A177316 Number of permutations of n copies of 1..4 with all adjacent differences <= 1 in absolute value.

1, 2, 26, 506, 11482, 284002, 7426610, 201922730, 5650739930, 161686253810, 4708709084026, 139111173397066, 4159013698117618, 125595645802182818, 3825428523179727266, 117382025506323434506, 3625185567639373456090, 112597953571519245194770

Offset: 0

Author

R. H. Hardin, May 06 2010

Keywords

• nonn

Comments

See A103882 and A177317 through A177328 for the number of permutations of n copies of 1..k (for different values of k) with adjacent differences restricted in size. We conjecture that all these sequences satisfy the congruences A(n*p^k) == A(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k and any prime p >= 5. - Peter Bala, Jan 16 2020

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..656 (terms n=1..31 from R. H. Hardin)

Crossrefs

Cf. A005259, A103882, A177317 - A177328, A352653.

Row n=4 of A331562.

Programs

Maple

a:= proc(n) option remember; `if`(n<3, [1, 2, 26][n+1],
       (3*((105*n^4-356*n^3+402*n^2-208*n+43)*a(n-1)
      -(105*n^4-904*n^3+2868*n^2-3932*n+1930)*a(n-2))
      +(9*n-11)*(n-3)^3*a(n-3))/((9*n-16)*n^3))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Jan 22 2020
A177316 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1):
seq(simplify(A177316(n)), n = 0..17); # Peter Luschny, Mar 27 2023

Mathematica

a[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, May 28 2023, after Peter Luschny *)

Python

def A177316(n):
    if n == 0: return 1
    m, g = 1, 0
    for k in range(n+1):
        g += m*n**2//(n+k)**2
        m *= ((n+k+1)*(n-k))**2
        m //= (k+1)**4
    return g # Chai Wah Wu, Oct 03 2022

Formula

From Peter Bala, Jan 14 2020: (Start)

Conjecture: a(n) = (1/3)*( A005259(n) + A005259(n-1) ).

Equivalently, a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2. Cf. A103882. If true, then the sequence satisfies the recurrence a(n) = (2*(102*n^6 - 612*n^5 + 1462*n^4 - 1768*n^3 + 1143*n^2 - 382*n+52) * a(n-1) - (2*n-1)*(3*n^2 - 3*n+1) * (n-2)^3 * a(n-2)) / (n^3*(2*n - 3) * (3*n^2 - 9*n+7)) and the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k and any prime p >= 5. [added Apr 18 2022: assuming the recurrence given in the Maple program below is correct then these conjectures are true.] (End)

a(n) = 2*A352653(n) for n >= 1. - Peter Bala, Apr 18 2022

a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1). - Peter Luschny, Mar 27 2023

a(n) ~ (1 + sqrt(2))^(4*n) / (2^(5/4) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Jan 20 2020