A378907 Number of permutations of [n] with at least one hit on both main diagonals.
0, 1, 0, 2, 10, 48, 270, 2004, 15406, 144656, 1399070, 15924940, 185817038, 2485431096, 33966603790, 522088434644, 8178526719550, 142034596036896, 2508925152633918, 48582127821078684, 955299461042098222, 20406401587894276040, 442067447198146300718
Offset: 0
Examples
For n = 3, the a(3) = 2 solutions are: X . . . . X . X . . X . . . X X . . For n = 4, one of a(4) = 10 solutions is: X . . . . . X . . X . . . . . X All a(4) = 10 permutations of 1..4 counted are: 1324, 1342, 1423, 2314, 2431, 3124, 3241, 4132, 4213, 4231.
Programs
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PARI
\\ here B(n) gives first terms of A003471 as vector. B(n)={my(a=vector(n+1)); a[1]=1; for(n=4, n, my(m=2-n%2); a[1+n] = (n-1)*a[n] + 2*(n-m)*a[1+n-2*m]); a} seq(n)={Vec(serlaplace((1 - 2*exp(-x + O(x*x^n)))/(1-x)) + Ser(B(n)), -n-1)} \\ Andrew Howroyd, Dec 11 2024
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Python
def generate_A378907(n): f, A000166, A378907 = 6, 2, [0, 1, 0, 2] A003471 = [1, 0, 0, 0] for i in range(4, n): f *= i A000166 = i*A000166 + (1 if i%2==0 else -1) d, e = (2,4) if i%2==0 else (1,2) A003471.append((i-1)*A003471[-1] + 2*(i-d)*A003471[-e]) A003471 = A003471[1:] s = f - 2*A000166 + A003471[-1] A378907.append(s) return A378907 A378907 = generate_A378907(23) # Jwalin Bhatt, Dec 27 2024
Comments