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a(n) = n^(n - 1) - 2*(n + 1)^(n - 2), by convention a(0) = 0.
1

%I #13 Dec 28 2024 09:32:00

%S 0,0,0,1,14,193,2974,52113,1034270,23046721,571282238,15617863897,

%T 467291386990,15198954783153,534222097472894,20185726770649633,

%U 816165851488045118,35167910642711951617,1609028732603454196606,77912950297911241532841,3981118415206568940420878

%N a(n) = n^(n - 1) - 2*(n + 1)^(n - 2), by convention a(0) = 0.

%H Steve Butler, Kimberly Hadaway, Victoria Lenius, Preston Martens, and Marshall Moats, <a href="https://arxiv.org/abs/2412.07873">Lucky cars and lucky spots in parking functions</a>, arXiv:2412.07873 [math.CO], 2024. See p. 7.

%F a(n) = A000169(n) - A007334(n+1) for n > 0. In the context of parking functions this is the difference between the main diagonals of A374756 and A379611. See corollary 3.1 and Table 2 in Butler et al.

%p a := n -> ifelse(n = 0, 0, n^(n-1) - 2*(n+1)^(n-2)): seq(a(n), n = 0..20);

%t {0}~Join~Table[n^(n - 1) - 2*(n + 1)^(n - 2), {n, 20}] (* _Michael De Vlieger_, Dec 27 2024 *)

%Y Cf. A000169, A007334, A374756, A379611.

%K nonn,new

%O 0,5

%A _Peter Luschny_, Dec 27 2024