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%I #13 Jun 28 2024 01:20:44
%S 1,4,27,257,3156,47442,843352,17300943,402210240,10448526896,
%T 299925224064,9426724628301,321959469056512,11872685912032350,
%U 470132249600142336,19895288956008203963,896055382220853362688,42793946679993786078108,2160123874888094765056000
%N a(n) = sum of lead terms of all parking functions of length n.
%H Andrew Howroyd, <a href="/A328694/b328694.txt">Table of n, a(n) for n = 1..200</a>
%H Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, <a href="https://arxiv.org/abs/2406.12941">Metered Parking Functions</a>, arXiv:2406.12941 [math.CO], 2024. See pp. 18, 22.
%F a(n) = Sum_{k=1..n} k*A298592(n,k).
%F a(n) = A318047(n) / n.
%e Case n = 2: There are 3 parking functions of length 2: [1, 1], [1, 2], [2, 1]. Summing up the initial values gives 1 + 1 + 2 = 4, so a(2) = 4.
%o (PARI) \\ here T(n,k) is A298592(n,k).
%o T(n, k)={sum(j=k, n, binomial(n-1, j-1)*j^(j-2)*(n+1-j)^(n-1-j))}
%o a(n)={sum(k=1, n, k*T(n, k))}
%Y Cf. A298592, A318047.
%K nonn
%O 1,2
%A _Andrew Howroyd_, Oct 25 2019
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