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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A002002 a(n) = Sum_{k=0..n-1} binomial(n,k+1) * binomial(n+k,k).

Original entry on oeis.org

0, 1, 5, 25, 129, 681, 3653, 19825, 108545, 598417, 3317445, 18474633, 103274625, 579168825, 3256957317, 18359266785, 103706427393, 586889743905, 3326741166725, 18885056428537, 107347191941249, 610916200215241
Offset: 0

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Keywords

Comments

From Benoit Cloitre, Jan 29 2002: (Start)
Array interpretation (first row and column are the natural numbers):
1 2 3 ..j ... if b(i,j) = b(i-1,j) + b(i-1,j-1) + b(i,j-1) then a(n+1) = b(n,n)
2 5 .........
.............
i........... b(i,j)
(End)
Number of ordered trees with 2n edges, having root of even degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
Coefficient of x^n in ((1-x)/(1-2x))^n, n>0. - Michael Somos, Sep 24 2003
Number of peaks in all Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0). Example: a(2)=5 because HH, HU*D, U*DH, UHD, U*DU*D, UU*DD contain 5 peaks (indicated by *). - Emeric Deutsch, Dec 06 2003
a(n) is the total number of HHs in all Schroeder (n+1)-paths. Example: a(2)=5 because UH*HD, H*H*H, UDH*H, H*HUD contain 5 HHs (indicated by *) and the other 18 Schroeder 3-paths contain no HHs. - David Callan, Jul 03 2006
a(n) is the total number of Hs in all Schroeder n-paths. Example: a(2)=5 as the Schroeder 2-paths are HH, DUH, DHU, HDU, DUDU and DDUU, and there are 5 H's. In general, a(n) is the total number of H..Hs (m+1 H's) in all Schroeder (n+m)-paths. - FUNG Cheok Yin, Jun 19 2021
a(n) is the number of points in Z^(n+1) that are L1 (Manhattan) distance <= n from the origin, or the number of points in Z^n that are L1 distance <= n+1 from the origin. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the (n+1)-dimensional cubic lattice as well as the (n+1)-st term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 25 2022 [Edited by Peter Munn, Jan 05 2023]

Examples

			G.f. = x + 5*x^2 + 25*x^3 + 129*x^4 + 681*x^5 + 3653*x^6 + 19825*x^7 + 108545*x^8 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisection of A002003, Cf. A047781, A001003.
a(n)=T(n, n+1), array T as in A050143.
a(n)=T(n, n+1), array T as in A064861.
Half the first differences of central Delannoy numbers (A001850).
a(n)=T(n, n+1), array T as in A008288.

Programs

  • Magma
    [&+[Binomial(n,k+1)*Binomial(n+k,k): k in [0..n]]: n in [0..21]];  // Bruno Berselli, May 19 2011
    
  • Maple
    A064861 := proc(n,k) option remember; if n = 1 then 1; elif k = 0 then 0; else A064861(n,k-1)+(3/2-1/2*(-1)^(n+k))*A064861(n-1,k); fi; end; seq(A064861(i,i+1),i=1..40);
  • Mathematica
    CoefficientList[Series[((1-x)/Sqrt[1-6x+x^2]-1)/2, {x,0,30}],x]  (* Harvey P. Dale, Mar 17 2011 *)
    a[ n_] := n Hypergeometric2F1[ n + 1, -n + 1, 2, -1] (* Michael Somos, Aug 09 2011 *)
    a[ n_] := With[{m = Abs@n}, Sign[n] Sum[ Binomial[ m, k] Binomial[ m + k - 1, m], {k, m}]]; (* Michael Somos, Aug 09 2011 *)
  • Maxima
    makelist(sum(binomial(n,k+1)*binomial(n+k,k), k, 0, n), n, 0, 21); /* Bruno Berselli, May 19 2011 */
    
  • PARI
    {a(n) = my(m = abs(n)); sign( n) * sum( k=0, m-1, binomial( m, k+1) * binomial( m+k, k))}; /* Michael Somos, Aug 09 2011 */
    
  • PARI
    /* L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1-x)^(-n)/n! */
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)/(1-x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, May 17 2015
  • Sage
    a = lambda n: hypergeometric([1-n, -n], [1], 2) if n>0 else 0
    [simplify(a(n)) for n in range(22)] # Peter Luschny, Nov 19 2014
    

Formula

G.f.: ((1-x)/sqrt(1-6*x+x^2)-1)/2. - Emeric Deutsch, Aug 02 2002
E.g.f.: exp(3*x)*(BesselI(0, 2*sqrt(2)*x)+sqrt(2)*BesselI(1, 2*sqrt(2)*x)). - Vladeta Jovovic, Mar 28 2004
a(n) = Sum_{k=0..n-1} binomial(n-1, k)*binomial(n+k, k+1). - Paul Barry, Sep 20 2004
a(n) = n * hypergeom([n + 1, -n + 1], [2], -1) = ((n+1)*LegendreP(n+1,3) - (5*n+3)*LegendreP(n,3))/(2*n) for n > 0. - Mark van Hoeij, Jul 12 2010
G.f.: x*d/dx log(1/(1-x*A006318(x))). - Vladimir Kruchinin, Apr 19 2011
a(n) = -a(-n) for all n in Z. - Michael Somos, Aug 09 2011
G.f.: -1 + 1 / ( 1 - x / (1 - 4*x / (1 - x^2 / (1 - 4*x / (1 - x^2 / (1 - 4*x / ...)))))). - Michael Somos, Jan 03 2013
a(n) = Sum_{k=0..n} A201701(n,k)^2 = Sum_{k=0..n} A124182(n,k)^2 for n > 0. - Philippe Deléham, Dec 05 2011
D-finite with recurrence: 2*(6*n^2-12*n+5)*a(n-1)-(n-2)*(2*n-1)*a(n-2)-n*(2*n-3)*a(n)=0. - Vaclav Kotesovec, Oct 04 2012
a(n) ~ (3+2*sqrt(2))^n/(2^(5/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 04 2012
D-finite (an alternative): n*a(n) = (6-n)*a(n-6) + (14*n-72)*a(n-5) + (264-63*n)*a(n-4) + 100*(n-3)*a(n-3) + (114-63*n)*a(n-2) + 2*(7*n-6)*a(n-1), n >= 7. - Fung Lam, Feb 05 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k*binomial(n-1,k)*binomial(n+k,k) and n^3*a(n) = Sum_{k=0..n-1} (4*k^3+4*k^2+4*k+1)*binomial(n-1,k)*binomial(n+k,k). For each of the two equalities, both sides satisfy the same recurrence -- this follows from the Zeilberger algorithm. - Zhi-Wei Sun, Aug 30 2014
a(n) = hypergeom([1-n, -n], [1], 2) for n >= 1. - Peter Luschny, Nov 19 2014
Logarithmic derivative of A001003 (little Schroeder numbers). - Paul D. Hanna, May 17 2015
L.g.f.: L(x) = Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1) * (1-x)^(-n) / n! = Sum_{n>=1} a(n)*x^n/n where exp(L(x)) = g.f. of A001003. - Paul D. Hanna, May 17 2015
a(n+1) = (1/2^(n+1)) * Sum_{k >= 0} (1/2^k) * binomial(n + k, n)*binomial(n + k, n + 1). - Peter Bala, Mar 02 2017
2*a(n) = A110170(n), n > 0. - R. J. Mathar, Feb 10 2022
a(n) = (LegendreP(n,3) - LegendreP(n-1,3))/2. - Mark van Hoeij, Jul 14 2022
D-finite with recurrence n*a(n) +(-7*n+5)*a(n-1) +(7*n-16)*a(n-2) +(-n+3)*a(n-3)=0. - R. J. Mathar, Aug 01 2022
From Peter Bala, Nov 08 2022: (Start)
a(n) = (-1)^(n+1)*hypergeom( [n+1, -n+1], [1], 2) for n >= 1.
The Gauss congruences hold: a(n*p^r) == a(n^p^(r-1)) (mod p^r) for all primes p and all positive integers n and r. (End)
From Peter Bala, Apr 18 2024: (Start)
G.f.: Sum_{n >= 1} binomial(2*n-1, n)*x^n/(1 - x)^(2*n) = x + 5*x^2 + 25*x^3 + 129*x^4 + ....
Row sums of A253283. (End)

Extensions

More terms from Clark Kimberling

A049600 Array T read by diagonals; T(i,j) is the number of paths from (0,0) to (i,j) consisting of nonvertical segments (x(k),y(k))-to-(x(k+1),y(k+1)) such that 0 = x(1) < x(2) < ... < x(n-1) < x(n)=i, 0 = y(1) <= y(2) <= ... <= y(n-1) <= y(n)=j, for i >= 0, j >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 4, 0, 1, 4, 8, 8, 0, 1, 5, 13, 20, 16, 0, 1, 6, 19, 38, 48, 32, 0, 1, 7, 26, 63, 104, 112, 64, 0, 1, 8, 34, 96, 192, 272, 256, 128, 0, 1, 9, 43, 138, 321, 552, 688, 576, 256, 0, 1, 10, 53, 190, 501, 1002, 1520, 1696, 1280, 512
Offset: 0

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Keywords

Comments

Essentially array A059576 divided by sequence A011782.
[Hetyei] calls a variant of this array (omitting the initial row of zeros) the asymmetric Delannoy numbers and shows how they arise in certain lattice path enumeration problems and a face enumeration problem associated to Jacobi polynomials. - Peter Bala, Oct 29 2008
Essentially triangle in A208341. - Philippe Deléham, Mar 23 2012
T(n+k,n) is the dot product of a vector from the n-th row of Pascal's triangle A007318 with a vector created by the first n+1 values evaluated from the cycle index of symmetry group S(k). Example: T(4+3,4) = T(7,4) = {1,4,6,4,1}.{1,4,10,20,35} = 192. - Richard Turk, Sep 21 2017
The formula T(n,k) = Sum_{r=0..n-1} C(k+r,r)*C(n-1,r) (Paul D. Hanna, Oct 06 2006) counts the paths of the title by number, r, of interior vertices in the path. - David Callan, Nov 25 2021

Examples

			Diagonals (each starting on row 1): {0}; {0,1}; {0,1,2}; ...
Array begins:
    0     0     0     0     0     0     0     0     0     0     0 ...
    1     1     1     1     1     1     1     1     1     1     1 ...
    2     3     4     5     6     7     8     9    10    11    12 ...
    4     8    13    19    26    34    43    53    64    76    89 ...
    8    20    38    63    96   138   190   253   328   416   518 ...
   16    48   104   192   321   501   743  1059  1462  1966  2586 ...
   32   112   272   552  1002  1683  2668  4043  5908  8378 11584 ...
   64   256   688  1520  2972  5336  8989 14407 22180 33028 47818 ...
Triangle begins:
  0;
  0, 1;
  0, 1, 2;
  0, 1, 3,  4;
  0, 1, 4,  8,  8;
  0, 1, 5, 13, 20,  16;
  0, 1, 6, 19, 38,  48,  32;
  0, 1, 7, 26, 63, 104, 112, 64;
  ...
(1, 0, -1/2, 1/2, 0, 0, 0, ...) DELTA (0, 2, 0, 0, 0, ...) where DELTA is the operator defined in A084938 begins:
  1;
  1, 0;
  1, 2,  0;
  1, 3,  4,  0;
  1, 4,  8,  8,   0;
  1, 5, 13, 20,  16,   0;
  1, 6, 19, 38,  48,  32,  0;
  1, 7, 26, 63, 104, 112, 64, 0;
		

Crossrefs

Diagonal sums are even-indexed Fibonacci numbers. Alternating (+-) diagonal sums are signed Fibonacci numbers.
T(n, n-1) = A001850(n) (Delannoy numbers). T(n, n)=A047781. Cf. A035028, A055587.

Programs

  • Haskell
    a049600 n k = a049600_tabl !! n !! k
    a049600_row n = a049600_tabl !! n
    a049600_tabl = [0] : map (0 :) a208341_tabl
    -- Reinhard Zumkeller, Apr 15 2014
  • Maple
    A049600 := proc(n,k)
        add(binomial(k+j,j)*binomial(n-1,j),j=0..n-1) ;
    end proc: # R. J. Mathar, Oct 26 2015
  • Mathematica
    t[n_, k_] := Hypergeometric2F1[ n-k+1, 1-k, 1, -1] // Floor; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2013 *)
    t[n_, k_] := Sum[LaguerreL[n-k, i, 0]* LaguerreL[k-i, i, 0], {i,0,k}] //Floor; Table[t[n,k], {n, 0, 16}, {k, -1, n}] (* Richard Turk, Sep 08 2017 *)
    T[n_, k_] := If[k == 0, 0, JacobiP[k - 1, 0, 1 - 2*k + n, 3]];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 25 2021 *)
  • PARI
    {A(i, j) = polcoeff( (x / (1 - 2*x)) * ((1 - x) / (1 - 2*x))^j + x * O(x^i), i)}; /* Michael Somos, Oct 01 2003 */
    
  • PARI
    T(n,k)=sum(j=0,n-1,binomial(k+j,j)*binomial(n-1,j)) \\ Paul D. Hanna, Oct 06 2006
    

Formula

T(n,k) = Sum_{j=0..n-1} C(k+j,j)*C(n-1,j). - Paul D. Hanna, Oct 06 2006
T(i,j) = 2*T(i-1,j) + T(i,j-1) - T(i-1,j-1) with T(0,0)=1 and T(i,j)=0 if one of i,j<0. - Theodore Kolokolnikov, Jul 05 2010
O.g.f.: t*x/(1 - (2*t+1)*x + t*x^2) = t*x + (t + 2*t^2)*x^2 + (t + 3*t^2 + 4*t^3)*x^3 + .... Taking the row reverse of this triangle (with an additional column of 1's) gives A055587. - Peter Bala, Sep 10 2012
T(i,0) = 2^(i-1) and for j>0, T(i,j) = T(i,j-1) + Sum_{k=0..i-1} T(k,j). - Glen Whitney, Aug 17 2021
T(n, k) = JacobiP(k - 1, 0, 1 - 2*k + n, 3) for k >= 1. - Peter Luschny, Nov 25 2021

A002003 a(n) = 2 * Sum_{k=0..n-1} binomial(n-1, k)*binomial(n+k, k).

Original entry on oeis.org

0, 2, 8, 38, 192, 1002, 5336, 28814, 157184, 864146, 4780008, 26572086, 148321344, 830764794, 4666890936, 26283115038, 148348809216, 838944980514, 4752575891144, 26964373486406, 153196621856192, 871460014012682, 4962895187697048, 28292329581548718
Offset: 0

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Keywords

Comments

a(n) is the number of order-preserving partial self maps of {1,...,n}. For example, a(2) = 8 because there are 8 order-preserving partial self maps of {1,2}: (1 2), (1 1), (2 2), (1 -), (2 -), (- 1), (- 2), (- -). Here for example (2 -) represents the partial map which maps 1 to 2 but does not include 2 in its domain. - James East, Oct 25 2005
From Peter Bala, Mar 02 2020: (Start)
For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(2*m,n)*b(n+1) + P(2*m,-n)*b(n-1) = Q(2*m,n)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial Q(2*m,n) is an even function of n; its 2*m zeros seem to belong to the interval [-1, 1] and 2*m - 2 of these zeros appear to lie close to the rational numbers of the form +-(2*k + 1)/(2*m), where 0 <= k <= m - 2. Cf. A103885. (End)
a(n), n>0, is the number of points at L1 distance = n from any given point in Z^n. The sequence is also the difference between the central diagonal (A001850) and +-1 diagonal (A002002) of the Delannoy number triangle (A008288). - Shel Kaphan, Feb 15 2023

Examples

			G.f. = 2*x + 8*x^2 + 38*x^3 + 192*x^4 + 1002*x^5 + 5336*x^6 + 28814*x^7 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A064861 := proc(n,k) option remember; if n = 1 then 1; elif k = 0 then 0; else A064861(n,k-1)+(3/2-1/2*(-1)^(n+k))*A064861(n-1,k); fi; end; seq(A064861(i,i-1),i=1..40);
  • Mathematica
    Flatten[{0,Table[SeriesCoefficient[((1+x)/Sqrt[1-6*x+x^2]-1)/2,{x,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Oct 04 2012 *)
    a[ n_] := If[ n < 1, 0, Hypergeometric2F1[ n, -n, 1, -1]]; (* Michael Somos, Aug 24 2014 *)
    Table[2*Sum[Binomial[n-1,k]Binomial[n+k,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Sep 18 2024 *)
  • PARI
    {a(n) = if( n<1, 0, polcoeff( ((1 - x^2) / (1 - x)^2 + x * O(x^n))^n, n))} /* Michael Somos, Sep 24 2003 */
    
  • Python
    from math import comb
    def A002003(n): return sum(comb(n,k)**2*k<Chai Wah Wu, Mar 22 2023

Formula

a(n) = 2*A047781(n).
From Vladeta Jovovic, Mar 28 2004: (Start)
G.f.: ((1+x)/sqrt(1-6*x+x^2)-1)/2.
E.g.f.: exp(3*x)*(2*BesselI(0, 2*sqrt(2)*x)+sqrt(2)*BesselI(1, 2*sqrt(2)*x)). (End)
a(n) = T(n, n-1), array T as in A064861.
a(n) = T(n, n-2), array T as in A049600.
a(n+1) = A110110(2n+1). - Tilman Neumann, Feb 05 2009
a(n) = 2 * JacobiP(n-1,0,1,3) = ((7*n+3)*LegendreP(n,3) - (n+1)*LegendreP(n+1,3)) /(2*n) for n > 0. - Mark van Hoeij, Jul 12 2010
Logarithmic derivative of A006318, the large Schroeder numbers. - Paul D. Hanna, Oct 25 2010
D-finite with recurrence: 4*(3*n^2-6*n+2)*a(n-1) - (n-2)*(2*n-1)*a(n-2) - n*(2*n-3)*a(n)=0. - Vaclav Kotesovec, Oct 04 2012
a(n) ~ (3+2*sqrt(2))^n/(2^(3/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 04 2012
Recurrence (an alternative): n*a(n) = (6-n)*a(n-6) + 2*(5*n-27)*a(n-5) + (84-15*n)*a(n-4) + 52*(3-n)*a(n-3) + 3*(2-5*n)*a(n-2) + 2*(5*n-3)*a(n-1), n>=7. - Fung Lam, Feb 05 2014
a(n) = Hyper2F1([-n, n], [1], -1) for n > 0. - Peter Luschny, Aug 02 2014
a(n) = [x^n] ((1+x)/(1-x))^n for n > 0. - Seiichi Manyama, Jun 07 2018
From Peter Bala, Mar 13 2020: (Start)
a(n) = 2 * Sum_{k = 0..n-1} 2^k*C(n,k+1)*C(n-1,k).
a(n) = 2 * (-1)^(n+1) * Sum_{k = 0..n-1} (-2)^k*C(n+k,n-1)*C(n-1,k).
a(n) = Sum_{k = 0..n} C(n,k)*C(2*n-k-1,n-1).
Conjecture: a(n) = - [x^n] (1 - F(x))^n, where F(x) = 2*x + 6*x^2 + 34*x^3 + 238*x^4 + ... is the o.g.f. of A108424. Equivalently, a(n) = -[x^n](G(x))^(-n), where G(x) = 1 + 2*x + 10*x^2 + 66*x^3 + 498*x^4 + ... is the o.g.f. of A027307.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
a(n) = Sum_{k = 1..n} C(n, k) * C(n-1, k-1) * 2^k. - Michael Somos, May 23 2021
a(n) = A001850(n) - A002002(n), for n > 0. - Shel Kaphan, Feb 15 2023

Extensions

More terms from Barbara Haas Margolius (b.margolius(AT)csuohio.edu), Oct 10 2001

A142978 Table of figurate numbers for the n-dimensional cross polytopes.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10
Offset: 1

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Author

Peter Bala, Jul 15 2008

Keywords

Comments

The n-th row entries for this array are the regular polytope numbers for the n-dimensional cross polytope as defined by [Kim]. The rows are the partial sums of the rows of the square array of Delannoy numbers A008288.
The odd numbered rows of this array form A142977. For a triangular version of this table see A104698. Cf. also A101603.
The n-th row of the array is the binomial transform of n-th row of triangle A081277, followed by zeros. Example: row 4 (1, 6, 19, 44, 85, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 17 2008
The main diagonal of the array T(n,k) is A047781 Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600. The link from A099193 to J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000, includes all n-D Hyperoctahedron (n-Cross Polytope) Numbers through 10-Cross(20) = 1669752016. - Jonathan Vos Post, Jul 16 2008

Examples

			The square array A(n, k) begins:
  n\k| 1   2    3     4     5       6
  ---+-------------------------------
   1 | 1   2    3     4      5      6    A000027
   2 | 1   4    9    16     25     36    A000290
   3 | 1   6   19    44     85    146    A005900
   4 | 1   8   33    96    225    456    A014820
   5 | 1  10   51   180    501   1182    A069038
   6 | 1  12   73   304    985   2668    A069039
   7 | 1  14   99   476   1765   5418    A099193
		

Crossrefs

Cf. A008288 (Delannoy numbers), A005900 (row 3), A014820 (row 4), A069038 (row 5), A069039 (row 6), A099193 (row 7), A099195 (row 8), A099196 (row 9), A099197 (row 10), A101603, A104698 (triangle version), A142977, A142983.

Programs

  • Haskell
    a142978 n k = a142978_tabl !! (n-1) !! (k-1)
    a142978_row n = a142978_tabl !! (n-1)
    a142978_tabl = map reverse a104698_tabl
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008
    A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2):
    seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023
  • Mathematica
    t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024

A108666 Number of (1,1)-steps in all Delannoy paths of length n.

Original entry on oeis.org

0, 1, 8, 57, 384, 2505, 16008, 100849, 628736, 3888657, 23900040, 146146473, 889928064, 5399971161, 32668236552, 197123362785, 1186790473728, 7131032334369, 42773183020296, 256161548120857, 1531966218561920, 9150330147133161, 54591847064667528, 325361790187810257
Offset: 0

Views

Author

Emeric Deutsch, Jul 07 2005

Keywords

Comments

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E =(1,0), N = (0,1) and D = (1,1).

Examples

			a(2)=8 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely, DD, DNE,DEN,NED,END,NDE,EDN,NENE,NEEN,ENNE,ENEN,NNEE and EENN, we have a total of eight D steps.
		

Crossrefs

a(n)/n = A047781(n) (for n >= 1).

Programs

  • Maple
    a := n -> add(k*binomial(n,k)*binomial(2*n-k,n),k=1..n): seq(a(n),n=0..24);
    # Alternative:
    a := n -> n^2*hypergeom([-n+1, -n+1], [2], 2):
    seq(simplify(a(n)), n=0..24); # Peter Luschny, Jan 20 2020
  • Mathematica
    CoefficientList[Series[x*(1-x)/(1-6*x+x^2)^(3/2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 18 2012 *)
  • PARI
    for(n=0,25, print1(sum(k=0,n, k*binomial(n,k)*binomial(2*n-k,n)), ", ")) \\ G. C. Greubel, Jan 31 2017
    
  • Python
    from math import comb
    def A108666(n): return sum(comb(n,k)**2*k<Chai Wah Wu, Mar 22 2023

Formula

a(n) = Sum_{k=0..n} k*A104684(k).
a(n) = Sum_{k=1..n} k*binomial(n, k)*binomial(2*n-k, n).
G.f.: x*(1-x)/(1-6*x+x^2)^(3/2).
D-finite with recurrence (n-1)*(2*n-3)*a(n) = 4*(3*n^2-6*n+2)*a(n-1) - (n-1)*(2*n-1)*a(n-2). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ (3+2*sqrt(2))^n*sqrt(n)/(2^(7/4)*sqrt(Pi)). - Vaclav Kotesovec, Oct 18 2012
a(n) = n^2*hypergeom([-n+1, -n+1], [2], 2). - Peter Luschny, Jan 20 2020
a(n) = Sum_{k=1..n} 2^(k-1)*k*binomial(n,k)^2. - Ridouane Oudra, Jun 15 2025

A113139 Number triangle, equal to half of Delannoy square array A008288.

Original entry on oeis.org

1, 3, 1, 13, 5, 1, 63, 25, 7, 1, 321, 129, 41, 9, 1, 1683, 681, 231, 61, 11, 1, 8989, 3653, 1289, 377, 85, 13, 1, 48639, 19825, 7183, 2241, 575, 113, 15, 1, 265729, 108545, 40081, 13073, 3649, 833, 145, 17, 1, 1462563, 598417, 224143, 75517, 22363, 5641
Offset: 0

Views

Author

Paul Barry, Oct 15 2005

Keywords

Comments

Row sums are A047781(n+1). Diagonal sums are A113140. Inverse is A113141.

Examples

			Triangle begins
     1;
     3,    1;
    13,    5,    1;
    63,   25,    7,   1;
   321,  129,   41,   9,  1;
  1683,  681,  231,  61, 11,  1;
  8989, 3653, 1289, 377, 85, 13, 1;
  ...
A113139 as a square array = A110171 * A008288:
  / 1   1   1   1 ... \   / 1         \ / 1 1  1  1 ...\
  | 3   5   7   9 ... |   | 2  1       || 1 3  5  7 ...|
  |13  25  41  61 ... | = | 8  4 1     || 1 5 13 25 ...|
  |63 129 231 377 ... |   |38 18 6 1   || 1 7 25 63 .. |
  |...                |   |...         || 1...         |
- _Peter Bala_, Dec 09 2015
		

Crossrefs

A001850 (column 0), A002002 (column 1), A026002 (column 2), A190666 (column 3), A047781 (row sums), A113140 (diagonal sums), A113141 (matrix inverse). Cf. A006318, A008288, A110171.

Programs

  • Maple
    T := (n,k) -> (-1)^(n-k)*hypergeom([n+1, -n+k], [1], 2):
    seq(seq(simplify(T(n,k)),k=0..n),n=0..8); # Peter Luschny, Mar 02 2017
  • Mathematica
    Table[Sum[Binomial[n - k, j] Binomial[n + j, k + j], {j, 0, n}], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 09 2015 *)

Formula

T(n, k) = Sum_{j=0..n} C(n-k, j)*C(n+j, k+j).
T(n, k) = Sum_{j=0..n} C(n, j)*C(n-k, j-k)*2^(n-j).
From Peter Bala, Dec 09 2015: (Start)
T(n,k) = A008288(n - k, n).
O.g.f.: 2/( sqrt(x^2 - 6*x + 1)*(t*sqrt(x^2 - 6*x + 1) + t*x - t + 2) ) = 1 + (3 + t)*x + (13 + 5*t + t^2)*x^2 + ....
Riordan array (f(x), x*g(x)), where f(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. for the central Delannoy numbers, A001850, and g(x) = 1/x* revert( x*(1 - x)/(1 + x) ) = 1 + 2*x + 6*x^2 + 22*x^3 + 90*x^4 + 394*x^5 + ... is the o.g.f. for the large Schroder numbers, A006318.
Read as a square array, this is the generalized Riordan array (f(x), g(x)) in the sense of the Bala link, which factorizes as (1 + x*g'(x)/g(x), x*g(x)) * (1/(1 - x), (1 + x)/(1 - x)) = A110171 * A008288. See the example below. (End)
T(n,k) = (-1)^(n-k)*hypergeom([n+1, -n+k], [1], 2). - Peter Luschny, Mar 02 2017
From Peter Bala, Feb 16 2020: (Start)
T(n,k) = P(n-k, k, 0, 3), where P(n, alpha, beta, x) is the n-th Jacobi polynomial with parameters alpha and beta.
T(n,k) = binomial(n,k) * hypergeom( [n + 1, k - n], [k + 1], -1 ).
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)^n/(1 - x)^(n+1) about 0. For example, for n = 4, (1 + x)^4/(1 - x)^5 = 1 + 9*x + 41*x^2 + 129*x^3 + 321*x^4 + O(x^5). Cf. A110171. (End)

A317057 a(n) is the number of time-dependent assembly trees satisfying the connected gluing rule for a cycle on n vertices.

Original entry on oeis.org

1, 1, 4, 23, 166, 1437, 14512, 167491, 2174746, 31374953, 497909380, 8619976719, 161667969646, 3265326093109, 70663046421208, 1631123626335707, 40004637435452866, 1038860856732399105, 28476428717448349996
Offset: 1

Views

Author

Keywords

Comments

A time-dependent assembly tree for a connected graph G = (V, E) on n vertices is a rooted tree, each node of which is labeled with a subset U of V and a nonnegative integer i such that:
1) each internal node has at least two children,
2) there are leaves labeled (v, 0) for each vertex v in V,
3) the label on the root is (V, m) for 1 <= m <= n-1,
4) for each node (U, i) with i < m, U is the union of the {u} for the children (u, 0) of (U, i),
5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i < i',
6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V.
A time-dependent assembly tree is said to satisfy the connected gluing rule if each internal vertex v of G, the graph induced by the vertices in the labels is connected.
Essentially the same as A053525. - R. J. Mathar, Aug 20 2018

Crossrefs

Programs

  • GAP
    a:=[1,1];; for n in [3..20] do a[n]:=1+Sum([2..n-1],j->Binomial(n,j)*a[j]); od; a; # Muniru A Asiru, Jul 25 2018
    
  • Maple
    A317057 := proc(n)
        option remember;
        if n <=2 then
            1;
        else
            1+add(binomial(n,j)*procname(j), j=2..n-1) ;
        end if;
    end proc:
    seq(A317057(n),n=1..30) ; # R. J. Mathar, Aug 08 2018
  • Mathematica
    Nest[Function[{a, n}, Append[a, 1 + Sum[Binomial[n, j] a[[j]], {j, 2, n - 1}]]][#, Length@ # + 1] &, {1, 1}, 17] (* Michael De Vlieger, Jul 26 2018 *)
  • PARI
    lista(nn) = my(v = vector(nn)); for (n=1, nn, if (n<=2, v[n] = 1, v[n] = 1 + sum(j=2, n-1, binomial(n, j)*v[j]))); v; \\ Michel Marcus, Aug 08 2018
  • Sage
    @cached_function
    def TimeDepenConCycle(n):
        if (n==1):
            return 1
        elif (n==2):
            return 1
        else:
            return sum([binomial(n, j)*TimeDepenConCycle(j) for j in range(2, n)])+1
    print(','.join(str(TimeDepenConCycle(i)) for i in range(1, 20)))
    

Formula

a(n) = 1 + Sum_{j = 2..n-1} binomial(n, j)*a(j), a(1) = a(2) = 1.
E.g.f.: (x - x*e^x + e^x - 1)/(2 - e^x).
a(n+1) = Sum_{k = 1..n} Stirling_2(n, k) * A142979(k). - Peter Bala, Dec 09 2024

A317059 a(n) is the number of time-dependent assembly trees satisfying the edge gluing rule for a complete graph on n vertices.

Original entry on oeis.org

1, 1, 3, 21, 255, 4815, 130095, 4763115, 226955925, 13646570175, 1010560060125, 90363456777825, 9599238270346725, 1194935000536101825, 172283712268118826375, 28481473075454845070625, 5351643310498951112521875, 1134140509146174954631081875, 269235074280949277622074328375
Offset: 1

Views

Author

Keywords

Comments

A time-dependent assembly tree for a connected graph G=(V, E) on n vertices is a rooted tree, each node of which is label a subset U of V and a nonnegative integer i such that:
1) each internal node has at least two children,
2) there are leaves labeled (v, 0) for each vertex v in V,
3) the label on the root is (V, m) for 1 <= m <= n-1,
4) for each node (U, i) with i
5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i
6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V.
A time-dependent assembly tree is said to satisfy the edge gluing rule if each internal vertex v of G has exactly two children and if U_1 and U_2 are the labels of the children of internal vertex v, then there is an edge (v_1,v_2) in the edge set of G such that v_1 is in U_1 and v_2 is in U_2.
a(n) is also the number of labeled histories possible for n leaves if simultaneous bifurcations are allowed. a(n) is also the number of single-elimination sports tournament schedules possible for n teams if matches involve pairs of teams, arbitrarily many arenas are available, and labeled teams have been specified, but the bracket of matches has not been specified. - Noah A Rosenberg, Feb 20 2025

Crossrefs

Programs

  • Mathematica
    Nest[Function[{a, n}, Append[a, Sum[(n!/((2^j) j! (n - 2 j)!)) a[[n - j]], {j, Floor[n/2]}]]][#, Length@ # + 1] &, {1, 1}, 17] (* Michael De Vlieger, Jul 26 2018 *)
  • PARI
    lista(nn) = my(v = vector(nn)); for (n=1, nn, if (n<=2, v[n] = 1, v[n] = sum(j=1, n\2, (n!/((2^j)*j!*(n-2*j)!))*v[n-j]))); v; \\ Michel Marcus, Aug 08 2018
  • Sage
    @cached_function
    def TimeDepenEdgeComp(n):
        if n==1:
            return 1
        elif n==2:
            return 1
        else:
            return sum((factorial(n)/((2^j)*factorial(j)*factorial(n-2*j)))*TimeDepenEdgeComp(n-j) for j in range(1, n//2+1))
    print(",".join(str(TimeDepenEdgeComp(i)) for i in range(1, 20)))
    

Formula

a(n) = Sum_{j=1..floor(n/2)}(n!/((2^j)j!(n-2j)!))*a(n-j), a(1)=a(2)=1.

A241023 Central terms of the triangle in A102413.

Original entry on oeis.org

1, 4, 16, 76, 384, 2004, 10672, 57628, 314368, 1728292, 9560016, 53144172, 296642688, 1661529588, 9333781872, 52566230076, 296697618432, 1677889961028, 9505151782288, 53928746972812, 306393243712384, 1742920028025364, 9925790375394096, 56584659163097436
Offset: 0

Author

Reinhard Zumkeller, Apr 15 2014

Keywords

Crossrefs

Programs

  • Haskell
    a241023 n = a102413 (2 * n) n
  • Mathematica
    a[0] = 1; a[n_] := 4 Hypergeometric2F1[1 - n, n + 1, 1, -1];
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jun 28 2019 *)

Formula

a(n) = 4 * A047781(n).
a(n) = A102413(2*n,n).
a(n) = 2*Hyper2F1([-n, n], [1], -1) for n>0. - Peter Luschny, Aug 02 2014
D-finite g.f. = (1+x)/sqrt(1-6*x+x^2), pairwise sums of A001850. - R. J. Mathar, Jan 15 2020
From Peter Bala, Apr 16 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*(2^k)*binomial(2*k, k)*binomial(n+k-1, n-k).
a(n) = (-1)^(n+1) * 4*n * hypergeom([n+1, -n+1], [2], 2).
n*(2*n - 3)*a(n) = 4*(3*n^2 - 6*n + 2)*a(n-1) - (2*n - 1)*(n - 2)*a(n-2) with a(0) = 1 and a(1) = 4.
O.g.f.: Sum_{n >= 0} (2^n)*binomial(2*n,n)*x^n/(1 + x)^(2*n) = 1 + 4*x + 16*x^2 + 76*x^3 + 384*x^4 + .... (End)
From Peter Bala, Sep 18 2024: (Start)
a(n) = [x^n] 1/S(-x)^(2*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318. Cf. A333481.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. (End)

A114656 Triangle read by rows: T(n,k) is the number of double rise-bicolored Dyck paths (double rises come in two colors; also called marked Dyck paths) of semilength n and having k peaks (1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 24, 12, 1, 16, 80, 80, 20, 1, 32, 240, 400, 200, 30, 1, 64, 672, 1680, 1400, 420, 42, 1, 128, 1792, 6272, 7840, 3920, 784, 56, 1, 256, 4608, 21504, 37632, 28224, 9408, 1344, 72, 1, 512, 11520, 69120, 161280, 169344, 84672, 20160, 2160, 90
Offset: 1

Author

Emeric Deutsch, Dec 23 2005

Keywords

Comments

Row sums are the little Schroeder numbers (A001003). Sum_{k=1..n} k*T(n,k) = A047781(n). T(n,k) = (1/2)A114655(n,k).
Triangle T(n,k), 1 <= k <= n, given by [0,2,0,2,0,2,0,2,0,2,0,2,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 02 2009

Examples

			T(3,2)=6 because we have (UD)Ub(UD)D, (UD)Ur(UD)D, Ub(UD)D(UD), Ur(UD)D(UD), Ub(UD)(UD)D and Ur(UD)(UD)D, where U=(1,1), D=(1,-1) and b (r) indicates a blue (red) double rise (the peaks are shown between parentheses).
Triangle begins:
   1;
   2,  1;
   4,  6,  1;
   8, 24, 12,  1;
  16, 80, 80, 20,  1;
  ....
Triangle T(n,k), 0 <= k <= n, given by [0,2,0,2,0,2,0,2,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] begins: 1; 0,1; 0,2,1; 0,4,6,1; 0,8,24,12,1; 0,16,80,80,20,1; ... - _Philippe Deléham_, Jan 02 2009
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->2^(n-k)*binomial(n,k)*binomial(n,k-1)/n: for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
  • Mathematica
    Table[2^(n - k) Binomial[n, k] Binomial[n, k - 1]/n, {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Apr 23 2019 *)

Formula

T(n, k) = 2^(n-k)*binomial(n, k)*binomial(n, k-1)/n.
G.f.: G = G(t, z) satisfies G = z(2G+t)(G+1).
T(n,k) = A001263(n,k)*2^(n-k). - Philippe Deléham, Apr 11 2007
G.f.: 1/(1-xy/(1-2x/(1-xy/(1-2x/(1-xy/(1-2x/(1-..... (continued fraction). - Paul Barry, Feb 06 2009
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