A361281 -id:A361281 - cp's OEIS frontend

A278070 a(n) = hypergeometric([n, -n], [], -1).

1, 2, 11, 106, 1457, 25946, 566827, 14665106, 438351041, 14862109042, 563501581931, 23624177026682, 1085079390005041, 54185293223976266, 2922842896378005707, 169366580127359119906, 10492171932362920604417, 691986726674000405367266, 48408260338825019327539531

Offset: 0

Peter Luschny, Nov 10 2016

From Peter Bala, Mar 12 2023: (Start)
We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 1, 1, 4, 2, 1, 2, 4, 1, 1, 4, 2, ...], apparently a periodic sequence of period 7.
More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x))*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k. The present sequence is the case F(x) = 1/(1 - x), G(x) = 1. Cf. A361281. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..370

Cf. A278069, A278071, A361281.

a := n -> hypergeom([n, -n], [], -1): seq(simplify(a(n)), n=0..18);
# Alternatively:
a := proc(n) option remember; `if`(n<2, n+1,
((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3)) end:
seq(a(n), n=0..18);

Table[HypergeometricPFQ[{n, -n}, {}, -1], {n, 0, 20}] (* Vaclav Kotesovec, Nov 10 2016 *)

a(n):=n!*sum(binomial(2*n-i-1,n-i)/i!,i,0,n); /* Vladimir Kruchinin, Nov 23 2016 */

def a():
    a, b, c, d, h, e = 1, 2, 1, 8, 4, 0
    yield a
    while True:
        yield b
        e = c; c += 2
        a, b = b, (c*a + h*b)//e
        d += 16; h += d
A278070 = a()
[next(A278070) for _ in range(19)]

a(-n) = a(n).
a(n) = n! [x^n] exp((1-h(x))/2)*(1+h(x))/(2*h(x)) with h(x) = sqrt(1-4*x).
a(n) = ((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3) for n>=2.
a(n) ~ 2^(2*n-1/2) * n^n / exp(n-1/2). - Vaclav Kotesovec, Nov 10 2016
a(n) = n!*Sum_{i=0..n}(binomial(2*n-i-1,n-i)/i!). - Vladimir Kruchinin, Nov 23 2016
a(n) = n! * [x^n] exp(x)/(1 - x)^n. - Ilya Gutkovskiy, Sep 21 2017

A293013 a(n) = n! * [x^n] exp(x/(1 - x)^n).

Original entry on oeis.org

1, 1, 5, 55, 961, 24101, 818821, 36053515, 1984670465, 132825475081, 10583425959301, 988018789759871, 106673677280748865, 13172700275176482925, 1842428769970603518341, 289406832942160060794451, 50677793314733587473331201, 9829328870566195730521433105

Offset: 0

Ilya Gutkovskiy, Sep 28 2017

Conjecture: a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. - Peter Bala, Mar 12 2023

Seiichi Manyama, Table of n, a(n) for n = 0..274

Main diagonal of A293012. Cf. A361281.

Table[n! SeriesCoefficient[Exp[x/(1 - x)^n] , {x, 0, n}], {n, 0, 17}]
(* or *)
nmax = 20; Join[{1}, Table[n!*Sum[Binomial[(n-1)*(k+1), k*n - 1]/k!, {k, 1, n}], {n, 1, nmax}]] (* Vaclav Kotesovec, Aug 24 2025 *)

a(n) = A293012(n,n).
For n > 0, a(n) = n! * Sum_{k=1..n} binomial((n-1)*(k+1), k*n - 1)/k!. - Vaclav Kotesovec, Aug 24 2025
log(a(n)) ~ n * (2*log(n) - log(log(n)) - 1 - log(2) + log(log(n))/log(n) + (1 + 2*log(2))/(2*log(n))). - Vaclav Kotesovec, Aug 25 2025

A361277 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..n} binomial(k*j,n-j)/j!.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 7, 1, 1, 1, 7, 19, 25, 1, 1, 1, 9, 37, 97, 81, 1, 1, 1, 11, 61, 241, 581, 331, 1, 1, 1, 13, 91, 481, 1981, 3661, 1303, 1, 1, 1, 15, 127, 841, 4881, 17551, 26335, 5937, 1, 1, 1, 17, 169, 1345, 10001, 55321, 171697, 202049, 26785, 1

Offset: 0

Seiichi Manyama, Mar 06 2023

			Square array begins:
  1,  1,   1,    1,    1,     1, ...
  1,  1,   1,    1,    1,     1, ...
  1,  3,   5,    7,    9,    11, ...
  1,  7,  19,   37,   61,    91, ...
  1, 25,  97,  241,  481,   841, ...
  1, 81, 581, 1981, 4881, 10001, ...

Columns k=0..4 give A000012, A047974, A361278, A361279, A361280.
Main diagonal gives A361281.
Cf. A293012.

T(n, k) = n!*sum(j=0, n, binomial(k*j, n-j)/j!);

E.g.f. of column k: exp(x * (1+x)^k).

T(0,k) = 1; T(n,k) = (n-1)! * Sum_{j=1..n} j * binomial(k,j-1) * T(n-j,k)/(n-j)!.

A361036 a(n) = n! * [x^n] (1 + x)^n * exp(x*(1 + x)^n).

Original entry on oeis.org

1, 2, 11, 124, 2225, 56546, 1928707, 85029596, 4687436609, 314255427490, 25077179715131, 2343489559096412, 253185531592066801, 31279831940279656514, 4376923336721600128115, 687815536092999747916156, 120491486068612766739548417, 23378730923206887237941740226

Offset: 0

Peter Bala, Mar 13 2023

We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, ...], apparently a periodic sequence of period 7.

More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x)^n)*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k.

Vaclav Kotesovec, Table of n, a(n) for n = 0..270

Cf. A047974, A278070, A361281.

seq( n!*add(add(binomial(n,i+j)*binomial(j*n,i)/j!, j = 0..n-i), i = 0..n), n = 0..20);

Table[n! * Sum[Sum[Binomial[n, i + j]*Binomial[j*n, i]/j!, {j, 0, n - i}], {i, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 27 2023 *)

a(n) = n!*Sum_{i = 0..n} Sum_{j = 0..n-i} binomial(n,i+j)*binomial(j*n,i)/j!.

a(n) ~ n! * exp(r*(1+r)^n) * (1+r)^(n/2 + 1) / (sqrt(2*Pi*n*(3 + n*r)) * r^(n+1)), where r = 2*LambertW(n/2)/n - (n + 2*LambertW(n/2)) * (n - 4*LambertW(n/2)^3) / (n^3 * (3 + 2*LambertW(n/2))). - Vaclav Kotesovec, Mar 28 2023

cp's OEIS Frontend

A278070 a(n) = hypergeometric([n, -n], [], -1).

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A293013 a(n) = n! * [x^n] exp(x/(1 - x)^n).

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A361277 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..n} binomial(k*j,n-j)/j!.

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A361036 a(n) = n! * [x^n] (1 + x)^n * exp(x*(1 + x)^n).

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