cp's OEIS Frontend

Also number of labeled pointed rooted trees (or vertebrates) on n nodes.

For n >= 1 a(n) is also the number of n X n (0,1) matrices in which each row contains exactly one entry equal to 1. - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001

Also the number of labeled rooted trees on (n+1) nodes such that the root is lower than its children. Also the number of alternating labeled rooted ordered trees on (n+1) nodes such that the root is lower than its children. - Cedric Chauve (chauve(AT)lacim.uqam.ca), Mar 27 2002

With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)} p(i, j)!)) * ((n!/(n - p(i)))!/(Product_{j=1..d(i)} m(i, j)!)). - Thomas Wieder, May 18 2005

All rational solutions to the equation x^y = y^x, with x < y, are given by x = A000169(n+1)/A000312(n), y = A000312(n+1)/A007778(n), where n = 1, 2, 3, ... . - Nick Hobson, Nov 30 2006

a(n) is the total number of leaves in all (n+1)^(n-1) trees on {0,1,2,...,n} rooted at 0. For example, with edges directed away from the root, the trees on {0,1,2} are {0->1,0->2},{0->1->2},{0->2->1} and contain a total of a(2)=4 leaves. - David Callan, Feb 01 2007

Limit_{n->infinity} A000169(n+1)/a(n) = exp(1). Convergence is slow, e.g., it takes n > 74 to get one decimal place correct and n > 163 to get two of them. - Alonso del Arte, Jun 20 2011

Also smallest k such that binomial(k, n) is divisible by n^(n-1), n > 0. - Michel Lagneau, Jul 29 2013

For n >= 2 a(n) is represented in base n as "one followed by n zeros". - R. J. Cano, Aug 22 2014

Number of length-n words over the alphabet of n letters. - Joerg Arndt, May 15 2015

Number of prime parking functions of length n+1. - Rui Duarte, Jul 27 2015

The probability density functions p(x, m=q, n=q, mu=1) = A000312(q)*E(x, q, q) and p(x, m=q, n=1, mu=q) = (A000312(q)/A000142(q-1))*x^(q-1)*E(x, q, 1), with q >= 1, lead to this sequence, see A163931, A274181 and A008276. - Johannes W. Meijer, Jun 17 2016

Satisfies Benford's law [Miller, 2015]. - N. J. A. Sloane, Feb 12 2017

A signed version of this sequence apart from the first term (1, -4, -27, 256, 3125, -46656, ...), has the following property: for every prime p == 1 (mod 2n), (-1)^(n(n-1)/2)*n^n = A057077(n)*a(n) is always a 2n-th power residue modulo p. - Jianing Song, Sep 05 2018

From Juhani Heino, May 07 2019: (Start)

n^n is both Sum_{i=0..n} binomial(n,i)*(n-1)^(n-i)

and Sum_{i=0..n} binomial(n,i)*(n-1)^(n-i)*i.

The former is the familiar binomial distribution of a throw of n n-sided dice, according to how many times a required side appears, 0 to n. The latter is the same but each term is multiplied by its amount. This means that if the bank pays the player 1 token for each die that has the chosen side, it is always a fair game if the player pays 1 token to enter - neither bank nor player wins on average.

Examples:

2-sided dice (2 coins): 4 = 1 + 2 + 1 = 1*0 + 2*1 + 1*2 (0 omitted from now on);

3-sided dice (3 long triangular prisms): 27 = 8 + 12 + 6 + 1 = 12*1 + 6*2 + 1*3;

4-sided dice (4 long square prisms or 4 tetrahedrons): 256 = 81 + 108 + 54 + 12 + 1 = 108*1 + 54*2 + 12*3 + 1*4;

5-sided dice (5 long pentagonal prisms): 3125 = 1024 + 1280 + 640 + 160 + 20 + 1 = 1280*1 + 640*2 + 160*3 + 20*4 + 1*5;

6-sided dice (6 cubes): 46656 = 15625 + 18750 + 9375 + 2500 + 375 + 30 + 1 = 18750*1 + 9375*2 + 2500*3 + 375*4 + 30*5 + 1*6.

(End)

For each n >= 1 there is a graph on a(n) vertices whose largest independent set has size n and whose independent set sequence is constant (specifically, for each k=1,2,...,n, the graph has n^n independent sets of size k). There is no graph of smaller order with this property (Ball et al. 2019). - David Galvin, Jun 13 2019

For n >= 2 and 1 <= k <= n, a(n)*(n + 1)/4 + a(n)*(k - 1)*(n + 1 - k)/2*n is equal to the sum over all words w = w(1)...w(n) of length n over the alphabet {1, 2, ..., n} of the following quantity: Sum_{i=1..w(k)} w(i). Inspired by Problem 12432 in the AMM (see links). - Sela Fried, Dec 10 2023

Also, dimension of the unique cohomology group of the smallest interval containing the poset of partitions decorated by Perm, i.e. the poset of pointed partitions. - Bérénice Delcroix-Oger, Jun 25 2025

Stirling's approximation for n! suggests that this should be about e^n/sqrt(pi*2n). Bill Gosper has noted that e^n/sqrt(pi*(2n+1/3)) is significantly better.

n^n/n! = A001142(n)/A001142(n-1), where A001142(n) is product{k=0 to n} C(n,k) (where C() is a binomial coefficient). - Leroy Quet, May 01 2004

There are n^n distinct functions from [n] to [n] or sequences on n symbols of length n, the number of those sequences having n distinct symbols is n!. So the probability P(n) of bijection is n!/n^n. The expected value of the number of functions that we pick until we found a bijection is the reciprocal of P(n), or n^n/n!. - Washington Bomfim, Mar 05 2012

The van der Waerden conjecture, now a theorem thanks to Egorycev, states that the permanent of any n X n doubly stochastic matrix is >= n!/n^n, with equality iff the matrix has all entries equal to 1/n.

Therefore the reciprocal of the permanent of any n X n doubly stochastic matrix is bounded from above by n^n/n! and this sequence.

n^n/n! = A001142(n)/A001142(n-1), where A001142(n) is product{k=0 to n} C(n,k) (where C() is a binomial coefficient). - Leroy Quet, May 01 2004