A331842 Number of positive integer solutions (x,y) to the equation x^y = y^(nx).

4, 4, 4, 4, 5, 4, 5, 4, 5, 3, 9, 3, 5, 6, 8, 3, 7, 3, 7, 5, 6, 4, 9, 4, 5, 7, 8, 3, 9, 3, 7, 5, 5, 6, 10, 3, 5, 5, 10, 3, 9, 3, 7, 8, 5, 4, 12, 4, 8, 5, 8, 3, 10, 5, 9, 5, 6, 3, 14, 4, 5, 8, 8, 5, 9, 3, 7, 6, 9, 3, 14, 3, 5, 9, 7, 6, 9, 4, 11, 6, 5, 3, 13

Offset

2,1

Comments

a(n) > d(n), where d(n) = A000005(n) is the number of divisors of n, because x = y = 1 is a solution for all n and for every divisor j of n, x = ((1 + 1/j)*n)^j, y = ((1 + 1/j)*n)^(j + 1) is a solution. The difference a(n) - d(n) can get arbitrarily large. The smallest n for which a(n) >= d(n) + 4 is n = 10800.

Links

Pontus von Brömssen, Table of n, a(n) for n = 2..16384

Example

For n = 15, the 1 + d(n) = 5 "standard" solutions are (1, 1), (30, 30^2), (20^3, 20^4), (18^5, 18^6), and (16^15, 16^16). In addition to these, (5^3, 5^5) is a solution, so a(15) = 6.
For n = 10800, the 3 "nonstandard" solutions are (180, 180^3), (30^2, 30^5), and (108^25, 108^27), so a(10800) = d(n) + 4 = 64.

Prog

(Python)
import sympy
def A331842(n):
  c=0
  d=sympy.divisors(n)
  i=2
  while 2**i<=n*(1+i):
    for j in d:
      if sympy.gcd(i, j)==1:
        e=sympy.perfect_power(n+i*n//j, [i])
        if e and e[1]%i==0: # The divisibility test is not necessary from version 1.5 of sympy.
          c+=1 # Count the solution (x, y)=(m^j, m^(j+i)), where m=e[0]**(e[1]//i).
    i+=1
  return c+1+len(d) # Add the number of "standard" solutions.

Crossrefs

Cf. A000005.

Sequence in context: A174444 A268237 A276866 * A006264 A134994 A138195

Adjacent sequences: A331839 A331840 A331841 * A331843 A331844 A331845

Keyword

nonn

Author

Pontus von Brömssen, Jan 29 2020

Status

approved