A122851 -id:A122851 - cp's OEIS frontend

A122852 Row sums of number triangle A122851.

1, 1, 2, 3, 6, 11, 24, 51, 122, 291, 756, 1979, 5526, 15627, 46496, 140451, 442194, 1414931, 4687212, 15785451, 54764846, 193129659, 698978136, 2570480147, 9672977706, 36967490691, 144232455524, 571177352091, 2304843053382, 9434493132011, 39289892366736

Offset: 0

Paul Barry, Sep 14 2006

Essentially the same as A072374. - R. J. Mathar, Jun 18 2008

Diagonal sums of A008279. - Paul Barry, Feb 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jonathan Fang, Zachary Hamaker, and Justin Troyka, On pattern avoidance in matchings and involutions, arXiv:2009.00079 [math.CO], 2020. See Theorem 1.6 (b).
Guo-Niu Han, Hankel Continued fractions and Hankel determinants of the Euler numbers, arXiv:1906.00103 [math.CO], 2019. See p. 27.
Qiong Qiong Pan and Jiang Zeng, The gamma-coefficients of Branden's (p,q)-Eulerian polynomials and André permutations, arXiv:1910.01747 [math.CO], 2019.

Cf. A072374, A008279.

Cf. A357532, A357533, A357570.

Table[Sum[Binomial[n-k,k]*k!,{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Feb 08 2014 *)

a(n) = sum(k=0, n, binomial(k,n-k)*(n-k)!); \\ Michel Marcus, Sep 02 2020

a(n) = Sum{k=0..n} C(k,n-k)*(n-k)!.
From Paul Barry, Feb 11 2009: (Start)
G.f.: 1/(1-x-x^2/(1-x^2/(1-x-2x^2/(1-2x^2/(1-x-3x^2/(1-3x^2/(1-x-4x^2/(1-4x^2/(1-... (continued fraction).
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*k!. (End)
D-finite with recurrence -2*a(n) + 3*a(n-1) + (n-1)*a(n-2) + (-n+1)*a(n-3) = 0. - R. J. Mathar, Nov 15 2012. Proof in [Han 2019]
a(n) ~ sqrt(Pi) * exp(sqrt(n/2) - n/2 + 1/8) * n^((n+1)/2) / 2^(n/2+1) * (1 + 37/(48*sqrt(2*n))). - Vaclav Kotesovec, Feb 08 2014
a(n) = (a(n-1) + n * a(n-2) + 1)/2 for n > 1. - Seiichi Manyama, Nov 19 2022

More terms from Vaclav Kotesovec, Jun 04 2019

A242653 Triangle read by rows: T(n,k) = ((n+k)/2)!/k! if n,k have same parity, otherwise 0.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 6, 0, 4, 0, 1, 6, 0, 12, 0, 5, 0, 1, 0, 24, 0, 20, 0, 6, 0, 1, 24, 0, 60, 0, 30, 0, 7, 0, 1, 0, 120, 0, 120, 0, 42, 0, 8, 0, 1, 120, 0, 360, 0, 210, 0, 56, 0, 9, 0, 1, 0, 720, 0, 840, 0, 336, 0, 72, 0, 10, 0, 1, 720, 0, 2520, 0, 1680, 0, 504, 0, 90, 0, 11, 0, 1

Offset: 0

N. J. A. Sloane, May 29 2014

			Triangle begins:
  1
  0  1
  1  0  1
  0  2  0  1
  2  0  3  0 1
  0  6  0  4 0 1
  6  0 12  0 5 0 1
  0 24  0 20 0 6 0 1
  ...

Robert Israel, Table of n, a(n) for n = 0..10010
Alexander Kreinin, Combinatorial Properties of Mills' Ratio, arXiv:1405.5852, 2014. See Table 4.

Cf. A122851.

N:= 1000; # to get a(0) to a(N)
count:= -1;
for n from 0 while count < N do
  for k from 0 to n while count  < N do
    count:= count+1;
    if type(n-k,even) then
       A[count]:= ((n+k)/2)!/k!
    else
       A[count]:= 0
    fi;
  od
od:
seq(A[i],i=0..N); # Robert Israel, Jun 10 2014

Table[If[EvenQ[n-k], ((n+k)/2)!/k!, 0], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 19 2018 *)

A344391 T(n, k) = binomial(n - k, k) * factorial(k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 2, 1, 4, 6, 1, 5, 12, 6, 1, 6, 20, 24, 1, 7, 30, 60, 24, 1, 8, 42, 120, 120, 1, 9, 56, 210, 360, 120, 1, 10, 72, 336, 840, 720, 1, 11, 90, 504, 1680, 2520, 720, 1, 12, 110, 720, 3024, 6720, 5040, 1, 13, 132, 990, 5040, 15120, 20160, 5040

Offset: 0

Peter Luschny, May 17 2021

The antidiagonal representation of the falling factorials (A008279).

			[ 0] [1]
[ 1] [1]
[ 2] [1,  1]
[ 3] [1,  2]
[ 4] [1,  3,  2]
[ 5] [1,  4,  6]
[ 6] [1,  5, 12,   6]
[ 7] [1,  6, 20,  24]
[ 8] [1,  7, 30,  60,  24]
[ 9] [1,  8, 42, 120, 120]
[10] [1,  9, 56, 210, 360, 120]
[11] [1, 10, 72, 336, 840, 720]

Cf. A122852 (row sums).

Cf. A008279, A122851.

T := (n, k) -> pochhammer(n + 1 - 2*k, k):
seq(print(seq(T(n, k), k=0..n/2)), n = 0..11);

def T(n, k): return rising_factorial(n + 1 - 2*k, k)
def T(n, k): return (-1)^k*falling_factorial(2*k - n - 1, k)
def T(n, k): return binomial(n - k, k) * factorial(k)
print(flatten([[T(n, k) for k in (0..n//2)] for n in (0..11)]))

T(n, k) = RisingFactorial(n + 1 - 2*k, k).

T(n, k) = (-1)^k*FallingFactorial(2*k - n - 1, k).

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A122852 Row sums of number triangle A122851.

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A242653 Triangle read by rows: T(n,k) = ((n+k)/2)!/k! if n,k have same parity, otherwise 0.

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A344391 T(n, k) = binomial(n - k, k) * factorial(k), for n >= 0 and 0 <= k <= floor(n/2). Triangle read by rows.

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