A278070 -id:A278070 - cp's OEIS frontend

A361281 a(n) = n! * Sum_{k=0..n} binomial(n*k,n-k)/k!.

1, 1, 5, 37, 481, 10001, 288901, 10820965, 511186817, 29843419681, 2106779832901, 176180844038981, 17165338119936865, 1924030148121500017, 245630480526435293381, 35409038825312233143301, 5719025066628373334423041, 1027649751647068260334391105

Offset: 0

Seiichi Manyama, Mar 06 2023

nonn

From  Peter Bala, Mar 12 2023: (Start)
It appears that a(n) == 1 (mod 4) and a(5*n+2) == 0 (mod 5) for all n. More generally 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.
Let F(x) and G(x) be power series with integer coefficients with G(0) = 1. Define b(n) = n! * [x^n] F(x)*exp(x*G(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, G(x) = 1 + x. Cf. A278070. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..275

Main diagonal of A361277.

Cf. A096131, A099237, A226391, A278070, A293013.

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

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

log(a(n)) ~ n*(2*log(n) - log(log(n)) - 1 - log(2) + log(log(n))/log(n) + 1/(2*log(n)) + log(2)/log(n) - 1/(8*log(n)^2)). - Vaclav Kotesovec, Mar 12 2023

A278069 a(n) = hypergeometric([n, -n], [], 1).

Original entry on oeis.org

1, 0, 3, -32, 465, -8544, 190435, -4996032, 150869313, -5155334720, 196677847971, -8286870547680, 382200680031313, -19152276311294112, 1036167879649219395, -60195061159370501504, 3737352803142621672705, -246970483156591884266112, 17306865588065164490357443

Offset: 0

Peter Luschny, Nov 10 2016

sign

Cf. A278070, A278071.

a := n -> hypergeom([n,-n], [], 1): seq(simplify(a(n)), n=0..18);
# Alternatively:
a := proc(n) option remember; `if`(n<2,1-n,
((2*n-1)*a(n-2)-8*(1+n*(n-2))*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={};For[n=0,n<19,n++,AppendTo[a,(-1)^n*Sum[(-1)^(j-n+1-Mod[n,2])*Product[(2*n-k)*k/(n-k+1),{k,j,n}],{j,1,n+1}]]]; a (* Detlef Meya, Sep 05 2023 *)

def a():
    a, b, c, d, h, e = 1, 0, 1, 8, 8, 0
    yield a
    while True:
        yield b
        e = c; c += 2
        a, b = b, (c*a - h*b)//e
        d += 16; h += d
A278069 = a()
[next(A278069) for _ in range(19)]

a(-n) = a(n).
a(n) = n! [x^n] (2*x*exp(h(x)/2))/(4*x-h(x)) with h(x) = sqrt(4*x+1)-1.
a(n) ~ (-1)^n * 2^(2*n-1/2) * n^n / exp(n+1/2). - Vaclav Kotesovec, Nov 10 2016
(-9-3*n)*a(n+1)+(12*n^2+53*n+52)*a(n+2)+(4*n^2+33*n+63)*a(n+3)+(n+4)*a(n+4) = 0. - Robert Israel, Nov 10 2016
a(n) = ((2*n-1)*a(n-2)-8*(1+n*(n-2))*a(n-1))/(2*n-3) for n>=2. - Peter Luschny, Nov 10 2016
a(n) = n! * [x^n] exp(x)/(1 + x)^n. - Ilya Gutkovskiy, Apr 07 2018

A278071 Triangle read by rows, coefficients of the polynomials P(n,x) = (-1)^n*hypergeom( [n,-n], [], x), powers in descending order.

Original entry on oeis.org

1, 1, -1, 6, -4, 1, 60, -36, 9, -1, 840, -480, 120, -16, 1, 15120, -8400, 2100, -300, 25, -1, 332640, -181440, 45360, -6720, 630, -36, 1, 8648640, -4656960, 1164240, -176400, 17640, -1176, 49, -1, 259459200, -138378240, 34594560, -5322240, 554400, -40320, 2016, -64, 1

Offset: 0

Peter Luschny, Nov 10 2016

			Triangle starts:
.       1,
.       1,      -1,
.       6,      -4,     1,
.      60,     -36,     9,    -1,
.     840,    -480,   120,   -16,   1,
.   15120,   -8400,  2100,  -300,  25,  -1,
.  332640, -181440, 45360, -6720, 630, -36, 1,
...

H. L. Krall and O. Fink, A New Class of Orthogonal Polynomials: The Bessel Polynomials, Trans. Amer. Math. Soc. 65, 100-115, 1949.
Herbert E. Salzer, Orthogonal Polynomials Arising in the Numerical Evaluation of Inverse Laplace Transforms, Mathematical Tables and Other Aids to Computation, Vol. 9, No. 52 (Oct., 1955), pp. 164-177, (see p.174 and footnote 7).

Cf. A278069 (x=1, row sums up to sign), A278070 (x=-1).
T(n,0) = Pochhammer(n, n) (cf. A000407).
T(n,1) = -(n+1)*(2n)!/n! (cf. A002690).
T(n,2) = (n+2)*(2n+1)*(2n-1)!/(n-1)! (cf. A002691).
T(n,n-1) = (-1)^(n+1)*n^2 for n>=1 (cf. A000290).
T(n,n-2) = n^2*(n^2-1)/2 for n>=2 (cf. A083374).

p := n -> (-1)^n*hypergeom([n, -n], [], x):
ListTools:-Flatten([seq(PolynomialTools:-CoefficientList(simplify(p(n)), x, termorder=reverse), n=0..8)]);
# Alternatively the polynomials by recurrence:
P := proc(n,x) if n=0 then return 1 fi; if n=1 then return x-1 fi;
((((4*n-2)*(2*n-3)*x+2)*P(n-1,x)+(2*n-1)*P(n-2,x))/(2*n-3));
sort(expand(%)) end: for n from 0 to 6 do lprint(P(n,x)) od;
# Or by generalized Laguerre polynomials:
P := (n,x) -> n!*(-x)^n*LaguerreL(n,-2*n,-1/x):
for n from 0 to 6 do simplify(P(n,x)) od;

row[n_] := CoefficientList[(-1)^n HypergeometricPFQ[{n, -n}, {}, x], x] // Reverse;
Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jul 12 2019 *)
(* T(n,k)= *) t={};For[n=8,n>-1,n--,For[j=n+1,j>0,j--,PrependTo[t,(-1)^(j-n+1-Mod[n,2])*Product[(2*n-k)*k/(n-k+1),{k,j,n}]]]];t (* Detlef Meya, Aug 02 2023 *)

The P(n,x) are orthogonal polynomials. They satisfy the recurrence
P(n,x) = ((((4*n-2)*(2*n-3)*x+2)*P(n-1,x)+(2*n-1)*P(n-2,x))/(2*n-3)) for n>=2.
In terms of generalized Laguerre polynomials (see the Krall and Fink link):
P(n,x) = n!*(-x)^n*LaguerreL(n,-2*n,-1/x).

A370706 Triangle read by rows: T(n, k) = binomial(n, k) * Pochhammer(n, k).

Original entry on oeis.org

1, 1, 1, 1, 4, 6, 1, 9, 36, 60, 1, 16, 120, 480, 840, 1, 25, 300, 2100, 8400, 15120, 1, 36, 630, 6720, 45360, 181440, 332640, 1, 49, 1176, 17640, 176400, 1164240, 4656960, 8648640, 1, 64, 2016, 40320, 554400, 5322240, 34594560, 138378240, 259459200

Offset: 0

Peter Luschny, Feb 28 2024

			Triangle starts:
  [0] 1;
  [1] 1,  1;
  [2] 1,  4,    6;
  [3] 1,  9,   36,    60;
  [4] 1, 16,  120,   480,    840;
  [5] 1, 25,  300,  2100,   8400,   15120;
  [6] 1, 36,  630,  6720,  45360,  181440,  332640;
  [7] 1, 49, 1176, 17640, 176400, 1164240, 4656960, 8648640;

Cf. A370707, A000407 (main diagonal), A278070 (row sums).

T := (n, k) -> binomial(n, k)*pochhammer(n, k):
seq(seq(T(n, k), k = 0..n), n = 0..8);

T[n_, k_] := Binomial[n, k] Pochhammer[n, k];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten

T(n, k) = A370707(n, k) / k!.

T(n, n) = Pochhammer(n, n) for n >= 0 (which is different from A000407(n)).

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

A361281 a(n) = n! * Sum_{k=0..n} binomial(n*k,n-k)/k!.

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A278069 a(n) = hypergeometric([n, -n], [], 1).

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A278071 Triangle read by rows, coefficients of the polynomials P(n,x) = (-1)^n*hypergeom( [n,-n], [], x), powers in descending order.

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A370706 Triangle read by rows: T(n, k) = binomial(n, k) * Pochhammer(n, k).

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

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