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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A047781 a(n) = Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600.

Original entry on oeis.org

0, 1, 4, 19, 96, 501, 2668, 14407, 78592, 432073, 2390004, 13286043, 74160672, 415382397, 2333445468, 13141557519, 74174404608, 419472490257, 2376287945572, 13482186743203, 76598310928096, 435730007006341, 2481447593848524, 14146164790774359
Offset: 0

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Author

Keywords

Comments

Also main diagonal of array: m(i,1)=1, m(1,j)=j, m(i,j)=m(i,j-1)+m(i-1,j-1)+m(i-1,j): 1 2 3 4 ... / 1 4 9 16 ... / 1 6 19 44 ... / 1 8 33 96 ... /. - Benoit Cloitre, Aug 05 2002
This array is now listed as A142978, where some conjectural congruences for the present sequence are given. - Peter Bala, Nov 13 2008
Convolution of central Delannoy numbers A001850 and little Schroeder numbers A001003. Hankel transform is 2^C(n+1,2)*A007052(n). - Paul Barry, Oct 07 2009
Define a finite triangle T(r,c) with T(r,0) = binomial(n,r) for 0 <= r <= n and the other terms recursively with T(r,c) = T(r-1,c-1) + 2*T(r,c-1). The sum of the last terms in the rows is Sum_{r=0..n} T(r,r) = a(n+1). Example: For n=4 the triangle has the rows 1; 4 9; 6 16 41; 4 14 44 129; 1 6 26 96 321 having sum of last terms 1 + 9 + 41 + 129 + 321 = 501 = a(5). - J. M. Bergot, Feb 15 2013
a(n) = A049600(2*n,n), when A049600 is seen as a triangle read by rows. - Reinhard Zumkeller, Apr 15 2014
a(n-1) for n > 1 is the number of assembly trees with the connected gluing rule for cycle graphs with n vertices. - Nick Mayers, Aug 16 2018

Crossrefs

Cf. A002003. Column 1 of A296129.

Programs

  • Haskell
    a047781 n = a049600 (2 * n) n  -- Reinhard Zumkeller, Apr 15 2014
    
  • Magma
    [n eq 0 select 0 else &+[Binomial(n-1, k)*Binomial(n+k, k): k in [0..n-1]]: n in [0..22]];  // Bruno Berselli, May 19 2011
    
  • Maple
    a := proc(n) local k; add(binomial(n-1,k)*binomial(n+k,k),k=0..n-1); end;
  • Mathematica
    Table[SeriesCoefficient[x*((1+x)-Sqrt[1-6*x+x^2])/(4*x*Sqrt[1-6*x+x^2]),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 08 2012 *)
    a[n_] := Hypergeometric2F1[1-n, n+1, 1, -1]; a[0] = 0; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Feb 26 2013 *)
    a[n_] := Sum[ Binomial[n - 1, k] Binomial[n + k, k], {k, 0, n - 1}]; Array[a, 25] (* Robert G. Wilson v, Aug 08 2018 *)
  • Maxima
    makelist(if n=0 then 0 else sum(binomial(n-1, k)*binomial(n+k, k), k, 0, n-1), n, 0, 22); /* Bruno Berselli, May 19 2011 */
    
  • PARI
    A047781(n)=polcoeff((1+x)/sqrt(1+(O(x^n)-6)*x+x^2),n)\4  \\ M. F. Hasler, Oct 09 2012
    
  • Python
    from sympy import binomial
    def a(n):
        return sum(binomial(n - 1, k) * binomial(n + k, k) for k in range(n))
    print([a(n) for n in range(51)]) # Indranil Ghosh, Apr 18 2017
    
  • Python
    from math import comb
    def A047781(n): return sum(comb(n,k)**2*k<Chai Wah Wu, Mar 22 2023

Formula

D-finite with recurrence n*(2*n-3)*a(n) - (12*n^2-24*n+8)*a(n-1) + (2*n-1)*(n-2)*a(n-2) = 0. - Vladeta Jovovic, Aug 29 2004
a(n+1) = Sum_{k=0..n} binomial(n, k)*binomial(n+1, k+1)*2^k. - Paul Barry, Sep 20 2004
a(n) = Sum_{k=0..n} T(n, k), array T as in A008288.
If shifted one place left, the third binomial transform of A098660. - Paul Barry, Sep 20 2004
G.f.: ((1+x)/sqrt(1-6x+x^2)-1)/4. - Paul Barry, Sep 20 2004, simplified by M. F. Hasler, Oct 09 2012
E.g.f. for sequence shifted left: Sum_{n>=0} a(n+1)*x^n/n! = exp(3*x)*(BesselI(0, 2*sqrt(2)*x)+BesselI(1, 2*sqrt(2)*x)/sqrt(2)). - Paul Barry, Sep 20 2004
a(n) = Sum_{k=0..n-1} C(n,k)*C(n-1,k)*2^(n-k-1); a(n+1) = 2^n*Hypergeometric2F1(-n,-n-1;1;1/2). - Paul Barry, Feb 08 2011
a(n) ~ 2^(1/4)*(3+2*sqrt(2))^n/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 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) = A241023(n) / 4. - Reinhard Zumkeller, Apr 15 2014
a(n) = Hyper2F1([-n, n], [1], -1)/2 for n > 0. - Peter Luschny, Aug 02 2014
n^2*a(n) = Sum_{k=0..n-1} (2*k^2+2*k+1)*binomial(n-1,k)*binomial(n+k,k). By the Zeilberger algorithm, both sides of the equality satisfy the same recurrence. - Zhi-Wei Sun, Aug 30 2014
a(n) = [x^n] (1/2) * ((1+x)/(1-x))^n for n > 0. - Seiichi Manyama, Jun 07 2018

A142992 Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 19, 25, 7, 1, 1, 33, 85, 49, 9, 1, 1, 51, 225, 231, 81, 11, 1, 1, 73, 501, 833, 489, 121, 13, 1, 1, 99, 985, 2471, 2241, 891, 169, 15, 1, 1, 129, 1765, 6321, 8361, 4961, 1469, 225, 17, 1
Offset: 0

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Author

Peter Bala, Jul 18 2008

Keywords

Comments

The lattice C_n consists of all integer lattice points v = (x_1,...,x_n) in Z^n such that the sum x_1 + ... + x_n is even. Let ||v|| = 1/2 * Sum_{i = 1..n} |x_i|; this defines a norm on C_n. The k-th term of the crystal ball sequence of C_n gives the number of lattice points v in C_n with ||v|| <= k [Bacher et al.]. The case n = 2 is illustrated in the Example section below.
This array has a remarkable relationship with the constant log(2). The row, column and (conjecturally) the diagonal entries of the array occur in series acceleration formulas for log(2) (see the Formula section below for some examples).
See A103884 for the table of coordination sequences of the C_n lattices. For the crystal ball sequences for the A_n and D_n lattices see A108625 and A108553 respectively. For the crystal ball sequences for the product lattices A_1 x ... x A_1(n copies) and A_n x A_n see A008288 and A143007 respectively.

Examples

			The square array begins
n\k|0...1....2.....3.....4......5
=================================
.0.|1...1....1.....1.....1......1
.1.|1...3....5.....7.....9.....11
.2.|1...9...25....49....81....121 A016754
.3.|1..19...85...231...489....891 A063496
.4.|1..33..225...833..2241...4961 A142993
.5.|1..51..501..2471..8361..22363 A142994
...
Triangular array begins
n\k|0...1...2...3...4...5
=========================
.0.|1
.1.|1...1
.2.|1...3...1
.3.|1...9...5...1
.4.|1..19..25...7...1
.5.|1..33..85..49...9...1
Case n = 2: The C_2 lattice consists of all integer lattice points v = (x,y) in Z x Z such that x + y is even, equipped with the taxicab type norm ||v|| = 1/2 * (|x| + |y|). There are 8 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 16 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the C_2 lattice (row 2 of the table) begins 1, 1+8 = 9, 1+8+16 = 25, ... .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 . 2 . . . .
. . . 2 . 1 . 2 . . .
. . 2 . 1 . 1 . 2 . .
. 2 . 1 . 0 . 1 . 2 .
. . 2 . 1 . 1 . 2 . .
. . . 2 . 1 . 2 . . .
. . . . 2 . 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
		

Crossrefs

Programs

  • Maple
    with combinat: T := (n,k) -> add(binomial(2n,2i)*binomial(k+i,n),i = 0..n): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;
  • Mathematica
    t[n_, k_] := Sum[ Binomial[2*n, 2*i]*Binomial[k+i, n], {i, 0, n}]; Table[t[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
O.g.f. for row n: 1/(1-x)^(n+1) * Sum_{k = 0..n} C(2*n,2*k)*x^k = 1/(1-x) * T(n,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
O.g.f. for the array: 1/(1-x) * {(1-t) - x*(1+t)}/{(1-t)^2 - x*(1+t)^2} = (1+x+x^2+x^3+...) + (1+3*x+5*x^2+7*x^3+...)*t + (1+9*x+25*x^2+49*x^3+...)*t^2 + ... .
Row n of the array has the form [p_n(0),p_n(1),p_n(2),...], where the polynomial function p_n(x) = Sum_{k = 0..n} C(2*n,2*k)*C(x+k,n). The first few are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = (2*x+1)^2, p_3(x) = (2*x+1)*(8*x^2+8*x+3)/3 and p_4(x) = (2*x+1)^2*(4*x^2+4*x+3)/3.
Alternative expressions for p_n(x) include p_n(x) = Sum_{k = 0..n} 2^(2*k)*n/(n+k)*C(n+k,2*k)*C(x,k) and p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(2*x+1,k).
The polynomials p_n(x) satisfy the 3-term recurrence relation n*p_n(x) = 2*(2*x+1)*p_(n-1)(x)+(n-2)*p(n-2)(x) for n >= 2; their generating function is 1/2*((1+t)/(1-t))^(2*x+1) = 1/2 + (2*x+1)*t + (2*x+1)^2*t^2 + (2*x+1)*(8*x^2+8*x+3)/3*t^3 + ... . Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(2*x+1;b,c) at b = 0, c = -1. Compare with A142979.
The polynomial p_n(x) is the unique polynomial solution to the difference equation (2*x+1)*{f(x+1/2) - f(x-1/2)} = 2*n*f(x), normalized so that f(0) = 1. The function p_n(x) is also the unique polynomial solution to the difference equation (2*x+1)*{(x+1)*f(x+1) + x*f(x-1)} = ((2*x+1)^2 + 2*n^2)*f(x), normalized so that f(0) = 1.
The zeros of p_n(x) lie on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p.4 of [BUMP et al.]).
For n > 0, the entries in row n of the array occur in series acceleration formulas for log(2): 2*log(2) = 1 + (1/2 - 1/6 +...+(-1)^n/(n*(n-1))) + (-1)^(n+1)*Sum_{k >= 1} 1/(k*T(n,k-1)*T(n,k)). For example, the fourth row of the table (n = 3) gives 2*log(2) = 4/3 + 1/(1*1*19) + 1/(2*19*85) + 1/(3*85*231) + ... .
The corresponding result for column k is 2*log(2) = 1 + (1/(1*3) + 1/(2*3*5) +...+ 1/(k*(2*k-1)*(2k+1)) + (2*k+1)*Sum_{n >= 1} (-1)^(n+1)/(n*(n+1)*T(n,k)* T(n+1,k)).
For example, the third column of the table (k = 2) gives 2*log(2) = 41/30 + 5*(1/(1*2*5*25) - 1/(2*3*25*85) + 1/(3*4*85*225) - ... ).
For the main diagonal calculation suggests the result: 2*log(2) = 4/3 + Sum_{n >= 1} (-1)^(n+1)*(5*n+3)/(n*(n+1)*T(n,n)*T(n+1,n+1)).
Similar series acceleration formulas for log(2) come from the row, column and diagonal entries of the square array of Delannoy numbers, A008288 (which may viewed as the array of crystal ball sequences for the product lattices A_1 x...x A_1). For corresponding results for the constants zeta(2) and zeta(3) see A108625 and A143007 respectively.

A144097 The 4-Schroeder numbers: a(n) = number of lattice paths (Schroeder paths) from (0,0) to (3n,n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 3x.

Original entry on oeis.org

1, 2, 14, 134, 1482, 17818, 226214, 2984206, 40503890, 561957362, 7934063678, 113622696470, 1646501710362, 24098174350986, 355715715691350, 5289547733908510, 79163575684710818, 1191491384838325474
Offset: 0

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Author

Joachim Schroeder (schroderjd(AT)qwa.uovs.ac.za), Sep 10 2008

Keywords

Comments

a(n) is also the number of lattice path from (0,0) to (4n,0) with unit steps (1,3), (2,2) and (1,-1) staying weakly above the x-axis.
Also, the number of planar rooted trees with n non-leaf vertices such that each non-leaf vertex has either 3 or 4 children. - Cameron Marcott, Sep 18 2013
a(n) equals the number of ordered complete 4-ary trees with 3*n + 1 leaves, where the internal vertices come in two colors and such that each vertex and its rightmost child have different colors. See Drake, Example 1.6.9. - Peter Bala, Apr 30 2023

Examples

			a(2)=14, because
  01: NNNENNNE,
  02: NNDNNNE,
  03: NNNENND,
  04: NNDNND,
  05: NNNDNNE,
  06: NNNDND,
  07: NNNNENNE,
  08: NNNNEND,
  09: NNNNDNE,
  10: NNNNDD,
  11: NNNNNENE,
  12: NNNNNED,
  13: NNNNNDE,
  14: NNNNNNEE
are all the paths from (0,0) to (2,6) with steps N,E and D weakly above y=3x.
		

References

  • Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.

Crossrefs

Cf. A027307 (the case y=2x), A008288 (Delannoy numbers), A008412 (4-dimensional coordination numbers).
This appears to equal 2*A243675. - N. J. A. Sloane, Mar 28 2021
The sequences listed in Yang-Jiang's Table 1 appear to be A006318, A001003, A027307, A034015, A144097, A243675, A260332, A243676. - N. J. A. Sloane, Mar 28 2021

Programs

  • Maple
    Schr:=proc(n,m,l)(n-l*m+1)/m*sum(2^v*binomial(m,v)*binomial(n,v-1),v=1..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(2^v*binomial(m-1,v-1)*binomial(n+1,v),v=0..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/m*sum(binomial(m,v)*binomial(n+v,m-1),v=0..m) end proc; where n=3m and l=3, also
    Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(binomial(n+1,v)*binomial(m-1+v,n),v=0..n+1) end proc; where n=3m and l=3.
    # alternative Maple program:
    a:= proc(n) option remember; `if`(n<2, n+1,
          ((15610*n^5 -67123*n^4 +106824*n^3 -77633*n^2
           +25514*n-3000)*a(n-1) -(3*(n-2))*(3*n-4)*
           (3*n-5)*(35*n^2-28*n+5)*a(n-2)) / ((3*(3*n-1))
           *(3*n+1)*n*(35*n^2-98*n+68)))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, May 26 2015
  • Mathematica
    d[n_, k_] := Binomial[n+k, k] Hypergeometric2F1[-k, -n, -n-k, -1]; a[0] = 1; a[n_] = d[3n, n] - 3d[3n+1, n-1] - 2d[3n, n-1]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 25 2017 *)
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k) * binomial(3*n+k+1, n)/(3*n+k+1))} \\ Seiichi Manyama, Jul 25 2020
    
  • PARI
    {a(n) = if(n==0, 1, sum(k=1, n, 2^k*binomial(n, k) * binomial(3*n, k-1)/n))} \\ Seiichi Manyama, Jul 25 2020

Formula

G.f. A(z) satisfies A(z) = 1 + z(A(z)^3 + A(z)^4) a(n)= S_{3n+1}(n) - 3S_n(3n + 1), where S_a(b) are coordination numbers, i.e., the number of points in the a-dimensional cubic lattice Z^a having distance b in the L_1 norm.
Also a(n) = D(3n,n) - 3D(3n + 1,n-1) - 2D(3n,n-1), where D(a,b) are the Delannoy numbers, i.e., the number of paths with N, E and D steps from (0,0) to (a,b).
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*(35*n^2-98*n+68) *a(n) +(-15610*n^5+67123*n^4-106824*n^3+77633*n^2-25514*n+3000)*a(n-1) +3*(n-2) *(3*n-4) *(3*n-5) *(35*n^2-28*n+5) *a(n-2)=0. - R. J. Mathar, Sep 06 2016
From Seiichi Manyama, Jul 25 2020: (Start)
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(3*n+k+1, n)/(3*n+k+1).
a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(3*n,k-1) for n > 0. (End)
a(n) ~ sqrt(12160 + 3853*sqrt(10)) * 3^(3*n - 9/2) / (2*sqrt(5*Pi) * n^(3/2) * (223 - 70*sqrt(10))^(n - 1/2)). - Vaclav Kotesovec, Jul 31 2021
Series reversion of x*(1 - x^3)/(1 + x^3) = x + 2*x^4 + 14*x^7 + 134*x^10 + ... = Sum_{n >= 0} a(n)*x^(3*n+1). - Peter Bala, Apr 30 2023
From Peter Bala, Jun 16 2023: (Start)
The g.f. A(x) = 1 + 2*x + 14*x^2 + 134*x^3 + ... satisfies A(x)^3 = (1/x) * the series reversion of ((1 - x)/(1 + x))^3.
Define b(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n = (1/3) * [x^n] ((1 + x)/(1 - x))^(3*n) = A333715(n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ).
a(n) = 2*hypergeom([1 - n, -3*n], [2], 2) for n >= 1. (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(4*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 09 2023

A266213 Square array A(n,r), the number of neighbors at a sharp Manhattan distance r in a finite n-hypercube lattice, read by upwards antidiagonals; A(n,r) = Sum_{k=0..min(n,r)} binomial(r-1,k-1)*binomial(n,k)* 2^k.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 2, 0, 1, 6, 8, 2, 0, 1, 8, 18, 12, 2, 0, 1, 10, 32, 38, 16, 2, 0, 1, 12, 50, 88, 66, 20, 2, 0, 1, 14, 72, 170, 192, 102, 24, 2, 0, 1, 16, 98, 292, 450, 360, 146, 28, 2, 0, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 0
Offset: 0

Views

Author

Dmitry Zaitsev, Dec 24 2015

Keywords

Comments

In an n-dimensional hypercube lattice, the array A(n,r) gives the number of nodes situated at a Manhattan distance equal to r, counting the current node. When counting coordinate offsets for neighboring nodes, binomial(n,k) chooses k nonzero coordinates from n coordinates, binomial(r-1,k-1) partitions the number r as the sum of exactly k nonzero numbers, and 2^k counts combinations of signs for coordinate offsets; starting indexing from 0 adds 1, which counts the current node.
In cellular automata theory, the cell surrounding with Manhattan distance less than or equal to r is called the von Neumann neighborhood of radius r or the diamond-shaped neighborhood to distinguish it from other generalizations of the von Neumann neighborhood for radius r>1, for instance, as a neighborhood having a difference in the range from -r to r in exactly one coordinate (the "narrow" von Neumann neighborhood of radius r).
The square array of partial sums of A(n,r) on rows gives us the Delannoy numbers A008288, which correspond to the number of nodes in the diamond-shaped neighborhood of radius r. - Dmitry Zaitsev, Dec 24 2015
For n >= 2, the term A(n,r) gives the number of polyominoes of bounding box 2 x (r+n-1) of area (r + 2(n-1)). Let A'(n,k) be the table A(n,k) without the first two rows. The sum of the terms in the i-th anti-diagonal of A'(n,k) gives the i-th term of A034182. - Louis Marin, Dec 11 2024

Examples

			The array A(n, k) begins:
n \ k  0  1   2   3    4     5     6      7      8      9
---------------------------------------------------------
0:     1  0   0   0    0     0     0      0      0      0
1:     1  2   2   2    2     2     2      2      2      2
2:     1  4   8  12   16    20    24     28     32     36
3:     1  6  18  38   66   102   146    198    258    326
4:     1  8  32  88  192   360   608    952   1408   1992
5:     1 10  50 170  450  1002  1970   3530   5890   9290
6:     1 12  72 292  912  2364  5336  10836  20256  35436
7:     1 14  98 462 1666  4942 12642  28814  59906 115598
8:     1 16 128 688 2816  9424 27008  68464 157184 332688
9:     1 18 162 978 4482 16722 53154 148626 374274 864146
...
For instance, in a 5-hypercube lattice there are 170 nodes situated at a Manhattan distance of 3 for a chosen node.
The triangle T(m, r) begins:
m\r 0  1   2   3   4    5   6   7  8 9 10 ...
0:  1
1:  1  0
2:  1  2   0
3:  1  4   2   0
4:  1  6   8   2   0
5:  1  8  18  12   2    0
6:  1 10  32  38  16    2   0
7:  1 12  50  88  66   20   2   0
8:  1 14  72 170 192  102  24   2  0
9:  1 16  98 292 450  360 146  28  2 0
10: 1 18 128 462 912 1002 608 198 32 2  0
... Formatted by _Wolfdieter Lang_, Jan 31 2016
		

Crossrefs

Other versions: A035607, A113413, A119800, A122542.
Partial sums on rows of A give A008288.
Cf. A001333 (row sums of T). A057077 (alternating row sums of T). - Wolfdieter Lang, Jan 31 2016

Programs

  • Maple
    # Prints the array by rows.
    gf := n -> ((1 + x)/(1 - x))^n: ser := n -> series(gf(n), x, 40):
    seq(lprint(seq(coeff(ser(n), x, k), k=0..6)), n=0..9); # Peter Luschny, Mar 20 2020
  • Mathematica
    Table[Sum[Binomial[r - 1, k - 1] Binomial[n - r, k] 2^k, {k, 0, Min[n - r, r]}], {n, 0, 10}, {r, 0, n}] // Flatten (* Michael De Vlieger, Dec 24 2015 *)
  • Python
    from sympy import binomial
    def T(n, r):
        if r==0: return 1
        return sum(binomial(r - 1, k - 1) * binomial(n - r, k) * 2**k for k in range(min(n - r, r) + 1))
    for n in range(11): print([T(n, r) for r in range(n + 1)]) # Indranil Ghosh, May 23 2017

Formula

A(n, 0)=1, n>=0, A(0, r)=0, r>0.
A(n, r) = A(n, r-1) + A(n-1, r-1) + A(n-1, r).
A(n, r) = Sum_{k=0..min(n,r)} binomial(r-1,k-1)*binomial(n,k)*2^k.
Triangle T(m, r) = A(m-r, r), n >= 0, 0 <= r <= n, otherwise 0. - Wolfdieter Lang, Jan 31 2016
A(n, k) = [x^k] ((1 + x)/(1 - x))^n. - Ilya Gutkovskiy, May 23 2017

A086764 Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 9, 11, 7, 3, 1, 44, 53, 32, 13, 4, 1, 265, 309, 181, 71, 21, 5, 1, 1854, 2119, 1214, 465, 134, 31, 6, 1, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 1, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 02 2003

Keywords

Comments

The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

Examples

			Formatted as a square array:
      1      3     7    13   21   31  43 57 ... A002061;
      2     11    32    71  134  227 356    ... A094792;
      9     53   181   465 1001 1909        ... A094793;
     44    309  1214  3539 8544             ... A094794;
    265   2119  9403 30637                  ... A023043;
   1854  16687 82508                        ... A023044;
  14833 148329                              ... A023045;
Formatted as a triangular array (mirror of A076731):
       1;
       0      1;
       1      1     1;
       2      3     2     1;
       9     11     7     3    1;
      44     53    32    13    4    1;
     265    309   181    71   21    5    1;
    1854   2119  1214   465  134   31    6   1;
   14833  16687  9403  3539 1001  227   43   7   1;
  133496 148329 82508 30637 8544 1909  356  57   8   1;
		

Crossrefs

Programs

  • Magma
    A086764:= func< n,k | (&+[(-1)^j*Binomial(n-k,j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
    [A086764(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Mathematica
    T[n_,k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k,j]*(n-j)!,{j,0,n}];Flatten[Table[T[n,k],{n,0,11},{k,0,n}]] (* Indranil Ghosh, Feb 20 2017 *)
    T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n},{-n},-1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)
  • SageMath
    def A086764(n,k): return sum((-1)^j*binomial(n-k,j)*factorial(n-j) for j in range(n+1))//factorial(k)
    flatten([[A086764(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
k!*T(n, k) = A068106(n, k). [corrected by Georg Fischer, Aug 13 2022]
Sum_{k>=0} T(n, k) = A003470(n+1).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
From Peter Bala, Aug 14 2008: (Start)
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017

Extensions

More terms from David Wasserman, Mar 28 2005
Additional comments from Zerinvary Lajos, Mar 30 2006
Edited by N. J. A. Sloane, Sep 24 2011

A112478 Expansion of (1 + x + sqrt(1 + 6*x + x^2))/2.

Original entry on oeis.org

1, 2, -2, 6, -22, 90, -394, 1806, -8558, 41586, -206098, 1037718, -5293446, 27297738, -142078746, 745387038, -3937603038, 20927156706, -111818026018, 600318853926, -3236724317174, 17518619320890, -95149655201962, 518431875418926, -2832923350929742, 15521467648875090
Offset: 0

Views

Author

Paul Barry, Sep 07 2005

Keywords

Comments

This is the A-sequence for the Delannoy triangle A008288. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. O.g.f. A(y) = y/Finv(y) = 2*y/(-(1 + y) + sqrt(y^2 + 6*y + 1)) = ((1 + y) + sqrt(1 + 6*y + y^2))/2 with Finv the inverse function of F(x) = x*(1 + x)/(1 - x). The o.g.f. of the Z-sequence is 1.

Examples

			G.f. = 1 + 2*x - 2*x^2 + 6*x^3 - 22*x^4 + 90*x^5 - 394*x^6 + 1806*x^7 + ...
		

Crossrefs

A minor variation of A006318. See A085403 for yet another version.
Row sums of number triangle A112477.
Cf. A366325.

Programs

  • Mathematica
    CoefficientList[Series[(1+x+Sqrt[1+6*x+x^2])/2, {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
  • PARI
    {a(n) = polcoeff((1 + x + sqrt(1 + 6*x + x^2 + x*O(x^n)))/2, n)}; /* Michael Somos, Jul 07 2020 */

Formula

G.f.: (1 + x + sqrt(1 + 6*x + x^2))/2. - Sergei N. Gladkovskii, Jan 04 2012
G.F.: G(0) where G(k)= 1 + x + x/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Jan 04 2012
D-finite with recurrence: n*a(n) + 3*(2*n-3)*a(n-1) + (n-3)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012
a(n) ~ (-1)^(n+1) * sqrt(3*sqrt(2) - 4) * (3 + 2*sqrt(2))^n / (2 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 12 2014
0 = a(n)*(a(n+1) + 15*a(n+2) + 4*a(n+3)) + a(n+1)*(-3*a(n+1) + 34*a(n+3) + 15*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all integer n > 0. - Michael Somos, Jul 07 2020
From Seiichi Manyama, Oct 08 2023: (Start)
G.f. satisfies A(x) = 1 + x + x/A(x).
a(n) = (-1)^(n-1) * Sum_{k=0..n} binomial(2*k-1,k) * binomial(n+k-2,n-k)/(2*k-1). (End)

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511
Offset: 0

Views

Author

Clark Kimberling, Aug 02 2011

Keywords

Comments

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).
Since degree(D(p))
Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.
D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14
D(D(p))=2(x+1)+7(1)+14=2x+23
D(D(D(p)))=2(1)+23=25;
the Q-residue of p is 25.
We may regard the sequence Q of polynomials as the triangular array formed by coefficients:
t(0,0)
t(1,0)....t(1,1)
t(2,0)....t(2,1)....t(2,2)
t(3,0)....t(3,1)....t(3,2)....t(3,3)
and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.
Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:
Q.....P...................Q-residue of P
1.....1...................A000079, 2^n
1....(x+1)^n..............A007051, (1+3^n)/2
1....(x+2)^n..............A034478, (1+5^n)/2
1....(x+3)^n..............A034494, (1+7^n)/2
1....(2x+1)^n.............A007582
1....(3x+1)^n.............A081186
1....(2x+3)^n.............A081342
1....(3x+2)^n.............A081336
1.....A040310.............A193649
1....(x+1)^n+(x-1)^n)/2...A122983
1....(x+2)(x+1)^(n-1).....A057198
1....(1,2,3,4,...,n)......A002064
1....(1,1,2,3,4,...,n)....A048495
1....(n,n+1,...,2n).......A087323
1....(n+1,n+2,...,2n+1)...A099035
1....p(n,k)=(2^(n-k))*3^k.A085350
1....p(n,k)=(3^(n-k))*2^k.A090040
1....A008288 (Delannoy)...A193653
1....A054142..............A101265
1....cyclotomic...........A193650
1....(x+1)(x+2)...(x+n)...A193651
1....A114525..............A193662
More examples:
Q...........P.............Q-residue of P
(x+1)^n...(x+1)^n.........A000110, Bell numbers
(x+1)^n...(x+2)^n.........A126390
(x+2)^n...(x+1)^n.........A028361
(x+2)^n...(x+2)^n.........A126443
(x+1)^n.....1.............A005001
(x+2)^n.....1.............A193660
A094727.....1.............A193657
(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)
(k+1).....(x+1)^n.........A112091
(x+1)^n...(k+1)...........A029761
(k+1)......A049310........A193663
(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)
A051162...(x+1)^n.........A193658
A094727...(x+1)^n.........A193659
A049310...(x+1)^n.........A193664
Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

Examples

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.
		

Crossrefs

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

Programs

  • Mathematica
    q[n_, k_] := 1;
    r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
    f[n_, x_] := Fibonacci[n + 1, x];
    p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
    v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
    Table[v[n], {n, 0, 24}]    (* A193649 *)
    TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
    Table[r[k], {k, 0, 8}]  (* 2^k *)
    TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Formula

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

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

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

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).

Original entry on oeis.org

1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000
Offset: 1

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 1 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Programs

  • Maple
    p := n -> 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)
  • Mathematica
    RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] (* Harvey P. Dale, May 20 2012 *)

Formula

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = (2*x - (1 + x)*log(1 + x))/(1 - x)^2 satisfies the differential equation 1 + (x + 3)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0.
Sum_{k = 1..n} Stirling_2(n, k) * a(k) = A317057(n+1). (End)

A113413 A Riordan array of coordination sequences.

Original entry on oeis.org

1, 2, 1, 2, 4, 1, 2, 8, 6, 1, 2, 12, 18, 8, 1, 2, 16, 38, 32, 10, 1, 2, 20, 66, 88, 50, 12, 1, 2, 24, 102, 192, 170, 72, 14, 1, 2, 28, 146, 360, 450, 292, 98, 16, 1, 2, 32, 198, 608, 1002, 912, 462, 128, 18, 1, 2, 36, 258, 952, 1970, 2364, 1666, 688, 162, 20, 1, 2, 40, 326
Offset: 0

Author

Paul Barry, Oct 29 2005

Keywords

Comments

Columns include A040000, A008574, A005899, A008412, A008413, A008414. Row sums are A078057(n)=A001333(n+1). Diagonal sums are A001590(n+3). Reverse of A035607. Signed version is A080246. Inverse is A080245.
For another version see A122542. - Philippe Deléham, Oct 15 2006
T(n,k) is the number of length n words on alphabet {0,1,2} with no two consecutive 1's and no two consecutive 2's and having exactly k 0's. - Geoffrey Critzer, Jun 11 2015
From Eric W. Weisstein, Feb 17 2016: (Start)
Triangle of coefficients (from low to high degree) of x^-n * vertex cover polynomial of the n-ladder graph P_2 \square p_n:
Psi_{L_1}: x*(2 + x) -> {2, 1}
Psi_{L_2}: x^2*(2 + 4 x + x^2) -> {2, 4, 1}
Psi_{L_3}: x^3*(2 + 8 x + 6 x^2 + x^3) -> {2, 8, 6, 1}
(End)
Let c(n, k), n > 0, be multiplicative sequences for some fixed integer k >= 0 with c(p^e, k) = T(e+k, k) for prime p and e >= 0. Then we have Dirichlet g.f.: Sum_{n>0} c(n, k) / n^s = zeta(s)^(2*k+2) / zeta(2*s)^(k+1). Examples: For k = 0 see A034444 and for k = 1 see A322328. Dirichlet convolution of c(n, k) and lambda(n) is Dirichlet inverse of c(n, k). - Werner Schulte, Oct 31 2022

Examples

			Triangle begins
  1;
  2,  1;
  2,  4,  1;
  2,  8,  6,  1;
  2, 12, 18,  8,  1;
  2, 16, 38, 32, 10,  1;
  2, 20, 66, 88, 50, 12,  1;
		

Crossrefs

Other versions: A035607, A119800, A122542, A266213.

Programs

  • Mathematica
    nn = 10; Map[Select[#, # > 0 &] &, CoefficientList[Series[1/(1 - 2 x/(1 + x) - y x), {x, 0, nn}], {x, y}]] // Grid (* Geoffrey Critzer, Jun 11 2015 *)
    CoefficientList[CoefficientList[Series[1/(1 - 2 x/(1 + x) - y x), {x, 0, 10}, {y, 0, 10}], x], y] (* Eric W. Weisstein, Feb 17 2016 *)
  • Sage
    T = lambda n,k : binomial(n, k)*hypergeometric([-k-1, k-n], [-n], -1).simplify_hypergeometric()
    A113413 = lambda n,k : 1 if n==0 and k==0 else T(n, k)
    for n in (0..12): print([A113413(n,k) for k in (0..n)]) # Peter Luschny, Sep 17 2014 and Mar 16 2016
    
  • Sage
    # Alternatively:
    def A113413_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return prec(n-1,k-1)+2*sum(prec(n-i,k-1) for i in (2..n-k+1))
        return [prec(n, k) for k in (1..n)]
    for n in (1..10): print(A113413_row(n)) # Peter Luschny, Mar 16 2016

Formula

From Paul Barry, Nov 13 2005: (Start)
Riordan array ((1+x)/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{i=0..n-k} C(k+1, i)*C(n-i, k).
T(n, k) = Sum_{j=0..n-k} C(k+j, j)*C(k+1, n-k-j).
T(n, k) = D(n, k) + D(n-1, k) where D(n, k) = Sum_{j=0..n-k} C(n-k, j)*C(k, j)*2^j = A008288(n, k).
T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k-1).
T(n, k) = Sum_{j=0..n} C(floor((n+j)/2), k)*C(k, floor((n-j)/2)). (End)
T(n, k) = C(n, k)*hypergeometric([-k-1, k-n], [-n], -1). - Peter Luschny, Sep 17 2014
T(n, k) = (Sum_{i=2..k+2} A137513(k+2, i) * (n-k)^(i-2)) / (k!) for 0 <= k < n (conjectured). - Werner Schulte, Oct 31 2022
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