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%I #47 May 28 2023 01:59:48
%S 1,2,26,506,11482,284002,7426610,201922730,5650739930,161686253810,
%T 4708709084026,139111173397066,4159013698117618,125595645802182818,
%U 3825428523179727266,117382025506323434506,3625185567639373456090,112597953571519245194770
%N Number of permutations of n copies of 1..4 with all adjacent differences <= 1 in absolute value.
%C 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
%H Alois P. Heinz, <a href="/A177316/b177316.txt">Table of n, a(n) for n = 0..656</a> (terms n=1..31 from R. H. Hardin)
%F From _Peter Bala_, Jan 14 2020: (Start)
%F Conjecture: a(n) = (1/3)*( A005259(n) + A005259(n-1) ).
%F 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)
%F a(n) = 2*A352653(n) for n >= 1. - _Peter Bala_, Apr 18 2022
%F a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1). - _Peter Luschny_, Mar 27 2023
%F a(n) ~ (1 + sqrt(2))^(4*n) / (2^(5/4) * Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Mar 29 2023
%p a:= proc(n) option remember; `if`(n<3, [1, 2, 26][n+1],
%p (3*((105*n^4-356*n^3+402*n^2-208*n+43)*a(n-1)
%p -(105*n^4-904*n^3+2868*n^2-3932*n+1930)*a(n-2))
%p +(9*n-11)*(n-3)^3*a(n-3))/((9*n-16)*n^3))
%p end:
%p seq(a(n), n=0..23); # _Alois P. Heinz_, Jan 22 2020
%p A177316 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1):
%p seq(simplify(A177316(n)), n = 0..17); # _Peter Luschny_, Mar 27 2023
%t a[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1];
%t Table[a[n], {n, 0, 17}] (* _Jean-François Alcover_, May 28 2023, after _Peter Luschny_ *)
%o (Python)
%o def A177316(n):
%o if n == 0: return 1
%o m, g = 1, 0
%o for k in range(n+1):
%o g += m*n**2//(n+k)**2
%o m *= ((n+k+1)*(n-k))**2
%o m //= (k+1)**4
%o return g # _Chai Wah Wu_, Oct 03 2022
%Y Cf. A005259, A103882, A177317 - A177328, A352653.
%Y Row n=4 of A331562.
%K nonn
%O 0,2
%A _R. H. Hardin_, May 06 2010
%E a(0)=1 prepended by _Alois P. Heinz_, Jan 20 2020
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