(1-4*z) = g.f. for central binomial coeffs (A000984); F = (1-sqrt(1-4*z))/(z*(3-sqrt(1-4*z))) = g.f. for (A000957).
B = 1 + 2*z + 6*z^2 + 20*z^3 + ... gives the number of nodes in all ordered trees with 0,1,2,3,... edges. On p. 288 of the Deutsch-Shapiro paper one finds that z*B*F = z + 2*z^2 + 7*z^3 + 24*z^4 + ... gives the number of nodes of odd outdegree in all ordered trees with 1,2,3,... edges (cf. A014300).
Consequently, B - z*B*F = 2/(3*sqrt(1-4*z)-1+4*z) = 1 + z + 4*z^2 + 13*z^3 + 46*z^4 + ... gives the total number of nodes of even degree in all ordered trees with 0,1,2,3,4,... edges. (End)
Main diagonal of the following array: first column is filled with 1's, first row is filled alternatively with 1's or 0's: m(i,j) = m(i-1,j) + m(i,j-1): 1 0 1 0 1 ... / 1 1 2 2 3 ... / 1 2 4 6 9 ... / 1 3 7 13 22 ... / 1 4 11 24 46 ... - Benoit Cloitre, Aug 05 2002
The Hankel transform of [1,1,4,13,46,166,610,2269,...] is 3^n. - Philippe Deléham, Mar 08 2007
Second binomial transform of A127361. - Philippe Deléham, Mar 14 2007
Starting with offset 1, generated from iterates of M * [1,1,1,...]; where M = a tridiagonal matrix with (0,2,2,2,...) in the main diagonal and (1,1,1,...) in the super and subdiagonals. - Gary W. Adamson, Jan 04 2009
Equals left border of triangle A158815. - Gary W. Adamson, Mar 27 2009
Equals the INVERTi transform of A101850: (1, 2, 7, 26, 100, ...). - Gary W. Adamson, Jan 10 2012
Diagonal of rational function 1/(1 - (x + x*y + y^2)). - Gheorghe Coserea, Aug 06 2018
Let A(i, j) denote the infinite array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i times to the function (-1)^(n+1) for n > 0. Then A(n, n) equals a(n-1) for all n > 0. - John M. Campbell, Jan 20 2019
These numbers have the same parity as the Catalan numbers A000108; that is, a(n) is odd if and only if n = 2^k - 1 for some nonnegative integer k. It appears that if a(n) is odd then a(n) == 1 (mod 4). - Peter Bala, Feb 07 2024
The number a(n)/(n+1) is the coefficient of x^(n+1) in log(1+(1-sqrt(1-4*x))/2), the generating series of the Sabinin operad. - F. Chapoton, Mar 14 2024

Examples

			From _Joerg Arndt_, Jul 01 2011: (Start)
The triangle of number of lattice paths from (0,0) to (n,k) using steps (1,0),(0,2),(1,1) begins
  1;
  1, 1;
  1, 2,  4;
  1, 3,  7, 13;
  1, 4, 11, 24,  46;
  1, 5, 16, 40,  86, 166;
  1, 6, 22, 62, 148, 314,  610;
  1, 7, 29, 91, 239, 553, 1163, 2269;
This sequence is the diagonal. (End)
G.f. = 1 + x + 4*x^2 + 13*x^3 + 46*x^4 + 166*x^5 + 610*x^6 + 2269*x^7 + ...
		

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), this sequence (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • GAP
    List([0..25],n->(-1)^n*Sum([0..n],k->(-1)^k*Binomial(n+k,k))); # Muniru A Asiru, Aug 06 2018
    
  • Magma
    [(-1)^n*(&+[(-1)^k*Binomial(n+k, k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 12 2019
    
  • Maple
    seq(add((binomial(k+n, n-k)*binomial(n-k, k)),k=0..floor(n/2)),n=0..30);
    # From Richard Choulet, Jan 22 2010: (Start)
    a:= n -> add(binomial(2*n-k, k)*binomial(k, n-k), k=floor(n/2)..n):
    a:= n -> `if`(n<2, 1, (3/(2))*binomial(2*n-1, n-1)-(1/2)*a(n-1)):
    a:= n -> (-1/2)^(n+2)+(2/3)*add(4^(n-k)*(binomial(2*k, k)*(1/(1-2*k))
            *(1-(-1/8)^(n-k+1))), k=0..n):
    a:= n -> (-1/2)^(n+2)+(3/4)*add(((-1/2)^(n-k))*(binomial(2*k, k)), k=0..n):
    seq(a(n), n=0..30); # (End)
    gf := log(1 + (1 - sqrt(1 - 4*x))/2) / x: ser := series(gf, x, 30):
    seq((n + 1)*coeff(ser, x, n), n = 0..24);  # Peter Luschny, Mar 16 2024
  • Mathematica
    f[n_]:= Sum[ Binomial[n+k, k]*Cos[Pi*(n+k)], {k, 0, n}]; Array[f, 25, 0] (* Robert G. Wilson v, Apr 02 2012 *)
    CoefficientList[Series[2/(3*Sqrt[1-4*x]-1+4*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
    a[ n_]:= SeriesCoefficient[ D[ Log[1+(1-Sqrt[1-4x])/2], x], {x, 0, n}]; (* Michael Somos, May 18 2015 *)
  • PARI
    a(n)=(-1)^n*sum(k=0,n,(-1)^k*binomial(n+k,k))
    
  • PARI
    /* same as in A092566 but use */
    steps=[[1,0], [0,2], [1,1]]; /* Joerg Arndt, Jun 30 2011 */
    
  • Sage
    [(-1)^n*sum((-1)^k*binomial(n+k, k) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Feb 12 2019

Formula

G.f. is logarithmic derivative of the generating function for the Catalan numbers A000108. So this sequence might be called the "log-Catalan" numbers. - Murray R. Bremner, Jan 25 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(k+n, n-k)*binomial(n-k, k). - Detlef Pauly (dettodet(AT)yahoo.de), Nov 15 2001
G.f.: 2/(3*sqrt(1-4*z)-1+4*z). - Emeric Deutsch, Jul 09 2002
a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{j=0..floor(n/2)} binomial(2*n-2*j-1, n-1). - Emeric Deutsch, Jan 28 2004
From Paul Barry, Dec 18 2004: (Start)
A Catalan transform of the Jacobsthal numbers A001045(n+1) under the mapping G(x)-> G(xc(x)), c(x) the g.f. of A000108. The inverse mapping is H(x)->H(x(1-x)).
a(n) = Sum_{k=0..n} (k/(2*n-k))*binomial(2*n-k, n-k)*A001045(k+1). (End)
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*binomial(k, n-k). - Paul Barry, Jul 25 2005
a(n) = Sum_{k=0..n-1} A126093(n,k). - Philippe Deléham, Mar 08 2007
a(n) = (-1/2)^(n+2) + (2/3)*Sum_{k=0..n} ( (4^n-k)*binomial(2*k,k)*(1/(1-2*k))*(1-(-1/8)^(n-k+1)) ). - Yalcin Aktar, Jul 06 2007
a(n) = (-1/2)^(n+2) + (3/4)*Sum_{k=0..n} (-1/2)^(n-k)*binomial(2*k,k). - Yalcin Aktar, Jul 06 2007
From Richard Choulet, Jan 22 2010: (Start)
a(n) = (3*binomial(2*n-1,n-1) - d(n-1))/2, where d(n) = Sum_{k=floor(n/2)..n} binomial(2*n-k, k)*binomial(k, n-k).
a(n) = a(n-1) + (3/2)*Sum_{k=2..n} (1/(2*k-1))*binomial(2*k,k)*a(n-k).
a(n) = (2/3)*binomial(2*n,n) + (2/9)*((-2)^n/n!)*Sum_{k>=0} ( Product_{p=0..n-1} (k-2*p) /3^k).
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n-k,n).
a(n) = ( Sum_{k=0..n} (1/2)^(n-k+1)*binomial(n+k,k) )^2*(-1/2)^(n+2). (End)
From Gary W. Adamson, Nov 22 2011: (Start)
a(n) is the upper left term of M^n, M = an infinite square production matrix as follows:
1, 3, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
...
Also, a(n+1) is the sum of top row terms of M^n; e.g. top row of M^3 = (13, 21, 9, 3), sum = 46 = a(4), a(3) = 13. (End)
D-finite with recurrence: 2n*a(n) + (4-7n)*a(n-1) + 2*(1-2n)*a(n-2) = 0. - R. J. Mathar, Dec 17 2011 [The recurrence is proved with the Wilf-Zeilberger (WZ) method applied to Sum_{k=0..floor(n/2)} binomial(k+n, n-k)*binomial(n-k, k). - T. Amdeberhan, Jul 23 2012]
a(n) = A035317(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jul 19 2012
a(n) ~ 2^(2*n+1) / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014
a(n) = binomial(2*n,n)*hypergeom([1, -n], [-2*n], -1). - Peter Luschny, May 22 2014
G.f. is the derivative of the logarithm of the g.f. for A120588. - Michael Somos, May 18 2015
a(n) = [x^n] 1/((1 - x^2)*(1 - x)^n). - Ilya Gutkovskiy, Oct 25 2017
From Peter Bala, Feb 25 2019: (Start)
a(n) = Sum_{k = 0..n} binomial(2*n + 1, n + k + 1)*(-2)^k.
a(n-1) = (1/2)*binomial(2*n,n)*( 1 - 2*(n-1)/(n+1) + 4*(n-1)*(n-2)/((n+1)*(n+2)) - 8*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...) = (1/2)*binomial(2*n,n)*hypergeom([1 - n, 1], [n + 1], 2). (End)
a(0)=1, a(1)=1, and a(n) = (2 - 1/n)*a(n-2) + (7/2 - 2/n)*a(n-1) for n > 1. - Reginald Robson, Nov 01 2022

A072547 Main diagonal of the array in which first column and row are filled alternatively with 1's or 0's and then T(i,j) = T(i-1,j) + T(i,j-1).

Original entry on oeis.org

1, 0, 2, 6, 22, 80, 296, 1106, 4166, 15792, 60172, 230252, 884236, 3406104, 13154948, 50922986, 197519942, 767502944, 2987013068, 11641557716, 45429853652, 177490745984, 694175171648, 2717578296116, 10648297329692, 41757352712480
Offset: 1

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Author

Benoit Cloitre, Aug 05 2002

Keywords

Comments

A Catalan transform of A078008 under the mapping g(x)->g(xc(x)). - Paul Barry, Nov 13 2004
Number of positive terms in expansion of (x_1 + x_2 + ... + x_{n-1} - x_n)^n. - Sergio Falcon, Feb 08 2007
Hankel transform is A088138(n+1). - Paul Barry, Feb 17 2009
Without the beginning "1", we obtain the first diagonal over the principal diagonal of the array notified by B. Cloitre in A026641 and used by R. Choulet in A172025, and from A172061 to A172066. - Richard Choulet, Jan 25 2010
Also central terms of triangles A108561 and A112465. - Reinhard Zumkeller, Jan 03 2014
With offset 0 and for p prime, the p-th term is divisible by p. - F. Chapoton, Dec 03 2021

Examples

			The array begins:
  1 0 1 0 1..
  0 0 1 1 2..
  1 1 2 3 5..
  0 1 3 6 11..
so sequence begins : 1, 0, 2, 6, ...
		

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Programs

  • Haskell
    a072547 n = a108561 (2 * (n - 1)) (n - 1)
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( x*(1 + Sqrt(1-4*x))/(Sqrt(1-4*x)*(3-Sqrt(1-4*x))) )); // G. C. Greubel, Feb 17 2019
    
  • Maple
    taylor( (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-1),z=0,42); for n from -1 to 40 do a(n):=sum('(-1)^(p)*binomial(2n-p+1,1+n-p)',p=0..n+1): od:seq(a(n),n=-1..40):od; # Richard Choulet, Jan 25 2010
  • Mathematica
    CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x]) /(2*x))^(-1), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
    a[n_] := Binomial[2 n - 2, n] Hypergeometric2F1[1, 2 - n, n + 1, 1/2] / 2 + (-2)^(1 - n); Table[a[n], {n, 1, 26}] (* Peter Luschny, Dec 03 2021 *)
  • PARI
    a(n) = (-1)^n*sum(k=0, n, binomial(-n, k));
    vector(100, n, a(n-1)) \\ Altug Alkan, Oct 02 2015
    
  • Sage
    a=(x*(1+sqrt(1-4*x))/(sqrt(1-4*x)*(3-sqrt(1-4*x)))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Feb 17 2019

Formula

If offset is 0, a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k). - Vladeta Jovovic, Feb 18 2003
G.f.: x*(1-x*C)/(1-2*x*C)/(1+x*C), where C = (1-sqrt(1-4*x))/(2*x) is g.f. for Catalan numbers (A000108). - Vladeta Jovovic, Feb 18 2003
a(n) = Sum_{j=0..floor((n-1)/2)} binomial(2*n-2*j-4, n-3). - Emeric Deutsch, Jan 28 2004
a(n) = A108561(2*(n-1),n-1). - Reinhard Zumkeller, Jun 10 2005
a(n) = (-1)^n*Sum_{k=0..n} binomial(-n,k) (offset 0). - Paul Barry, Feb 17 2009
Other form of the G.f: f(z) = (2/(3*sqrt(1-4*z) -1 +4*z))*((1 -sqrt(1-4*z))/(2*z))^(-1). - Richard Choulet, Jan 25 2010
D-finite with recurrence 2*(-n+1)*a(n) + (9*n-17)*a(n-1) + (-3*n+19)*a(n-2) + 2*(-2*n+7)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
From Peter Bala, Oct 01 2015: (Start)
a(n) = [x^n] ((1 - x)^2/(1 - 2*x))^n.
Exp( Sum_{n >= 1} a(n+1)*x^n/n ) = 1 + x^2 + 2*x^3 + 6*x^4 + 18*x^5 + ... is the o.g.f for A000957. (End)
a(n) = binomial(2*n-2, n)*hypergeom([1, 2-n], [n+1], 1/2) / 2 + (-2)^(1-n). - Peter Luschny, Dec 03 2021
a(n) = 2 * A014301(n-1) for n>=3. - Alois P. Heinz, Dec 27 2023

Extensions

Corrected and extended by Vladeta Jovovic, Feb 17 2003

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

Views

Author

N. J. A. Sloane, Jan 23 2001

Keywords

Comments

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

Examples

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33
		

Crossrefs

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001

Programs

  • Maple
    read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
  • PARI
    T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
    for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
    for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
  • Sage
    def A059260_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
    for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Extensions

Formula corrected by Philippe Deléham, Jan 11 2014

A035317 Pascal-like triangle associated with A000670.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 2, 1, 4, 7, 6, 3, 1, 5, 11, 13, 9, 3, 1, 6, 16, 24, 22, 12, 4, 1, 7, 22, 40, 46, 34, 16, 4, 1, 8, 29, 62, 86, 80, 50, 20, 5, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 1, 11, 56, 174, 367, 553, 610, 496, 295, 125
Offset: 0

Views

Author

Keywords

Comments

From Johannes W. Meijer, Jul 20 2011: (Start)
The triangle sums, see A180662 for their definitions, link this "Races with Ties" triangle with several sequences, see the crossrefs. Observe that the Kn4 sums lead to the golden rectangle numbers A001654 and that the Fi1 and Fi2 sums lead to the Jacobsthal sequence A001045.
The series expansion of G(x, y) = 1/((y*x-1)*(y*x+1)*((y+1)*x-1)) as function of x leads to this sequence, see the second Maple program. (End)
T(2n,k) = the number of hatted frog arrangements with k frogs on the 2xn grid. See the linked paper "Frogs, hats and common subsequences". - Chris Cox, Apr 12 2024

Examples

			Triangle begins:
  1;
  1,  1;
  1,  2,  2;
  1,  3,  4,   2;
  1,  4,  7,   6,   3;
  1,  5, 11,  13,   9,   3;
  1,  6, 16,  24,  22,  12,   4;
  1,  7, 22,  40,  46,  34,  16,   4;
  1,  8, 29,  62,  86,  80,  50,  20,  5;
  1,  9, 37,  91, 148, 166, 130,  70, 25,  5;
  1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6;
  ...
		

Crossrefs

Row sums are A000975, diagonal sums are A080239.
Central terms are A014300.
Similar to the triangles A059259, A080242, A108561, A112555.
Cf. A059260.
Triangle sums (see the comments): A000975 (Row1), A059841 (Row2), A080239 (Kn11), A052952 (Kn21), A129696 (Kn22), A001906 (Kn3), A001654 (Kn4), A001045 (Fi1, Fi2), A023435 (Ca2), Gi2 (A193146), A190525 (Ze2), A193147 (Ze3), A181532 (Ze4). - Johannes W. Meijer, Jul 20 2011
Cf. A181971.

Programs

  • Haskell
    a035317 n k = a035317_tabl !! n !! k
    a035317_row n = a035317_tabl !! n
    a035317_tabl = map snd $ iterate f (0, [1]) where
       f (i, row) = (1 - i, zipWith (+) ([0] ++ row) (row ++ [i]))
    -- Reinhard Zumkeller, Jul 09 2012
    
  • Maple
    A035317 := proc(n,k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(A035317(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
    A035317 := proc(n,k): coeff(coeftayl(1/((y*x-1)*(y*x+1)*((y+1)*x-1)), x=0, n), y, k) end: seq(seq(A035317(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
  • Mathematica
    t[n_, k_] := (-1)^k*(((-1)^k*(n+2)!*Hypergeometric2F1[1, n+3, k+2, -1])/((k+1)!*(n-k+1)!) + 2^(k-n-2)); Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Dec 14 2011, after Johannes W. Meijer *)
  • PARI
    {T(n,k)=if(n==k,(n+2)\2,if(k==0,1,if(n>k,T(n-1,k-1)+T(n-1,k))))}
    for(n=0,12,for(k=0,n,print1(T(n,k),","));print("")) \\ Paul D. Hanna, Jul 18 2012
    
  • Sage
    def A035317_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^k*prec(n+2, k) for k in (1..n)]
    for n in (1..11): print(A035317_row(n)) # Peter Luschny, Mar 16 2016

Formula

T(n,k) = Sum_{j=0..floor(n/2)} binomial(n-2j, k-2j). - Paul Barry, Feb 11 2003
From Johannes W. Meijer, Jul 20 2011: (Start)
T(n, k) = Sum_{i=0..k}((-1)^(i+k) * binomial(i+n-k+1,i)). (Mendelson)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = floor(n/2) + 1. (Mendelson)
Sum_{k = 0..n}((-1)^k * (n-k+1)^n * T(n, k)) = A000670(n). (Mendelson)
T(n, n-k) = A128176(n, k); T(n+k, n-k) = A158909(n, k); T(2*n-k, k) = A092879(n, k). (End)
T(2*n+1,n) = A014301(n+1); T(2*n+1,n+1) = A026641(n+1). - Reinhard Zumkeller, Jul 19 2012

Extensions

More terms from James Sellers

A119259 Central terms of the triangle in A119258.

Original entry on oeis.org

1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889
Offset: 0

Views

Author

Reinhard Zumkeller, May 11 2006

Keywords

Comments

The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 06 2022

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Haskell
    a119259 n = a119258 (2 * n) n  -- Reinhard Zumkeller, Aug 06 2014
    
  • Mathematica
    Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
  • Python
    from itertools import count, islice
    def A119259_gen(): # generator of terms
        yield from (1,3)
        a, c = 2, 1
        for n in count(1):
            yield (a<>1
    A119259_list = list(islice(A119259_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n,k). - Seiichi Manyama, Jul 31 2025

A172061 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=4.

Original entry on oeis.org

1, 5, 22, 91, 367, 1461, 5776, 22748, 89402, 350974, 1377174, 5403193, 21201211, 83211277, 326703424, 1283211208, 5042294926, 19822108582, 77958648604, 306739666198, 1207433301046, 4754874514690, 18732340230592, 73827134976216
Offset: 0

Views

Author

Richard Choulet, Jan 24 2010

Keywords

Comments

This sequence is the 4th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The MAPLE programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+4 that are 2 steps higher than the preceding peak. - David Scambler, Apr 22 2013

Examples

			a(4) = C(12,4)-C(11,3)+C(10,2)-C(9,1)+C(8,0)=55*9-55*3+45-9+1=367.
		

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • Magma
    k:=4; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 16 2019
    
  • Maple
    a:= n-> add((-1)^(p)*binomial(2*n+4-p, n-p), p=0..n):
    seq(a(n), n=0..30);
    # second Maple program:
    gf:= (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^4:
    a:= n-> coeff(series(gf, z, n+10), z, n):
    seq(a(n), n=0..30);
  • Mathematica
    CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^4, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)
  • PARI
    k=4; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 16 2019
    
  • Sage
    k=4; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 16 2019

Formula

G.f.: (2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k with k=4.
a(n) = Sum_{p=0..n} (-1)^(p)*binomial(2*n+k-p, n-p), with k=4.
a(n) ~ 2^(2*n+5)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
D-finite with recurrence: +2*(n+4)*a(n) +(-13*n-36)*a(n-1) +(15*n+16)*a(n-2) +(19*n+14)*a(n-3) +2*(2*n-1)*a(n-4)=0. - R. J. Mathar, Feb 21 2020

A091526 Coefficient of x^n in 1/((1+x)*(1-x)^(n-1)).

Original entry on oeis.org

1, -1, 1, 2, 9, 34, 130, 496, 1897, 7274, 27966, 107788, 416394, 1611908, 6251596, 24287212, 94499689, 368202778, 1436458486, 5610483532, 21936442894, 85852554748, 336300861436, 1318441228432, 5172792817834, 20309402206084
Offset: 0

Views

Author

Michael Somos, Jan 18 2004

Keywords

Comments

Number of positive terms in expansion of (x_1+x_2+...+x_{n-1}-x_n)^(n+1). - Sergio Falcon, Feb 08 2007
Without the beginning "1" and "-1", we obtain the second diagonal over the principal diagonal of the array notified by B. Cloitre in A026641 and used by R. Choulet in A172025, and from A172061 to A172066. - Richard Choulet, Jan 25 2010

Crossrefs

Cf. A172025, A172061-A172066. - Richard Choulet, Jan 25 2010
Cf. A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • Magma
    k:=-2; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 18 2019
    
  • Maple
    for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+2,2+n-p)',p=0..n+2): od:seq(a(n),n=0..40):od; taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-2),z=0,42); # Richard Choulet, Jan 25 2010
  • Mathematica
    Table[Sum[Binomial[n+i-2, i]*(-1)^(n-i),{i,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 19 2014 *)
    Table[(-1)^n 2^(1-n)+Binomial[-1+2 n,1+n] Hypergeometric2F1[1,2 n,2+n,-1],{n,0,20}] (* Vaclav Kotesovec, Apr 19 2014 *)
    With[{k = -2}, CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1 - Sqrt[1-4*x])/(2*x))^k, {x, 0, 30}], x]] (* G. C. Greubel, Feb 18 2019 *)
  • PARI
    a(n)=sum(i=0,n,binomial(n+i-2,i)*(-1)^(n-i));
    
  • PARI
    k=-2; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 18 2019
    
  • Sage
    k=-2; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019

Formula

From Richard Choulet, Jan 25 2010: (Start)
G.f: f such as: f(z)=(2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^(-2).
a(n) = Sum_{j=0..n+2} (-1)^j*binomial(2*n-j+2, 2+n-j). (End)
Recurrence: 2*n*(3*n-7)*a(n) = (21*n^2 - 61*n + 48)*a(n-1) + 2*(2*n-3)*(3*n-4)*a(n-2). - Vaclav Kotesovec, Apr 19 2014
a(n) ~ 2^(2*n-1)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014

A172025 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=3.

Original entry on oeis.org

1, 4, 16, 62, 239, 920, 3544, 13672, 52834, 204528, 793092, 3080226, 11980667, 46662704, 181971248, 710454896, 2776717742, 10863073784, 42537035408, 166704021596, 653827252022, 2566222449104, 10079023179536, 39611016586832
Offset: 0

Views

Author

Richard Choulet, Jan 23 2010

Keywords

Comments

This sequence is the third diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows:
1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7,
1, 2, 4, 6, 9, 12, 16, 20, 25, 30,
1, 3, 7, 13, 22, 34, 50, 70, 95.
The Maple programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+3 that are 1 step higher than the preceding peak. - David Scambler, Apr 22 2013

Examples

			a(4) = C(11,4) - C(10,3) + C(9,2) - C(8,1) + C(7,0) = 330 - 120 + 36 - 8 + 1 = 239.
		

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • Magma
    k:=3; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 16 2019
    
  • Maple
    a:= n-> add((-1)^(p)*binomial(2*n+3-p,n-p), p=0..n):
    seq(a(n), n=0..30);
    # second Maple program:
    gf:= (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^3:
    a:= n-> coeff(series(gf,z,n+10),z,n):
    seq(a(n), n=0..30);
  • Mathematica
    a[n_] := Binomial[2*n+3, n+3]*Hypergeometric2F1[1, -n, -3-2*n, -1]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Dec 17 2013 *)
  • PARI
    k=3; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 16 2019
    
  • Sage
    k=3; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 16 2019

Formula

G.f.: (2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k with k=3.
a(n) = Sum_{p=0..n} (-1)^(p)*binomial(2*n+k-p,n-p), with k=3.
a(n) ~ 2^(2*n+4)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+3)*a(n) + (-7*n^2 - 17*n - 8)*a(n-1) -2*(n+2)*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Feb 19 2016
a(n) = [x^n] 1/((1 - x^2)*(1 - x)^(n+3)). - Ilya Gutkovskiy, Oct 25 2017

A172062 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=5.

Original entry on oeis.org

1, 6, 29, 128, 541, 2232, 9076, 36568, 146446, 584082, 2322967, 9220544, 36548573, 144732176, 572756312, 2265577184, 8959034798, 35421613196, 140035644602, 553606049024, 2188652065586, 8653317051056, 34216118389384
Offset: 0

Views

Author

Richard Choulet, Jan 24 2010

Keywords

Comments

This sequence is the 5th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+5 that are 3 steps higher than the preceding peak. - David Scambler, Apr 22 2013
Apparently half the sum of all height differences between adjacent peaks in all Dyck paths of semilength n+3. - David Scambler, Apr 22 2013

Examples

			a(4) = C(13,4) - C(12,3) + C(11,2) - C(10,1) + C(9,0) = 13*11*5 - 20*11 + 55 - 10 + 1 = 541.
		

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • Magma
    k:=5; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019
    
  • Maple
    for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
    # 2nd program
    for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;
  • Mathematica
    CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^5, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)
  • PARI
    k=5; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019
    
  • Sage
    k=5; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

Formula

a(n) = Sum_{j=0..n} (-1)^j*binomial(2*n+k-j, n-j), with k=5.
a(n) ~ 2^(2*n+6)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+5)*(3*n+7)*a(n) - (n+3)*(21*n^2+79*n+80)*a(n-1) - 2*(3*n+10)*(2*n+3)*(n+2)*a(n-2) = 0. - R. J. Mathar, Feb 19 2016

A172063 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=6.

Original entry on oeis.org

1, 7, 37, 174, 771, 3300, 13820, 57044, 233108, 945793, 3817351, 15347362, 61520899, 246052888, 982365976, 3916739872, 15599504614, 62076995998, 246866382826, 981218764540, 3898442536366, 15483778158792, 61482966826992
Offset: 0

Views

Author

Richard Choulet, Jan 24 2010

Keywords

Comments

This sequence is the 6th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.
Apparently the number of peaks in all Dyck paths of semilength n+6 that are 4 steps higher than the preceding peak. - David Scambler, Apr 22 2013

Examples

			a(4) = C(14,4) - C(13,3) + C(12,2) - C(11,1) + C(10,0) = 7*13*11 - 26*11 + 66 - 11 + 1 = 771.
		

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

  • Magma
    k:=6; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019
    
  • Maple
    for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
    # 2nd program
    for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;
  • Mathematica
    CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^6, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)
  • PARI
    k=6; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019
    
  • Sage
    k=6; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

Formula

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=6.
a(n) ~ 2^(2*n+7)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+6)*(n+3)*a(n) -(7*n^3+59*n^2+166*n+160)*a(n-1) -2*(2*n+5)*(n+4)*(n+2)*a(n-2)=0. - R. J. Mathar, Feb 19 2016
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