A298593 -id:A298593 - cp's OEIS frontend

A298592 Triangle read by rows: T(n,k) = number of parking functions of length n whose lead number is k.

1, 2, 1, 8, 5, 3, 50, 34, 25, 16, 432, 307, 243, 189, 125, 4802, 3506, 2881, 2401, 1921, 1296, 65536, 48729, 40953, 35328, 30208, 24583, 16807, 1062882, 800738, 683089, 601441, 531441, 461441, 379793, 262144, 20000000, 15217031, 13119879, 11708091, 10546875, 9453125, 8291909, 6880121, 4782969

Offset: 1

Rui Duarte, Jan 22 2018

			Triangle begins:
        1;
        2,      1;
        8,      5,      3;
       50,     34,     25,     16;
      432,    307,    243,    189,    125;
     4802,   3506,   2881,   2401,   1921,   1296;
    65536,  48729,  40953,  35328,  30208,  24583,  16807;
  1062882, 800738, 683089, 601441, 531441, 461441, 379793, 262144;
  ...

Steve Butler, Kimberly Hadaway, Victoria Lenius, Preston Martens, and Marshall Moats, Lucky cars and lucky spots in parking functions, arXiv:2412.07873 [math.CO], 2024. See p. 6.
D. Foata and J. Riordan, Mappings of acyclic and parking functions, J. Aeq. Math., 10 (1974) 10-22.

Cf. A000272, A298593, A298594, A298597.

Table[Sum[Binomial[n - 1, j - 1] j^(j - 2)*(n + 1 - j)^(n - 1 - j), {j, k, n}], {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Jan 22 2018 *)

T(n,k) = Sum_{j=k..n} binomial(n-1, j-1)*j^(j-2)*(n+1-j)^(n-1-j).
T(n,k) = A298593(n,k)/n.
T(n,k) = Sum_{j=k..n} A298594(n,j).
T(n,k) = (Sum_{j=k..n} A298597(n,j))/n.
Sum_{k=1..n} T(n,k) = A000272(n+1).

A298594 Triangle read by rows: T(n,k) = number of parking functions a of length n such that a(1) = k and if we replace a(1) = k with k+1 we don't get a parking function.

Original entry on oeis.org

1, 1, 1, 3, 2, 3, 16, 9, 9, 16, 125, 64, 54, 64, 125, 1296, 625, 480, 480, 625, 1296, 16807, 7776, 5625, 5120, 5625, 7776, 16807, 262144, 117649, 81648, 70000, 70000, 81648, 117649, 262144, 4782969, 2097152, 1411788, 1161216, 1093750, 1161216, 1411788, 2097152, 4782969

Offset: 1

Rui Duarte, Jan 22 2018

			Triangle begins:
       1;
       1,      1;
       3,      2,     3;
      16,      9,     9,    16;
     125,     64,    54,    64,   125;
    1296,    625,   480,   480,   625,  1296;
   16807,   7776,  5625,  5120,  5625,  7776,  16807;
  262144, 117649, 81648, 70000, 70000, 81648, 117649, 262144;
  ...

Steve Butler, Kimberly Hadaway, Victoria Lenius, Preston Martens, and Marshall Moats, Lucky cars and lucky spots in parking functions, arXiv:2412.07873 [math.CO], 2024. See p. 6.

Cf. A000272, A298592, A298593, A298597.

Table[Binomial[n - 1, k - 1] k^(k - 2)*(n + 1 - k)^(n - 1 - k), {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Jan 22 2018 *)

T(n,k) = binomial(n-1, k-1)*k^(k-2)*(n+1-k)^(n-1-k).
T(n,k) = A298592(n,k) - A298592(n,k+1).
T(n,k) = (A298593(n,k) - A298593(n,k+1))/n.
T(n,k) = A298597(n,k)/n.
T(n,1) = A000272(n+2).
T(n,n) = A000272(n+2).
T(n,k) = T(n,n-k).

A298597 Number T(n,k) of times the value k appears on the parking functions of length n and such that if we replace that value k with k+1 we don't get a parking function.

Original entry on oeis.org

1, 2, 2, 9, 6, 9, 64, 36, 36, 64, 625, 320, 270, 320, 625, 7776, 3750, 2880, 2880, 3750, 7776, 117649, 54432, 39375, 35840, 39375, 54432, 117649, 2097152, 941192, 653184, 560000, 560000, 653184, 941192, 2097152, 43046721, 18874368, 12706092, 10450944, 9843750, 10450944, 12706092, 18874368, 43046721

Offset: 1

Rui Duarte, Jan 22 2018

			Triangle begins:
        1;
        2,      2;
        9,      6,      9;
       64,     36,     36,     64;
      625,    320,    270,    320,    625;
     7776,   3750,   2880,   2880,   3750,   7776;
   117649,  54432,  39375,  35840,  39375,  54432, 117649;
  2097152, 941192, 653184, 560000, 560000, 653184, 941192, 2097152;
  ...

Cf. A000169, A000272, A298592, A298593, A298594.

Table[n Binomial[n - 1, k - 1] k^(k - 2)*(n + 1 - k)^(n - 1 - k), {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Jan 22 2018 *)

T(n,k) = n*binomial(n-1, k-1)*k^(k-2)*(n+1-k)^(n-1-k).
T(n,k) = n*A298594(n,k).
T(n.k) = A298593(n,k)-A298593(n,k+1).
T(n,k) = n*(A298592(n,k)-A298592(n,k+1)).
T(n,1) = n*A000272(n+2).
T(n,n) = n*A000272(n+2).
T(n,1) = A000169(n).
T(n,n) = A000169(n).
T(n,k) = T(n,n-k).

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

Original entry on oeis.org

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

Offset: 1

Yukun Yao, Aug 13 2018

nonn

			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.

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

Cf. A000272, A298593.

#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)

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 *)

\\ 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

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

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

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A298592 Triangle read by rows: T(n,k) = number of parking functions of length n whose lead number is k.

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A298594 Triangle read by rows: T(n,k) = number of parking functions a of length n such that a(1) = k and if we replace a(1) = k with k+1 we don't get a parking function.

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A298597 Number T(n,k) of times the value k appears on the parking functions of length n and such that if we replace that value k with k+1 we don't get a parking function.

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

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