A365108 - OEIS

%I #50 Apr 04 2024 11:10:36

%S 1,2,9,4,625,672,117649,32,2187,5941760,25937424601,1397760,

%T 23298085122481,308548739072,29192926025390625,4096,

%U 48661191875666868481,3817734144,104127350297911241532841,174339220,209430786243,24639156314201655345152,907846434775996175406740561329

%N a(n) is the smallest integer value of (p^n - q^n)/n for all choices of integers p > q >= 0.

%C (p^n - q^n)/n has the integer value n^(n - 1) for p = n and q = 0. For p = n + k + 1 (k: nonnegative integer) the term has its minimum for q = n + k. With the binomial theorem follows ((n + k + 1)^n - (n + k)^n)/n >= ((n + k)^n - n*(n + k)^(n - 1) - (n + k)^n)/n = (n + k)^(n - 1) >= n^(n - 1). Therefore, for p > n, there is no smaller value of (p^n - q^n)/n than n^(n - 1). Thus a(n) <= n^(n - 1) exists with 1 <= p <= n and 0 <= q <= p - 1.

%C a(n) is also the smallest integer value that the integral over f(x) = x^(n - 1) between the nonnegative integer integration limits q and p (p > q) can have.

%H Felix Huber, <a href="/A365108/b365108.txt">Table of n, a(n) for n = 1..388</a>

%F a(n) is the integer minimum of (p^n - q^n)/n for 1 <= p <= n and 0 <= q <= p - 1.

%e For n = 5, a(5) = 672 with p = 4 and q = 2.

%p A365108 := proc(n) local q, p, s, a_n; a_n := n^(n - 1); for p to n do for q from 0 to p - 1 do s := (p^n - q^n)/n; if s = floor(s) and s < a_n then a_n := s; end if; end do; end do; return a_n; end proc;

%p seq(A365108(n), n = 1 .. 23);

%o (Python)

%o from sympy.ntheory.residue_ntheory import nthroot_mod

%o def A365108(n):

%o     c, qdict = n**(n-1), {}

%o     for p in range(1,n+1):

%o         r, m = pow(p,n,n), p**n

%o         if r not in qdict:

%o             qdict[r] = tuple(nthroot_mod(r,n,n,all_roots=True))

%o         c = min(c,min(((m-q**n)//n for q in qdict[r] if q<p),default=c))

%o     return int(c) # _Chai Wah Wu_, Sep 23 2023

%Y Cf. A000169, A055860, A020725.

%K nonn

%O 1,2

%A _Felix Huber_, Aug 21 2023