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 A177316 Number of permutations of n copies of 1..4 with all adjacent differences <= 1 in absolute value. 3

%I

%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 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. - _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. (End)

%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

%Y Cf. A005259, A103882, A177317 - A177328.

%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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Last modified July 29 06:21 EDT 2021. Contains 346340 sequences. (Running on oeis4.)