A234839 -id:A234839 - cp's OEIS frontend

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)(1 - x^2))^(nk) for n, k >= 1.

1, 1, 5, 1, 7, 19, 1, 9, 46, 85, 1, 11, 82, 327, 376, 1, 13, 127, 793, 2376, 1715, 1, 15, 181, 1547, 7876, 17602, 7890, 1, 17, 244, 2653, 19376, 79686, 132056, 36693, 1, 19, 316, 4175, 40001, 247205, 816684, 1000263, 171820, 1, 21, 397, 6177, 73501, 614389, 3195046, 8450585, 7632433, 809380

Offset: 0

Peter Bala, Jun 13 2023

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.
We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.
The table can be extended to negative values of n, and the row sequences also appear to satisfy the above supercongruences.

			The square array begins
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  1   5   19     85     376      1715       7890   ...     (A348410)
  2  |  1   7   46    327    2376     17602     132056   ...
  3  |  1   9   82    793    7876     79686     816684   ...
  4  |  1  11  127   1547   19376    247205    3195046   ...
  5  |  1  13  181   2653   40001    614389    9560097   ...
  6  |  1  15  244   4175   73501   1318236   23952720   ...
  7  |  1  17  316   6177  124251   2546288   52867620   ...
  8  |  1  19  397   8723  197251   4544407  106076867   ...
  9  |  1  21  487  11877  298126   7624551  197571088   ...
 10  |  1  23  586  15703  433126  12172550  346618308   ...
Array extended to negative values of n:
 n\k |  1   2    3      4       5         6          7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
 -5  |  1  -7   46   -247     626      8642    -194480   ...
 -4  |  1  -5   19     -5    -874     11569    -105300   ...
 -3  |  1  -3    1     77    -749      4641     -19893   ...
 -2  |  1  -1   -8     63    -249       440       1716   ...
 -1  |  1   1   -8     17       1      -116        344   ...     (-A234839)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Row 1 A348410. Cf. A362724 - A362733, A363418.

# display as a square array
T := (n,k) -> (1/n)*add( (-1)^(k+j) * binomial(-n*k,j)*binomial(-n*k, k-2*j) , j = 0..floor(k/2)): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
# display as a sequence
seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

T(n,k) = (1/n)*Sum_{j = 0..floor(k/2)} binomial(n*k + j - 1, j)*binomial((n+1)*k - 2*j - 1, k - 2*j).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)(1 - x^2))^n.
T(n,k) = (1/n)*binomial(n*k+k-1,k) * hypergeom([n*k, -k/2, (1 - k)/2], [(1 - (n+1)*k)/2, (2 - (n+1)*k)/2], 1) except when n = 1 and k = 1 or 2.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = 1/((1 - x)*(1 - x^2)), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 21, 84, 84, 1, 36, 270, 660, 495, 1, 55, 660, 2860, 5005, 3003, 1, 78, 1365, 9100, 27300, 37128, 18564, 1, 105, 2520, 23800, 107100, 244188, 271320, 116280, 1, 136, 4284, 54264, 339150, 1139544, 2089164, 1961256, 735471, 1, 171, 6840, 111720, 921690, 4239774, 11306064

Offset: 0

Peter Bala, Feb 13 2024

Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).

			Triangle begins
n\k| 0    1     2      3       4       5       6       7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 0 | 1
 1 | 1    3
 2 | 1   10    15
 3 | 1   21    84     84
 4 | 1   36   270    660     495
 5 | 1   55   660   2860    5005    3003
 6 | 1   78  1365   9100   27300   37128   18564
 7 | 1  105  2520  23800  107100  244188  271320  116280
 ...

A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).

Cf. A026000, A063007, A102537, A110608, A339710.

seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);

n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
(x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
Analog of Rodrigues' formula for the shifted Legendre polynomials:
R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
Analog of Rodrigues' formula for the Legendre polynomials:
R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
Orthogonality properties:
Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
Recurrence for row polynomials:
2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
(2^n)*R(n, -1/2) = A234839(n).

cp's OEIS Frontend

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)(1 - x^2))^(nk) for n, k >= 1.

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.

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cp's OEIS Frontend

A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)*(1 - x^2))^(n*k) for n, k >= 1.

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.

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A363419 Square array read by ascending antidiagonals: T(n,k) = 1/n * [x^k] 1/((1 - x)(1 - x^2))^(nk) for n, k >= 1.

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.