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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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

A099601 Quotient of de Bruijn sums S(4,n)/S(2,n).

Original entry on oeis.org

1, 7, 131, 3067, 79459, 2181257, 62165039, 1818812387, 54257991011, 1642977121597, 50344383988381, 1557608560147757, 48577698917598031, 1525245771206644117, 48165918788138198759, 1528611371067309862067
Offset: 0

Views

Author

Michael Somos, Oct 24 2004

Keywords

Comments

The de Bruijn sum S(s,n) = Sum_{k=0..2n} (-1)^(k+n) binomial(2n,k)^s.
Also the n-th term of the crystal ball sequence for the A_{2n} lattice.

Examples

			A003215(1)=7=a(1), A008384(2)=131=a(2), A008388(3)=3067=a(3), ...
		

References

  • G. E. Andrews, Application of SCRATCHPAD to Problems in Special Functions and Combinatorics, in Trends in Computer Algebra, Springer-Verlag, 1988, pp. 158-166 MR0935413 (89c:05010).

Crossrefs

Programs

  • Maple
    a := n -> hypergeom([-2*n, -n, 2*n+1], [1, 1], 1):
    seq(simplify(a(n)), n=0..15); # Peter Luschny, Feb 13 2018
  • Mathematica
    Table[Sum[(-1)^k*Binomial[2*n,k]^4,{k,0,2*n}] / Sum[(-1)^k*Binomial[2*n, k]^2,{k,0,2*n}],{n,0,20}] (* Vaclav Kotesovec, Mar 07 2014 *)
  • PARI
    a(n)=if(n<0,0, sum(k=0,2*n, (-1)^k*binomial(2*n,k)^4)/ sum(k=0,2*n, (-1)^k*binomial(2*n,k)^2))

Formula

Recurrence: n^2*(2*n-1)^2*(48*n^2 - 126*n + 83)*a(n) = (6528*n^6 - 30192*n^5 + 55040*n^4 - 50406*n^3 + 24465*n^2 - 5985*n + 585)*a(n-1) - (n-1)^2*(2*n-3)^2*(48*n^2 - 30*n + 5)*a(n-2). - Vaclav Kotesovec, Mar 07 2014
a(n) ~ (1+sqrt(2))^(4*n+3/2) / (2^(9/4)*Pi*n). - Vaclav Kotesovec, Mar 07 2014
a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k,2*n-k)*binomial(2*k,k). - Ilya Gutkovskiy, Nov 24 2017
a(n) = hypergeom([-2*n, -n, 2*n+1], [1, 1], 1). - Peter Luschny, Feb 13 2018
From Peter Bala, Dec 21 2020: (Start)
a(n) = Sum_{k = 0..n} C(2*n,n-k)^2 * C(2*n+k,k). Cf. A005258.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*C(2*n,n-k)*C(2*n+k,k)^2.
a(n) = C(2*n,n)^2 * hypergeom([-n, -n, 2*n+1], [n+1, n+1], 1).
a(n) = (-1)^n* C(2*n,n) * hypergeom([-n, 2*n+1, 2*n+1], [1, n+1], 1). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(m*n,(1 + x)/(1 - x)) ) at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 23 2020
a(n) = A108625(2*n, n). - Peter Bala, Jun 20 2023

A142995 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1), n >= 1.

Original entry on oeis.org

0, 1, 7, 89, 1836, 56164, 2390832, 135213840, 9809203968, 888117094656, 98167241088000, 13010123816064000, 2036436482119680000, 371699564417796096000, 78251077775510986752000
Offset: 0

Views

Author

Peter Bala, Jul 18 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + m^2 + m + 1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^2 for the constant zeta(2). For other cases see A001819 (m=0), A142996 (m=2), A142997 (m=3) and A142998 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} 1/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_{k = 0..m} C(m,k)^2*C(x+k,m) = Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k) is the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m (equivalently, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.]).
The first few are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 3*x^2 + 3*x + 1 and p_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. The o.g.f. for the p_m(x) is ((1-t^2)^x/(1-t)^(2x+1))*Legendre_P(x,(1+t^2)/(1-t^2)) = 1 + (2*x+1)*t + (3*x^2+3*x+1)*t^2 + ... [Gogin & Hirvensalo, Theorem 1 with N = -1].
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^2*f(x+1) + x^2*f(x-1) = (2*x^2 + 2*x + m^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 (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^2*p_m(n) with initial conditions b(0) = 1, b(1) = m^2 + m + 1. Hence the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = Sum_{k>=1} 1/(k^2*p_m(k-1)*p_m(k)) = 1/((m^2 + m + 1) - 1^4/((m^2 + m + 5) - 2^4/((m^2 + m + 13) - ... - n^4/((2*n^2 + 2*n + m^2 + m + 1) - ...)))) = 2*Sum_{k>=1} (-1)^(k+1)/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
For related results see A142999. For corresponding results for the constants e, log(2) and zeta(3) see A000522, A142979 and A143003 respectively.

References

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

Crossrefs

Cf. A000522, A001819, A003215 (A_2 lattice), A005902 (A_3 lattice), A008384 (A_4 lattice), A008386 (A_5 lattice), A108625, A142979, A142996, A142997, A142998, A143003.

Programs

  • Maple
    p := n -> 2*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);
  • Mathematica
    a[n_] := -1/6*n!^2*(2*n*(Pi^2-12) + Pi^2 - 6*(2*n+1)*PolyGamma[1, n+1]) // Simplify; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 06 2013 *)

Formula

a(n) = n!^2*p(n)*Sum_{k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 2*n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - (n-1)^4/(2*n^2 - 2*n + 3))))), for n >= 2. Lim_{n -> infinity} a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - n^4/((2*n^2 + 2*n + 3) - ...))))) = Sum_{k>=1} 1/(k^2*(4*k^2 - 1)) = 2 - zeta(2).

A145905 Square array read by antidiagonals: Hilbert transform of triangle A060187.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 27, 25, 7, 1, 1, 81, 125, 49, 9, 1, 1, 243, 625, 343, 81, 11, 1, 1, 729, 3125, 2401, 729, 121, 13, 1, 1, 2187, 15625, 16807, 6561, 1331, 169, 15, 1, 1, 6561, 78125, 117649, 59049, 14641, 2197, 225, 17, 1, 1, 19683, 390625, 823543
Offset: 0

Views

Author

Peter Bala, Oct 27 2008

Keywords

Comments

Definition of the Hilbert transform of a triangular array:
For many square arrays in the database the entries in a row are polynomial in the column index, of degree d say and hence the row generating function has the form P(x)/(1-x)^(d+1), where P is some polynomial function. Often the array whose rows are formed from the coefficients of these P polynomials is of independent interest. This suggests the following definition.
Let [L(n,k)]n,k>=0 be a lower triangular array and let R(n,x) := sum {k = 0 .. n} L(n,k)*x^k, denote the n-th row generating polynomial of L. Then we define the Hilbert transform of L, denoted Hilb(L), to be the square array whose n-th row, n >= 0, has the generating function R(n,x)/(1-x)^(n+1).
In this particular case, L is the array A060187, the array of Eulerian numbers of type B, whose row polynomials are the h-polynomials for permutohedra of type B. The Hilbert transform is an infinite Vandermonde matrix V(1,3,5,...).
We illustrate the Hilbert transform with a few examples:
(1) The Delannoy number array A008288 is the Hilbert transform of Pascal's triangle A007318 (view as the array of coefficients of h-polynomials of n-dimensional cross polytopes).
(2) The transpose of the array of nexus numbers A047969 is the Hilbert transform of the triangle of Eulerian numbers A008292 (best viewed in this context as the coefficients of h-polynomials of n-dimensional permutohedra of type A).
(3) The sequence of Eulerian polynomials begins [1, x, x + x^2, x + 4*x^2 + x^3, ...]. The coefficients of these polynomials are recorded in triangle A123125, whose Hilbert transform is A004248 read as square array.
(4) A108625, the array of crystal ball sequences for the A_n lattices, is the Hilbert transform of A008459 (viewed as the triangle of coefficients of h-polynomials of n-dimensional associahedra of type B).
(5) A142992, the array of crystal ball sequences for the C_n lattices, is the Hilbert transform of A086645, the array of h-vectors for type C root polytopes.
(6) A108553, the array of crystal ball sequences for the D_n lattices, is the Hilbert transform of A108558, the array of h-vectors for type D root polytopes.
(7) A086764, read as a square array, is the Hilbert transform of the rencontres numbers A008290.
(8) A143409 is the Hilbert transform of triangle A073107.

Examples

			Triangle A060187 (with an offset of 0) begins
1;
1, 1;
1, 6, 1;
so the entries in the first three rows of the Hilbert transform of
A060187 come from the expansions:
Row 0: 1/(1-x) = 1 + x + x^2 + x^3 + ...;
Row 1: (1+x)/(1-x)^2 = 1 + 3*x + 5*x^2 + 7*x^3 + ...;
Row 2: (1+6*x+x^2)/(1-x)^3 = 1 + 9*x + 25*x^2 + 49*x^3 + ...;
The array begins
n\k|..0....1.....2.....3......4
================================
0..|..1....1.....1.....1......1
1..|..1....3.....5.....7......9
2..|..1....9....25....49.....81
3..|..1...27...125...343....729
4..|..1...81...625..2401...6561
5..|..1..243..3125.16807..59049
...
		

Crossrefs

Cf. A008292, A039755, A052750 (first superdiagonal), A060187, A114172, A145901.

Programs

  • Maple
    T:=(n,k) -> (2*k + 1)^n: seq(seq(T(n-k,k),k = 0..n),n = 0..10);

Formula

T(n,k) = (2*k + 1)^n, (see equation 4.10 in [Franssens]). This array is the infinite Vandermonde matrix V(1,3,5,7, ....) having a LDU factorization equal to A039755 * diag(2^n*n!) * transpose(A007318).

A108628 n-th term of the crystal ball sequence for A_{n+1} lattice for n >= 0.

Original entry on oeis.org

1, 7, 55, 471, 4251, 39733, 380731, 3716695, 36808723, 368750757, 3728940249, 38003358693, 389866749975, 4022124746409, 41697566691555, 434124925278807, 4536783726146499, 47569453938399445, 500266519237489357, 5275183203229043221, 55760274296452936741
Offset: 0

Views

Author

Paul D. Hanna, Jun 14 2005

Keywords

Comments

Equals the secondary diagonal of square array A108625, in which row n equals the crystal ball sequence for A_n lattice. Main diagonal of square array A108625 equals the Apery numbers (A005258).

Crossrefs

Programs

  • Magma
    A108628:= func< n | (&+[Binomial(n+1,k)^2*Binomial(n+k,k-1): k in [0..n+1]]) >;
    [A108628(n): n in [0..30]]; // G. C. Greubel, Oct 06 2023
    
  • Maple
    seq(add(binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k), k = 0..n), n = 0..20); # Peter Bala, Apr 14 2022
  • Mathematica
    Table[Sum[Binomial[n+1,k]^2 Binomial[n+k,k-1],{k,0,n+1}],{n,0,20}] (* Harvey P. Dale, Apr 01 2013 *)
  • PARI
    a(n)=sum(k=0,n+1,binomial(n+1,k)^2*binomial(n+k,k-1))
    
  • Python
    def A108628(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= (n+k+2)*(n-k)*(n-k+1)
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 03 2022
    
  • SageMath
    def A108628(n): return sum(binomial(n+1,k)^2*binomial(n+k,k-1) for k in range(n+2))
    [A108628(n) for n in range(31)] # G. C. Greubel, Oct 06 2023

Formula

a(n) = Sum_{k = 0..n+1} C(n+1, k)^2 * C(n+k, k-1).
a(n) = A108625(n+1, n).
exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + 144*x^4 + 1048*x^5 + 8189*x^6 + 67325*x^7 + 574999*x^8 + ... appears to have integer coefficients. Cf. A208675. - Peter Bala, Jan 12 2016
Recurrence: (n+1)^2*(5*n^2 - 6*n + 2)*a(n) = (55*n^4 - 11*n^3 - 26*n^2 + 5*n + 5)*a(n-1) + (n-1)^2*(5*n^2 + 4*n + 1)*a(n-2). - Vaclav Kotesovec, Jan 13 2016
a(n) ~ sqrt(89/8 + 199/(8*sqrt(5))) * ((1+sqrt(5))/2)^(5*n) / (Pi*n). - Vaclav Kotesovec, Jan 13 2016
Equivalently, a(n) ~ phi^(5*n + 11/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
From Peter Bala, Mar 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k).
Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+1,k)*binomial(-n+k+1,k) = Sum_{k} (-1)^k* binomial(n+k-1,k)*binomial(n+k-2,k)*binomial(n-2,k) = (-1)^n*A208675(n-1) for n >= 2.
a(n-1) = (1/2)*Sum_{k = 0..floor(n/2)} binomial(n,k)^2 * binomial(3*n-2*k-1,n-2*k) for n >= 1. Cf. A103882.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,k)^2.
Equivalently, a(n) = [(x*z)^(n+1)*y^n] ( (x + y + z)^(n+1) * (x + y)^n * (y + z)^(n+1) ).
a(n) = (1/5)*( 2*A005258(n+1) - A005258(n) ).
a(n) = hypergeometric3F2([n + 2, -n, -n - 1], [1, 1], 1).
a(n) = (-1)^n*(n+1)*hypergeom([n + 2, n + 2, -n ], [1, 2], 1).
a(n) = [x^n] 1/(1 - x)*P(n+1,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A208675(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1 and A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n+1,n,n+1) in the notation of Straub, equation 24. Hence
a(n) = [(x*z)^(n+1)*y^n] 1/(1 - x - y - z + x*z + y*z - x*y*z).
The following takes the sequence offset to be 1: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Conjectures: a((p-1)/2) == 0 (mod p) and a((p^3-1)/2) == 0 (mod p^3) for primes p of the form 4*m + 1; a((p^3-1)/2) == 0 (mod p^2) for primes p of the form 4*m + 3; a((p^2-1)/2) == 0 (mod p^2) for primes p >= 5. (End)
From Peter Bala, Sep 11 2024: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(n+k+1)*binomial(n+1, k)*binomial(n+k, k)* binomial(n+k+1, k).
a(n) = (-1)^(n+1) * hypergeom([n+1, n+2, -n-1], [1, 1], 1). (End)
a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n-1, k). Cf. A005258(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Sep 25 2024

A143409 Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 16, 11, 4, 1, 65, 49, 19, 5, 1, 326, 261, 106, 29, 6, 1, 1957, 1631, 685, 193, 41, 7, 1, 13700, 11743, 5056, 1457, 316, 55, 8, 1, 109601, 95901, 42079, 12341, 2721, 481, 71, 9, 1, 986410, 876809, 390454, 116125, 25946, 4645, 694, 89, 10, 1
Offset: 0

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Author

Peter Bala, Aug 14 2008

Keywords

Comments

The Euler-Seidel matrix for the sequence {k!} is array A076571 read as a square, whose k-th column entries have a common factor of k!. Removing these common factors gives the current table.
This table is closely connected to the constant 1/e. The row, column and diagonal entries of this table occur in series acceleration formulas for 1/e.
For a similar table based on the differences of the sequence {k!} and related to the constant e, see A086764. For other arrays similarly related to constants see A143410 (for sqrt(e)), A143411 (for 1/sqrt(e)), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

Examples

			The Euler-Seidel matrix for the sequence {k!} begins
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....2.....6....24...120...720
1..|.....2.....3.....8....30...144...840
2..|.....5....11....38...174...984
3..|....16....49...212..1158
4..|....65...261..1370
5..|...326..1631
6..|..1957
...
Dividing the k-th column by k! gives
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....1.....1.....1.....1.....1
1..|.....2.....3.....4.....5.....6.....7
2..|.....5....11....19....29....41
3..|....16....49...106...193
4..|....65...261...685
5..|...326..1631
6..|..1957
...
Examples of series formula for 1/e:
Row 2: 1/e = 2*(1/5 - 1/(1!*5*11) + 1/(2!*11*19) - 1/(3!*19*29) + ...).
Column 4: 24/e = 9 - (0!/(1*6) + 1!/(6*41) + 2!/(41*316) + ...).
...
Displayed as a triangle:
0 |     1
1 |     2,     1
2 |     5,     3,    1
3 |    16,    11,    4,    1
4 |    65,    49,   19,    5,   1
5 |   326,   261,  106,   29,   6,  1
6 |  1957,  1631,  685,  193,  41,  7, 1
7 | 13700, 11743, 5056, 1457, 316, 55, 8, 1
		

Crossrefs

Cf. A008288, A076571, A086764, A108625, A143007, A143410, A143411, A143413, A001517 (main diagonal), A028387 (row 2), A000522 (column 0), A001339 (column 1), A082030 (column 2), A095000 (column 3), A095177 (column 4).

Programs

  • Maple
    T := (n, k) -> 1/k!*add(binomial(n,j)*(k+j)!, j = 0..n):
    for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
    # Alternate:
    T:= proc(n,k) option remember;
      if n = 0 then return 1 fi;
      (n+k)*procname(n-1,k) + procname(n-1,k-1);
    end proc:
    seq(seq(T(s-n,n),n=0..s),s=0..10); # Robert Israel, Jul 07 2017
    # Or:
    A143409 := (n,k) -> hypergeom([k+1, k-n], [], -1):
    seq(seq(simplify(A143409(n,k)),k=0..n),n=0..9); # Peter Luschny, Oct 05 2017
  • Mathematica
    T[n_, k_] := HypergeometricPFQ[{k+1,k-n}, {}, -1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

Formula

T(n,k) = (1/k!)*Sum_{j = 0..n} binomial(n,j)*(k+j)!.
T(n,k) = ((n+k)!/k!)*Num_Pade(n,k), where Num_Pade(n,k) denotes the numerator of the Padé approximation for the function exp(x) of degree (n,k) evaluated at x = 1.
Recurrence relations:
T(n,k) = T(n-1,k) + (k+1)*T(n-1,k+1);
T(n,k) = (n+k)*T(n-1,k) + T(n-1,k-1).
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 + ...) + (2 + 3*x + 4*x^2 + ...)*y + (5 + 11*x + 19*x^2 + ...)*y^2/2! + ... .
Row n lists the values of the Poisson-Charlier polynomial x^(n) + C(n,1)*x^(n-1) + C(n,2)*x^(n-2) + ... + C(n,n) for x = 1,2,3,..., where x^(m) denotes the rising factorial x*(x+1)*...*(x+m-1).
Main diagonal is A001517.
Series formulas for 1/e:
Row n: 1/e = n!*[1/T(n,0) - 1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) - 1/(3!*T(n,2)*T(n,3)) + ...].
Column k: k!/e = A000166(k) + (-1)^(k+1)*[0!/(T(0,k)*T(1,k)) + 1!/(T(1,k)*T(2,k)) + 2!/(T(2,k)*T(3,k)) + ...].
Main diagonal: 1/e = 1 - 2*Sum_{n>=0} (-1)^n/(T(n,n)*T(n+1,n+1)) = 1 - 2*[1/(1*3) - 1/(3*19) + 1/(19*193) - ...].
Second subdiagonal: 1/e = 2*(1^2/(1*5) - 2^2/(5*49) + 3^2/(49*685) - ...).
Compare with A143413.
From Peter Luschny, Oct 05 2017: (Start)
T(n, k) = hypergeom([k+1, k-n], [], -1).
When seen as a triangular array then the row sums are A273596 and the alternating row sums are A003470. (End)

A208675 Number of words, either empty or beginning with the first letter of the ternary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

Original entry on oeis.org

1, 1, 5, 37, 309, 2751, 25493, 242845, 2360501, 23301307, 232834755, 2349638259, 23905438725, 244889453043, 2523373849701, 26132595017037, 271826326839477, 2838429951771795, 29740725671232119, 312573076392760183, 3294144659048391059, 34802392680979707121
Offset: 0

Views

Author

Alois P. Heinz, Feb 29 2012

Keywords

Comments

Also the number of (3*n-1)-step walks on 3-dimensional cubic lattice from (1,0,0) to (n,n,n) with positive unit steps in all dimensions such that the absolute difference of the dimension indices used in consecutive steps is <= 1.

Examples

			a(2) = 5 = |{aabbcc, aabcbc, aabccb, ababcc, abccba}|.
a(3) = 37 = |{aaabbbccc, aaabbcbcc, aaabbccbc, aaabbcccb, aaabcbbcc, aaabcbcbc, aaabcbccb, aaabccbbc, aaabccbcb, aaabcccbb, aababbccc, aababcbcc, aababccbc, aababcccb, aabbabccc, aabbcccba, aabcbabcc, aabcbccba, aabccbabc, aabccbcba, aabcccbab, aabcccbba, abaabbccc, abaabcbcc, abaabccbc, abaabcccb, abababccc, ababcccba, abbaabccc, abbcccbaa, abcbaabcc, abcbccbaa, abccbaabc, abccbcbaa, abcccbaab, abcccbaba, abcccbbaa}|.
		

Crossrefs

Programs

  • Magma
    A208675:= func< n | (&+[Binomial(n,j)*Binomial(n-1,j)*Binomial(n+j-1,j): j in [0..2*n]]) >;
    [A208675(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    a:= n-> add(binomial(n-1, k)^2 *binomial(2*n-1-k, n-k), k=0..n):
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 26 2012
  • Mathematica
    a[n_]:= HypergeometricPFQ[{1-n,-n,n}, {1,1}, 1] (* Michael Somos, Jun 03 2012 *)
  • SageMath
    def A208675(n): return sum(binomial(n,j)*binomial(n-1,j)*binomial(n+j-1,j) for j in range(n+1))
    [A208675(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

From Michael Somos, Jun 03 2012: (Start)
a(n) = A108625(n-1, n).
a(n) = Hypergeometric3F2([1-n, -n, n], [1, 1], 1).
(n+1)^2 * (1 -4*n +5*n^2) * a(n+1) = (5 -5*n -26*n^2 +11*n^3 +55*n^4) * a(n) + (n-1)^2 * (2 +6*n +5*n^2) * a(n-1). (End)
a(n) ~ sqrt((5-sqrt(5))/10)/(2*Pi*n) * ((1+sqrt(5))/2)^(5*n). - Vaclav Kotesovec, Dec 06 2012. Equivalently, a(n) ~ phi^(5*n - 1/2) / (2 * 5^(1/4) * Pi * n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 15*x^3 + 94*x^4 + 668*x^5 + 5144*x^6 + 41884*x^7 + 355307*x^8 + ... appears to have integer coefficients. Cf. A108628. - Peter Bala, Jan 12 2016
From Peter Bala, Apr 05 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n-1,k)*binomial(n+k-1,k).
Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have:
a(-n) = Sum_{k = 0..n} binomial(-n,k)*binomial(-n-1,k)*binomial(-n+k-1,k).
a(-n) = Sum_{k = 0..n} (-1)^k* binomial(n+k-1,k)*binomial(n+k,k)*binomial(n,k)
a(-n) = (-1)^n*A108628(n-1), for n >= 1.
a(n) = Sum_{k = 1..n} binomial(n,k)*binomial(n-1,k-1)*binomial(n+k-1,k-1) for n >= 1.
Equivalently, a(n) = [(x^n)*(y*z)^(n-1)] (x + y + z)^n*(x + y)^(n-1)*(y + z)^(n-1) for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)*binomial(2*n-k-1,n-k)^2.
a(n) = (1/5)*(A005258(n) + 2*A005258(n-1)) for n >= 1.
a(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Compare with A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n,n-1,n-1) in the notation of Straub, equation 24. Hence
a(n) = [(x^n)*(y*z)^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 1.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Conjectures:
1) a(n) = [(x*y)^n*z^(n-1)] 1/(1 - x - y - z + x*y + x*y*z) for n >= 1.
2) a(n) = - [(x*z)^(n-1)*(y^n)] 1/(1 + y + z + x*y + y*z + x*z + x*y*z) for n >= 1.
3) a(n) = [x^(n-1)*(y*z)^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1. (End)
From Peter Bala, Mar 17 2023: (Start)
For n >= 1:
a(n) = Sum_{k = 0..n} ((n-k)/(n+k))*binomial(n,k)^2*binomial(n+k,k).
a(n) = Sum_{k = 0..n} (-1)^(n+k-1) * ((n-k)/(n+k)) * binomial(n,k) * binomial(n+k,k)^2. (End)

A001451 a(n) = (5*n)!/((3*n)!*n!*n!).

Original entry on oeis.org

1, 20, 1260, 100100, 8817900, 823727520, 79919739900, 7962100660800, 808906548235500, 83426304143982800, 8707404737345073760, 917663774856743842200, 97491279924241456098300, 10427604345391237790688000, 1121786259855036145008408000
Offset: 0

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Author

Keywords

Comments

From Peter Bala, Jul 15 2016: (Start)
This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
Note the similar results:
Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - Bradley Klee, Aug 12 2018

Examples

			G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018
		

Crossrefs

Programs

  • GAP
    List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # Muniru A Asiru, Aug 12 2018
  • Magma
    [Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // Vincenzo Librandi, May 22 2011
    
  • Maple
    f := n->(5*n)!/((3*n)!*n!*n!);
  • Mathematica
    Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* Vincenzo Librandi, Sep 04 2012 *)

Formula

a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 13 2016
G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - Bradley Klee, Aug 12 2018
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 12 2024

A363864 a(n) = A143007(2*n,n).

Original entry on oeis.org

1, 13, 661, 46705, 3833941, 342981013, 32443313449, 3191377294153, 323158664855125, 33461619685494025, 3526747995762849661, 377103695616260979313, 40807220545026078297961, 4460530114068960956304865, 491780450707942086338993761, 54624008737670717933342875705
Offset: 0

Views

Author

Peter Bala, Jun 25 2023

Keywords

Comments

a(n) = A(2*n,n,2*n,n) in the notation of Straub, equation 8. It follows from Straub, Theorem 1.2, that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and all positive integers n and k.
More generally, for positive integers r and s the sequence {A143007(r*n, s*n) : n >= 0} satisfies the same supercongruences. For cases, see A005259(r = s = 1), A363865 (r = 3, s = 1) and A363866 (r = 3, s = 2).

Crossrefs

Programs

  • Maple
    A143007 := proc(n, k); add(binomial(n+j, 2*j)*binomial(2*j, j)^2*binomial(k+j, 2*j), j = 0..n) end:
    seq(A143007(2*n, n), n = 0..20);
    # alternative program
    seq(simplify(hypergeom([2*n+1, -2*n, n+1, -n], [1, 1, 1], 1)), n = 0..20);

Formula

a(n) = Sum_{k = 0..n} binomial(2*n,n-k)^2*binomial(2*n+k,k)^2.
a(n) = Sum_{k = 0..n} binomial(2*n+k,2*k)*binomial(2*k,k)^2*binomial(n+k,2*k).
a(n) = hypergeom([2*n+1, -2*n, n+1, -n], [1, 1, 1], 1)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(2*n,(1 + x)/(1 - x)) )^2 = [x^(2*n)] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^2.
P-recursive: 2*(440*n^3 - 1782*n^2 + 2412*n - 1091)*(2*n - 1)^3*n^3*a(n) = (865920*n^9 - 6104736*n^8 + 18475432*n^7 - 31464562*n^6 + 33227280*n^5 - 22586875*n^4 + 9902182*n^3 - 2707173*n^2 + 420336*n - 28350)*a(n-1) - 2*(440*n^3 - 462*n^2 + 168*n - 21)*(n - 1)^3*(2*n - 3)^3*a(n-2) with a(0) = 1 and a(1) = 13.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n, k)*binomial(2*n+k, k)*A108625(n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 16 2024

A364113 Square array read by ascending antidiagonals: T(n,k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^n for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 19, 1, 1, 7, 73, 147, 1, 1, 9, 163, 1445, 1251, 1, 1, 11, 289, 5623, 33001, 11253, 1, 1, 13, 451, 14409, 235251, 819005, 104959, 1, 1, 15, 649, 29531, 908001, 11009257, 21460825, 1004307, 1, 1, 17, 883, 52717, 2511251, 65898009, 554159719, 584307365, 9793891, 1
Offset: 0

Views

Author

Peter Bala, Jul 07 2023

Keywords

Comments

The two types of Apéry numbers A005258 and A005259 are related to the Legendre polynomials by A005258(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x)) and A005259(k) = [x^k] 1/(1 - x) * Legendre_P(k, (1 + x)/(1 - x))^2 and thus form rows 1 and 2 of the present array.
Both types of Apéry numbers satisfy the supercongruences
1) u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
and the shifted supercongruences
2) u(n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
for all primes p >= 5 and positive integers n and r.
We conjecture that each row sequence of the present table satisfies the same pair of supercongruences.

Examples

			Square array begins
 n\k|  0   1    2      3        4          5             6               7
  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1   1    1      1        1          1             1               1
  1 |  1   3   19    147     1251      11253        104959         1004307
  2 |  1   5   73   1445    33001     819005      21460825       584307365
  3 |  1   7  163   5623   235251   11009257     554159719     29359663991
  4 |  1   9  289  14409   908001   65898009    5246665201    445752724041
  5 |  1  11  451  29531  2511251  251831261   28224521263   3423024241627
  6 |  1  13  649  52717  5665001  730485013  106898093065  17144295476461
		

Crossrefs

Cf. A005258 (row 1), A005259 (row 2), A364114 (row 3), A364115 (row 4), A364116 (main diagonal), A364117 (first subdiagonal).

Programs

  • Maple
    T(n,k) := coeff(series(1/(1-x)* LegendreP(k,(1+x)/(1-x))^n, x, 11), x, k):
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);
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