A324152 -id:A324152 - cp's OEIS frontend

A324465 Exponent of highest power of 2 that divides A324152(n).

0, 0, 1, 3, 2, 2, 3, 5, 2, 3, 4, 6, 5, 4, 5, 7, 2, 3, 4, 6, 5, 5, 6, 8, 5, 6, 7, 9, 8, 6, 7, 9, 2, 3, 4, 6, 5, 5, 6, 8, 5, 6, 7, 9, 8, 7, 8, 10, 5, 6, 7, 9, 8, 8, 9, 11, 8, 9, 10, 12, 11, 8, 9, 11, 2, 3, 4, 6, 5, 5, 6, 8, 5, 6, 7, 9, 8, 7, 8, 10, 5, 6, 7, 9

Offset: 0

N. J. A. Sloane, Mar 01 2019

nonn

First occurrence of k=0,1,2,...: 0, 2, 4, 3, 10, 7, 11, 15, 23, 27, 47, 55, 59, 111, 119, 123, 239, 247, 251, 495, 503, 507, 1007, 1015, 1019, 2031, 2039, 2043, 4079, 4087, 4091, 8175, 8183, 8187, 16367, 16375, 16379, 32751, 32759, 32763, 65519, 65527, 65531, 131055, 131063, 131067, ..., . Robert G. Wilson v, Mar 01 2019

Robert G. Wilson v, Table of n, a(n) for n = 0..10000 (first 501 terms from N. J. A. Sloane)

Cf. A000120 (binary weight), A007814, A324152, A324467.

f[n_] := IntegerExponent[(3/((n + 1)(n + 2)(n + 3)))*Multinomial[n, n, n, n], 2]; f[0] = 0; Array[f, 84, 0] (* Robert G. Wilson v, Mar 01 2019 *)

a(n) = 3*hammingweight(n) - valuation((n+1)*(n+2)*(n+3), 2); \\ Michel Marcus, Jul 10 2022

a(n) = 3*wt(n) - (2-adic valuation of (n+1)*(n+2)*(n+3))
= 3*A000120(n) - (A007814(n+1)+A007814(n+2)+A007814(n+3)).
E.g. if n = 14 = 1110_2, with weight 3, we get a(14) = 3*3 - 2-adic valuation of 15*16*17 = 9 - 4 = 5.

A324151 a(n) = (2/((n+1)(n+2)))multinomial(3*n;n,n,n).

Original entry on oeis.org

1, 2, 15, 168, 2310, 36036, 612612, 11085360, 210344706, 4143153300, 84106011990, 1750346095680, 37194854533200, 804553314979680, 17671438882589400, 393345439598342880, 8858467087621013610, 201578121034100464500, 4629577513083174001350, 107211268724031397926000

Offset: 0

Michael De Vlieger and N. J. A. Sloane, Mar 01 2019

nonn

a(n) is an integer, because as Fredes and Sepulveda show, it gives the number of spanning tree decorated quadrangulations rooted in the tree.

For a direct proof, a(n) may also be written as (binomial(3*n,n)/(2*n+1))*(binomial(2*n+2,n)/(n+1)) = A001764(n)*A000108(n+1), and so is an integer. - N. J. A. Sloane, Mar 01 2019

Michael De Vlieger, Table of n, a(n) for n = 0..704
Luis Fredes and Avelio Sepulveda, Tree-decorated planar maps, arXiv:1901.04981 [math.CO], 2019. See Remark 4.6.

Cf. A000108, A001764, A324152.

a:= n-> (2/((n+1)*(n+2)))*combinat[multinomial](3*n, n$3):
seq(a(n), n=0..20);  # Alois P. Heinz, Jan 25 2022

c[m_, n_] := m Product[1/(n + i), {i, m}] (Multinomial @@ ConstantArray[n, m + 1]); Array[c[2, #] &, 20, 0] (* Michael De Vlieger, Mar 01 2019 *)

from sympy.ntheory import multinomial_coefficients
def A324151(n): return 2*multinomial_coefficients(3,3*n)[(n,n,n)]//(n+1)//(n+2) # Chai Wah Wu, Jan 25 2022

A324478 a(n) = (6/((n+1)(n+2)(n+3))) * multinomial(4*n;n,n,n,n).

Original entry on oeis.org

1, 6, 252, 18480, 1801800, 209513304, 27485041584, 3937652896320, 603400560305400, 97512510301206000, 16452310738019476320, 2876570958459008603520, 518262201015698050067520, 95794174581229987212924000, 18101994022606737439599480000

Offset: 0

N. J. A. Sloane, Mar 10 2019, following a suggestion from Luis Fredes and Avelio Sepulveda

nonn

Theorem (Luis Fredes, Mar 04 2019): (Start)
a(n) is an integer for all n >= 0.
Proof:
a(n) = (6/((n+1)*(n+2)*(n+3)))*multinomial(4*n;n,n,n,n) = multinomial(4*n;n,n,n,n) - multinomial(4*n;n+3,n-3,n,n) + 3*multinomial(4*n;n+2,n-2,n,n) + 15*multinomial(4*n;n+3,n-2,n-1,n) - 18*multinomial(4*n;n+3,n-1,n-1,n-1).
The right hand side is equal (after some manipulation) to
    (f(n)/((n+1)*(n+2)*(n+3)))*multinomial(4*n;n,n,n,n)
where f(n) = (n+1)*(n+2)*(n+3) - (n-1)*(n-2)*n + 3*(n-1)*n*(n+3) + 15*(n-1)*n^2 - 18*n^3 = 6. QED (End)
(11!/12)*a(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^2 for all n. More generally, we conjecture that there is a constant C(r) such that C(r)*(4*n)!/(n!*(n+r)!^3) is an integer for all n. Calculation suggests that we may take C(r) = (1/8)*(4*r)!/r! for r >= 1. - Peter Bala, Feb 28 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Luis Fredes and Avelio Sepulveda, Tree-decorated planar maps, arXiv:1901.04981 [math.CO], 2019. See Remark 4.6.

Cf. A324152.

a:= n-> 6*combinat[multinomial](4*n, n$4)/((n+1)*(n+2)*(n+3)):
seq(a(n), n=0..20);  # Alois P. Heinz, Mar 11 2019

c[m_,n_]:=2m Product[1/(n+i), {i, m}] (Multinomial@@ConstantArray[n, m+1]);{1}~Join~Array[c[3, #]&, 20] (* Vincenzo Librandi, Mar 11 2019 *)
Flatten[{1, Table[6*(4*n)! / ((n!)^3 * (n+3)!), {n, 1, 15}]}] (* Vaclav Kotesovec, Jul 21 2019 *)

From Vaclav Kotesovec, Jul 21 2019: (Start)
For n>0, a(n) = 6*(4*n)! / ((n!)^3 * (n+3)!).
a(n) ~ 6 * 2^(8*n - 1/2) / (Pi^(3/2) * n^(9/2)). (End)

cp's OEIS Frontend

A324465 Exponent of highest power of 2 that divides A324152(n).

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A324151 a(n) = (2/((n+1)(n+2)))multinomial(3*n;n,n,n).

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A324478 a(n) = (6/((n+1)(n+2)(n+3))) * multinomial(4*n;n,n,n,n).

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cp's OEIS Frontend

A324465 Exponent of highest power of 2 that divides A324152(n).

Views

Author

Keywords

Comments

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Crossrefs

Programs

Mathematica

PARI

Formula

A324151 a(n) = (2/((n+1)*(n+2)))*multinomial(3*n;n,n,n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

A324478 a(n) = (6/((n+1)*(n+2)*(n+3))) * multinomial(4*n;n,n,n,n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A324151 a(n) = (2/((n+1)(n+2)))multinomial(3*n;n,n,n).

A324478 a(n) = (6/((n+1)(n+2)(n+3))) * multinomial(4*n;n,n,n,n).