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 A331842 Number of positive integer solutions (x,y) to the equation x^y = y^(nx). 1
 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 (list; graph; refs; listen; history; text; internal format)
 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%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**(e//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

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Last modified July 4 20:08 EDT 2022. Contains 355086 sequences. (Running on oeis4.)