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.

Showing 1-10 of 12 results. Next

A108554 Main diagonal of square array A108553, in which row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 3, 13, 147, 1681, 18733, 204205, 2195399, 23429185, 249258777, 2650436845, 28209301211, 300745906449, 3212758605189, 34392151363021, 368908132170511, 3964622452867329, 42682164804034609, 460242445806795853
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Limit a(n+1)/a(n) ~ 11.08...
Limit is equal to A001622^5 = A244593 = 11.090169943749474241... - Vaclav Kotesovec, Aug 31 2025

Crossrefs

Programs

  • Mathematica
    Join[{1, 3}, Table[Sum[Binomial[2*n - j, n - j]*(Binomial[2*n, 2*j] - 2*j*(n - j)*Binomial[n, j]/(n - 1)), {j, 0, n}], {n, 2, 20}]] (* Vaclav Kotesovec, Aug 31 2025 *)
  • PARI
    {a(n)=if(n<0,0,if(n==0,1,if(n==1,3,sum(j=0,n,binomial(2*n-j,n-j)* (binomial(2*n,2*j)-2*j*(n-j)*binomial(n,j)/(n-1))))))}

Formula

a(n) ~ phi^(5*n + 1) / (2*5^(1/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 31 2025

A108555 Antidiagonal sums of square array A108553, in which row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 2, 5, 12, 35, 116, 411, 1482, 5333, 19030, 67265, 235704, 819863, 2834600, 9752951, 33428054, 114228713, 389419114, 1325155261, 4503031332, 15285291851, 51842203804, 175719341971, 595316455842, 2016131645245, 6826076021310
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Limit a(n+1)/a(n) ~ 3.3829757679..., real root of cubic (1+x+3*x^2-x^3).

Crossrefs

Programs

  • PARI
    {a(n)=if(n<0,0,if(n==0,1,sum(k=0,n,sum(j=0,k,binomial(n-j,k-j)* (binomial(2*(n-k),2*j)-2*j*(n-k-j)*binomial(n-k,j)/if(n==k+1,1,(n-k-1)))))))}

Formula

Empirical G.f.: (x^3+x^2+x-1)*(2*x^5+5*x^4-4*x^3+10*x^2-6*x+1) / ((x-1)^2*(x^2+2*x-1)^2*(x^3+x^2+3*x-1)). [Colin Barker, Nov 21 2012]

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 12 2005

Keywords

Comments

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

Examples

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...
		

Crossrefs

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).

Programs

  • Magma
    T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
    A108625:= func< n,k | T(n-k,k) >; // antidiagonals
    [A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
    seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)
  • PARI
    T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))
    
  • SageMath
    def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
    def A108625(n,k): return T(n-k, k) # antidiagonals
    flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A047969 Square array of nexus numbers a(n,k) = (n+1)^(k+1) - n^(k+1) (n >= 0, k >= 0) read by upwards antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 19, 15, 1, 1, 9, 37, 65, 31, 1, 1, 11, 61, 175, 211, 63, 1, 1, 13, 91, 369, 781, 665, 127, 1, 1, 15, 127, 671, 2101, 3367, 2059, 255, 1, 1, 17, 169, 1105, 4651, 11529, 14197, 6305, 511, 1, 1, 19, 217, 1695, 9031
Offset: 0

Views

Author

Keywords

Comments

If each row started with an initial 0 (i.e., a(n,k) = (n+1)^k - n^k) then each row would be the binomial transform of the preceding row. - Henry Bottomley, May 31 2001
a(n-1, k-1) is the number of ordered k-tuples of positive integers such that the largest of these integers is n. - Alford Arnold, Sep 07 2005
From Alford Arnold, Jul 21 2006: (Start)
The sequences in A047969 can also be calculated using the Eulerian Array (A008292) and Pascal's Triangle (A007318) as illustrated below: (cf. A101095).
1 1 1 1 1 1
1 1 1 1 1 1
-----------------------------------------
1 2 3 4 5 6
1 2 3 4 5
1 3 5 7 9 11
-----------------------------------------
1 3 6 10 15 21
4 12 24 40 60
1 3 6 10
1 7 19 37 61 91
-----------------------------------------
1 4 10 20 35 56
11 44 110 220 385
11 44 110 220
1 4 10
1 15 65 175 369 671
----------------------------------------- (End)
From Peter Bala, Oct 26 2008: (Start)
The above remarks of Alford Arnold may be summarized by saying that (the transpose of) this array is the Hilbert transform of the triangle of Eulerian numbers A008292 (see A145905 for the definition of the Hilbert transform). In this context, A008292 is best viewed as the array of h-vectors of permutohedra of type A. See A108553 for the Hilbert transform of the array of h-vectors of type D permutohedra. Compare this array with A009998.
The polynomials n^k - (n-1)^k, k = 1,2,3,..., which give the nonzero entries in the columns of this array, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re s = 1/2 in the complex plane. See A019538 for the connection between the polynomials n^k - (n-1)^k and the Stirling polynomials of the simplicial complexes dual to the type A permutohedra.
(End)
Empirical: (n+1)^(k+1) - n^(k+1) is the number of first differences of length k+1 arrays of numbers in 0..n, k > 0. - R. H. Hardin, Jun 30 2013
a(n-1, k-1) is the number of bargraphs of width k and height n. Examples: a(1,2) = 7 because we have [1,1,2], [1,2,1], [2,1,1], [1,2,2], [2,1,2], [2,2,1], and [2,2,2]; a(2,1) = 5 because we have [1,3], [2,3], [3,1], [3,2], and [3,3] (bargraphs are given as compositions). This comment is equivalent to A. Arnold's Sep 2005 comment. - Emeric Deutsch, Jan 30 2017

Examples

			Array a begins:
  [n\k][0  1   2    3    4   5  6  ...
  [0]   1  1   1    1    1   1  1  ...
  [1]   1  3   7   15   31  63  ...
  [2]   1  5  19   65  211  ...
  [3]   1  7  37  175  ...
  ...
Triangle T begins:
  n\m   0   1    2     3     4      5      6      7      8     9  10 ...
  0:    1
  1:    1   1
  2:    1   3    1
  3:    1   5    7     1
  4:    1   7   19    15     1
  5:    1   9   37    65    31      1
  6:    1  11   61   175   211     63      1
  7:    1  13   91   369   781    665    127      1
  8:    1  15  127   671  2101   3367   2059    255      1
  9:    1  17  169  1105  4651  11529  14197   6305    511     1
  10:   1  19  217  1695  9031  31031  61741  58975  19171  1023   1
  ...  - _Wolfdieter Lang_, May 07 2021
		

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 54.

Crossrefs

Cf. A047970.
Cf. A009998, A108553 (Hilbert transform of array of h-vectors of type D permutohedra), A145904, A145905.
Row n sequences of array a: A000012, A000225(k+1), A001047(k+1), A005061(k+1), A005060(k+1), A005062(k+1), A016169(k+1), A016177(k+1), A016185(k+1), A016189(k+1), A016195(k+1), A016197(k+1).
Column k sequences of array a: (nexus numbers): A000012, A005408, A003215, A005917(n+1), A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528.
Cf. A343237 (row reversed triangle).

Programs

  • Mathematica
    Flatten[Table[n = d - e; k = e; (n + 1)^(k + 1) - n^(k + 1), {d, 0, 100}, {e, 0, d}]] (* T. D. Noe, Feb 22 2012 *)
  • Maxima
    T(n,m):=if m=0 then 1 else sum(k!*(-1)^(m+k)*stirling2(m,k)*binomial(n+k-1,n),k,0,m); /* Vladimir Kruchinin, Jan 28 2018 */

Formula

From Vladimir Kruchinin: (Start)
O.g.f. of e.g.f of rows of array: ((1-x)*exp(y))/(1-x*exp(y))^2.
T(n,m) = Sum_{k=0..m} k!*(-1)^(m+k)*Stirling2(m,k)*C(n+k-1,n), T(n,0)=1.(End)
From Wolfdieter Lang, May 07 2021: (Start)
T(n,m) = a(n-m,m) = (n-m+1)^(m+1) - (n-m)^(m+1), n >= 0, m = 0, 1,..., n.
O.g.f. column k of the array: polylog(-(k+1), x)*(1-x)/x. See the Peter Bala comment above, and the Eulerian triangle A008292 formula by Vladeta Jovovic, Sep 02 2002.
E.g.f. of e.g.f. of row of the array: exp(y)*(1 + x*(exp(y) - 1))*exp(x*exp(y)).
O.g.f. of triangle's exponential row polynomials R(n, y) = Sum_{m=0} T(n, m)*(y^m)/m!: G(x, y) = exp(x*y)*(1 - x)/(1 - x*exp(x*y))^2. (End)

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

Views

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.

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).

A108558 Symmetric triangle, read by rows, where row n equals the (n+1)-th differences of the crystal ball sequence for D_n lattice, for n>=0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 9, 9, 1, 1, 20, 54, 20, 1, 1, 35, 180, 180, 35, 1, 1, 54, 447, 852, 447, 54, 1, 1, 77, 931, 2863, 2863, 931, 77, 1, 1, 104, 1724, 7768, 12550, 7768, 1724, 104, 1, 1, 135, 2934, 18186, 43128, 43128, 18186, 2934, 135, 1, 1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Row n equals the (n+1)-th differences of row n of the square array A108553. G.f. of row n equals: (1-x)^(n+1)*CBD_n(x), where CBD_n denotes the g.f. of the crystal ball sequence for D_n lattice.
From Peter Bala, Oct 23 2008: (Start)
Let D_n be the root lattice generated as a monoid by {+-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(D_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(D_n) [Ardila et al.]. See A108556 for the corresponding array of f-vectors for these type D_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A086645 for the array of h-vectors associated with type C_n polytopes.
The Hilbert transform of this array (as defined in A145905) equals A108553.
(End)

Examples

			G.f.s of initial rows of square array A108553 are:
  (1)/(1-x),
  (1 + x)/(1-x)^2,
  (1 + 2*x + x^2)/(1-x)^3,
  (1 + 9*x + 9*x^2 + x^3)/(1-x)^4,
  (1 + 20*x + 54*x^2 + 20*x^3 + x^4)/(1-x)^5,
  (1 + 35*x + 180*x^2 + 180*x^3 + 35*x^4 + x^5)/(1-x)^6.
Triangle begins:
  1;
  1,   1;
  1,   2,    1;
  1,   9,    9,     1;
  1,  20,   54,    20,      1;
  1,  35,  180,   180,     35,      1;
  1,  54,  447,   852,    447,     54,      1;
  1,  77,  931,  2863,   2863,    931,     77,     1;
  1, 104, 1724,  7768,  12550,   7768,   1724,   104,    1;
  1, 135, 2934, 18186,  43128,  43128,  18186,  2934,  135,   1;
  1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1;
  ...
		

Crossrefs

Cf. A108553, A008353, A108558, A008459, A086645, A108556. Row n equals (n+1)-th differences of: A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).
T(2n,n) gives A305693.

Programs

  • Mathematica
    T[1, 0] = T[1, 1]=1; T[n_, k_] := Binomial[2n, 2k] - 2n Binomial[n-2, k-1];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 25 2018 *)
  • PARI
    T(n,k)=if(n
    				

Formula

T(n, k) = C(2*n, 2*k) - 2*n*C(n-2, k-1) for n>1, with T(0, 0)=1, T(1, 0)=T(1, 1)=1. Row sums equal A008353: 2^(n-1)*(2^n-n) for n>1.
From Peter Bala, Oct 23 2008: (Start)
O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + x)*z + (3 + 2*x + 3*x^2)*z^2 - (1 + x)*(1 - 8*x + x^2)z^3 - 8*x*(1 + x^2)*z^4 + 2*x*(1 + x)*(1 - x)^2*z^5 and D(x,z) = ((1 - z)^2 - 2*x*z*(1 + z) + x^2*z^2)*(1 - z*(1 + x))^2.
For n >= 2, the row n generating polynomial equals 1/2*[(1 + sqrt(x))^(2n) + (1 - sqrt(x))^(2n)] - 2*n*x*(1 + x)^(n-2).
(End)

A108556 Triangle, read by rows, where row n equals the inverse binomial transform of the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 1, 12, 30, 20, 1, 24, 120, 192, 96, 1, 40, 330, 940, 1080, 432, 1, 60, 732, 3200, 6240, 5568, 1856, 1, 84, 1414, 8708, 25200, 37184, 27104, 7744, 1, 112, 2480, 20352, 80960, 173824, 206080, 126976, 31744, 1, 144, 4050, 42588, 221544, 643824, 1096032, 1085760, 579456, 128768
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Row n equals the inverse binomial transform of row n of the square array A108553.
Array of f-vectors for type D root polytopes [Ardila et al.]. See A063007 and A127674 for the arrays of f-vectors for type A and type C root polytopes respectively. - Peter Bala, Oct 23 2008

Examples

			Triangle begins:
1;
1,2;
1,4,4;
1,12,30,20;
1,24,120,192,96;
1,40,330,940,1080,432;
1,60,732,3200,6240,5568,1856;
1,84,1414,8708,25200,37184,27104,7744;
1,112,2480,20352,80960,173824,206080,126976,31744; ...
		

Crossrefs

Cf. A108553, A108557 (row sums), A108558, Rows are inverse binomial transforms of: A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).

Programs

  • Mathematica
    T[n_, k_] := Module[{A}, A = Table[Table[If[r - 1 == 0 || c - 1 == 0, 1, If[r - 1 == 1, 2c - 1, Sum[Binomial[r + c - j - 2, c - j - 1] (Binomial[2r - 2, 2j] - 2(r - 1) Binomial[r - 3, j - 1]), {j, 0, c - 1}]]], {c, 1, n + 1}], {r, 1, n + 1}]; SeriesCoefficient[((A[[n + 1]]. x^Range[0, n]) /. x -> x/(1 + x))/(1 + x), {x, 0, k}]];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018, from PARI *)
  • PARI
    T(n,k)=local(A=vector(n+1,r,vector(n+1,c,if(r-1==0 || c-1==0,1,if(r-1==1,2*c-1, sum(j=0,c-1,binomial(r+c-j-2,c-j-1)*(binomial(2*r-2,2*j)-2*(r-1)*binomial(r-3,j-1)))))))); polcoeff(subst(Ser(A[n+1]),x,x/(1+x))/(1+x),k)

Formula

Main diagonal equals A008353: 2^(n-1)*(2^n-n) for n>1.
O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + 2*x)*z + (3 + 8*x + 8*x^2)*z^2 - (1 + 2*x)*(1 - 6*x - 6*x^2)z^3 - 8*x*(1 + x)(1 + 2*x + 2*x^2)*z^4 + 2*x*(1 + x)*(1 + 2*x)*z^5 and D(x,z) = ((1-z)^2 - 4*x*z)*(1 - z*(1 + 2*x))^2. - Peter Bala, Oct 23 2008

A108998 Square array, read by antidiagonals, where row n equals the coordination sequence of B_n lattice, for n >= 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 8, 2, 0, 1, 18, 16, 2, 0, 1, 32, 74, 24, 2, 0, 1, 50, 224, 170, 32, 2, 0, 1, 72, 530, 768, 306, 40, 2, 0, 1, 98, 1072, 2562, 1856, 482, 48, 2, 0, 1, 128, 1946, 6968, 8130, 3680, 698, 56, 2, 0, 1, 162, 3264, 16394, 28320, 20082, 6432, 954, 64, 2, 0
Offset: 0

Views

Author

Paul D. Hanna, Jun 17 2005

Keywords

Comments

Compare with A108553, where row n equals the crystal ball sequence for D_n lattice.

Examples

			Square array begins:
  1,  0,    0,     0,     0,      0,      0,      0, ...
  1,  2,    2,     2,     2,      2,      2,      2, ...
  1,  8,   16,    24,    32,     40,     48,     56, ...
  1, 18,   74,   170,   306,    482,    698,    954, ...
  1, 32,  224,   768,  1856,   3680,   6432,  10304, ...
  1, 50,  530,  2562,  8130,  20082,  42130,  78850, ...
  1, 72, 1072,  6968, 28320,  85992, 214864, 467544, ...
  1, 98, 1946, 16394, 83442, 307314, 907018, ...
Product of the g.f. of row n and (1-x)^n generates the rows of triangle A109001:
  1;
  1,  1;
  1,  6,   1;
  1, 15,  23,    1;
  1, 28, 102,   60,    1;
  1, 45, 290,  402,  125,   1;
  1, 66, 655, 1596, 1167, 226, 1; ...
		

Crossrefs

Cf. A108999 (main diagonal), A109000 (antidiagonal sums), A109001, A022144 (row 2), A022145 (row 3), A022146 (row 4), A022147 (row 5), A022148 (row 6), A022149 (row 7), A022150 (row 8), A022151 (row 9), A022152 (row 10), A022153 (row 11), A022154 (row 12).

Programs

  • PARI
    T(n,k)=if(n<0 || k<0,0,sum(j=0,k, binomial(n+k-j-1,k-j)*(binomial(2*n+1,2*j)-2*n*binomial(n-1,j-1))))

Formula

T(n, k) = Sum_{j=0..k} C(n+k-j-1, k-j)*(C(2*n+1, 2*j)-2*n*C(n-1, j-1)) for n >= k >= 0.
G.f. for coordination sequence of B_n lattice: ((Sum_{i=0..n} binomial(2*n+1, 2*i)*z^i)-2*n*z*(1+z)^(n-1))/(1-z)^n. [Bacher et al.]

A108999 Main diagonal of square array A108998, in which row n equals the coordination sequence of B_n lattice.

Original entry on oeis.org

1, 2, 16, 170, 1856, 20082, 214864, 2282394, 24165120, 255708578, 2708805776, 28752157898, 305908697152, 3262741154194, 34882914424528, 373781033269306, 4013444615232512, 43174945822078530, 465247083731404048
Offset: 0

Views

Author

Paul D. Hanna, Jun 17 2005

Keywords

Comments

Compare to diagonal A108554 of square array A108553, in which row n equals the crystal ball sequence for D_n lattice.

Crossrefs

Programs

  • GAP
    List([0..20],n->Sum([0..n],j->Binomial(2*n-j-1,n-j)*(Binomial(2*n+1,2*j)-2*n*Binomial(n-1,j-1)))); # Muniru A Asiru, Nov 21 2018
  • Mathematica
    a[n_]:= Sum[Binomial[2*n-j-1, n-j]*(Binomial[2*n+1, 2*j] - 2*n*Binomial[n-1, j-1]), {j,0,n}]; Array[a, 20, 0] (* Stefano Spezia, Nov 21 2018 *)
  • PARI
    {a(n)=sum(j=0,n, binomial(2*n-j-1,n-j)*(binomial(2*n+1,2*j)-2*n*binomial(n-1,j-1)))}
    

Formula

a(n) = Sum_{j=0..n} C(2*n-j-1, n-j)*( C(2*n+1, 2*j) - 2*n*C(n-1, j-1) ).
a(n) ~ phi^(5*n+1) / (2*5^(1/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 31 2025
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