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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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

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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)

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

Views

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
Showing 1-2 of 2 results.