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

1, 8, 81, 1028, 15780, 284652, 5903464, 138407544, 3619892160, 104485268960, 3299177464704, 113120695539612, 4185473097734656, 166217602768452900, 7051983744002135040, 318324623296131263408, 15232941497754507165696, 770291040239888149405944, 41042353622873800536064000, 2298206207793743728251532020

Offset: 1

Author

Yukun Yao, Aug 13 2018

Keywords

• nonn

Examples

			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.
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.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..100
• Y. Yao and D. Zeilberger, An Experimental Mathematics Approach to the Area Statistics of Parking Functions, arXiv 1806.02680, 2018

Crossrefs

Cf. A000272, A298593.

Programs

Maple

#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.
Pnax:=proc(n,a,x) local k:
option remember:
if n=0 then
  return 1:
fi:
if n>0 and a=0 then
  return 0:
fi:
return expand(x^n*add(binomial(n,k)*Pnax(n-k,a+k-1,x),k=0..n)):
end:
seq(subs(x = 1, diff(Pnax(n, 1, x), x)), n = 1 .. 20)

Mathematica

T[n_, k_] := n Sum[Binomial[n-1, j-1] j^(j-2) (n-j+1)^(n-j-1), {j, k, n}];
a[n_] := Sum[k T[n, k], {k, 1, n}];
Array[a, 20] (* Jean-François Alcover, Aug 29 2018, after Andrew Howroyd *)

PARI

\\ here T(n,k) is A298593.
T(n,k)={n*sum(j=k, n, binomial(n-1, j-1)*j^(j-2)*(n+1-j)^(n-1-j))}
a(n)={sum(k=1, n, k*T(n,k))} \\ Andrew Howroyd, Aug 17 2018

Formula

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.

a(n) = Sum_{k=1..n} k*A298593(n, k). - Andrew Howroyd, Aug 17 2018

Extensions

Edited by Andrew Howroyd and N. J. A. Sloane, Aug 19 2018