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A318047 a(n) = sum of values taken by all parking functions of length n.

%I A318047 #21 Aug 29 2018 16:06:54
%S A318047 1,8,81,1028,15780,284652,5903464,138407544,3619892160,104485268960,
%T A318047 3299177464704,113120695539612,4185473097734656,166217602768452900,
%U A318047 7051983744002135040,318324623296131263408,15232941497754507165696,770291040239888149405944,41042353622873800536064000,2298206207793743728251532020
%N A318047 a(n) = sum of values taken by all parking functions of length n.
%H A318047 Andrew Howroyd, <a href="/A318047/b318047.txt">Table of n, a(n) for n = 1..100</a>
%H A318047 Y. Yao and D. Zeilberger, <a href="https://arxiv.org/abs/1806.02680">An Experimental Mathematics Approach to the Area Statistics of Parking Functions</a>, arXiv 1806.02680, 2018
%F A318047 a(n) is the first derivative of P(n,1,x) evaluated at x = 1 where P(n,m,x) satisfies P(n,m,x) = x^n*Sum_{k=0..n} binomial(n,k)*P(n-k, m+k-1, x) with P(0,m,x) = 1 and P(n,0,x) = 0 for n > 0.
%F A318047 a(n) = Sum_{k=1..n} k*A298593(n, k). - _Andrew Howroyd_, Aug 17 2018
%e A318047 Case n = 2: There are 3 parking functions of length 2: [1, 1], [1, 2], [2, 1]. Summing up all values gives 2 + 3 + 3 = 8, so a(2) = 8.
%e A318047 Case n = 3: There are 16 parking functions of length 3: [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 3, 1], [1, 3, 2], [2, 1, 1], [2, 1, 2], [2, 1, 3], [2, 2, 1], [2, 3, 1], [3, 1, 1], [3, 1, 2], [3, 2, 1]. Summing up all values gives a total of 81, so a(3) = 81.
%p A318047 #Pnax(n,a,x): the sum of x^(sum of all entries in the parking function) over the set of a-parking functions of length n by recurrence relation.
%p A318047 Pnax:=proc(n,a,x) local k:
%p A318047 option remember:
%p A318047 if n=0 then
%p A318047   return 1:
%p A318047 fi:
%p A318047 if n>0 and a=0 then
%p A318047   return 0:
%p A318047 fi:
%p A318047 return expand(x^n*add(binomial(n,k)*Pnax(n-k,a+k-1,x),k=0..n)):
%p A318047 end:
%p A318047 seq(subs(x = 1, diff(Pnax(n, 1, x), x)), n = 1 .. 20)
%t A318047 T[n_, k_] := n Sum[Binomial[n-1, j-1] j^(j-2) (n-j+1)^(n-j-1), {j, k, n}];
%t A318047 a[n_] := Sum[k T[n, k], {k, 1, n}];
%t A318047 Array[a, 20] (* _Jean-François Alcover_, Aug 29 2018, after _Andrew Howroyd_ *)
%o A318047 (PARI) \\ here T(n,k) is A298593.
%o A318047 T(n,k)={n*sum(j=k, n, binomial(n-1, j-1)*j^(j-2)*(n+1-j)^(n-1-j))}
%o A318047 a(n)={sum(k=1, n, k*T(n,k))} \\ _Andrew Howroyd_, Aug 17 2018
%Y A318047 Cf. A000272, A298593.
%K A318047 nonn
%O A318047 1,2
%A A318047 _Yukun Yao_, Aug 13 2018
%E A318047 Edited by _Andrew Howroyd_ and _N. J. A. Sloane_, Aug 19 2018