A362087 -id:A362087 - cp's OEIS frontend

A350383 a(n) = [x^n] 1/(1 + x + x^2)^n.

1, -1, 1, 2, -15, 49, -98, 48, 561, -2860, 8151, -12948, -9282, 149226, -594320, 1428952, -1448655, -5538975, 37450900, -122995950, 239589735, -37528755, -1886983020, 8939152560, -24579514050, 35197176924, 51580335366, -541312482256, 2033695030128, -4624358661240

Offset: 0

Seiichi Manyama, Dec 29 2021

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A002426, A027907, A103779, A110556, A165817, A213684, A228960, A350406, A350407, A362087.

a := n -> (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1): seq(simplify(a(n)), n = 0..30); # Peter Bala, Apr 17 2023

a[n_] := Coefficient[Series[1/(1 + x + x^2)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 3*k));

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,3*k).
Recurrence: 3*(n-1)*n*(4*n - 7)*a(n) = -2*(n-1)*(28*n^2 - 63*n + 27)*a(n-1) - 3*(3*n - 5)*(3*n - 4)*(4*n - 3)*a(n-2). - Vaclav Kotesovec, Mar 18 2023
From Peter Bala, Apr 15 2023: (Start)
a(n) = (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. Cf. A228960.
More generally, let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where g(x) and h(x) are both finite products of cyclotomic polynomials. Then we conjecture that the same supercongruences hold, except for a finite number of primes p depending on f(x), for the sequence {a_(k,m,f)(n): n >= 0} defined by a_(k,m,f)(n) = [x^(k*n)] f(x)^(m*n). (End)
From Peter Bala, Mar 11 2025: (Start)
G.f.: A(x) = 1 + x*d/dx(log(G(x)/x)), where G(x) = x - x^2 + x^3 - 4*x^5 + 14*x^6 - 30*x^7 + ... is the g.f. of A103779.
The following formulas hold for n >= 1:
a(n) = [x^n] T(2*n, (1 - x)/2), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = Sum_{k = 0..n} (-1)^(n+k) * n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).
a(n) = (1/2)*(-1)^n*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], 4). Cf. A213684. (End)

A362078 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = [x^n] 1/(1 - x*(1+x)^k)^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 5, 10, 1, 1, 7, 22, 35, 1, 1, 9, 37, 105, 126, 1, 1, 11, 55, 215, 511, 462, 1, 1, 13, 76, 369, 1271, 2534, 1716, 1, 1, 15, 100, 571, 2526, 7651, 12720, 6435, 1, 1, 17, 127, 825, 4401, 17577, 46614, 64449, 24310, 1, 1, 19, 157, 1135, 7026, 34412, 123810, 286599, 328900, 92378

Offset: 0

Seiichi Manyama, Apr 08 2023

			Square array begins:
    1,   1,    1,    1,    1,    1, ...
    1,   1,    1,    1,    1,    1, ...
    3,   5,    7,    9,   11,   13, ...
   10,  22,   37,   55,   76,  100, ...
   35, 105,  215,  369,  571,  825, ...
  126, 511, 1271, 2526, 4401, 7026, ...

Columns k=0..3 give A088218, A213684, A362087, A362088.
Main diagonal gives A362080.
Cf. A362079.

T(n, k) = sum(j=0, n, binomial(n+j-1, j)*binomial(k*j, n-j));

T(n,k) = Sum_{j=0..n} (-1)^j * binomial(-n,j) * binomial(k*j,n-j) = Sum_{j=0..n} binomial(n+j-1,j) * binomial(k*j,n-j).

A378565 a(n) = Sum_{k=0..n} binomial(n+k-1,k) * binomial(n+k-1,n-k).

Original entry on oeis.org

1, 1, 7, 43, 271, 1746, 11425, 75615, 504799, 3392953, 22930282, 155664356, 1060710457, 7250779238, 49700101101, 341474150583, 2351032782783, 16216401440106, 112035931072915, 775163096510445, 5370301986029066, 37249469056575504, 258648802856972348

Offset: 0

Seiichi Manyama, Dec 01 2024

nonn

Cf. A002002, A378566, A378567.

Cf. A011270, A362087.

Table[Sum[Binomial[n+k-1, k] * Binomial[n+k-1, 2*k-1], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Dec 01 2024 *)

a(n) = sum(k=0, n, binomial(n+k-1, k)*binomial(n+k-1, n-k));

a(n) = [x^n] 1/(1 - x/(1 - x)^2)^n.

a(n) ~ (525 - 32*210^(2/3)/(157*sqrt(105) - 1575)^(1/3) + 4*(210*(157*sqrt(105) - 1575))^(1/3))^(1/6) * ((36 + (1208682 - 28350*sqrt(105))^(1/3)/3 + (6*(7461 + 175*sqrt(105)))^(1/3))^n / (2^(2/3) * 7^(1/3) * sqrt(Pi*n) * 3^(n + 1/6) * 5^(n + 1/3))). - Vaclav Kotesovec, Dec 01 2024

cp's OEIS Frontend

A350383 a(n) = [x^n] 1/(1 + x + x^2)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A362078 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = [x^n] 1/(1 - x*(1+x)^k)^n.

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A378565 a(n) = Sum_{k=0..n} binomial(n+k-1,k) * binomial(n+k-1,n-k).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

PARI

Formula