A173515 Consider positive integer solutions to x^3+ y^3 = z^3 - n or 'Fermat near misses' of 1, 2, 3 ... Arrange known solutions by increasing values of n. Sequence gives value of lowest z for a given n.

9, 7, 2, 812918, 18, 217, 4, 3, 9730705, 332, 14, 135, 3, 19, 156, 16, 15584139827, 3, 139643, 6, 1541, 4, 2220422932, 5, 14, 4, 445, 12205, 9, 8, 16234, 815, 31, 4

Offset: 1

Author

Andy Martin, Feb 20 2010

Keywords

• nonn

Comments

The submitted values are for z when 0 < n < 51. There is no solution for any n congruent to 4 or 5 mod 9. This eliminates 4,5,13,14,22,23,31,32,40,41,49 and 50 in the 0-to-50 range.

Per the Elsenhans and Jahnel link there are no solutions found for 3, 33, 39 and 42 in the 0-to-50 range, with a search bound of 10^14.

If sequences could contain 'nil' for no solution, and '?' for cases where a solution is not known, but might exist, then a more concise definition is possible: Least positive integer such that a(n)^3 - n is the sum of two positive cubes. The sequence would then start: 9, 7, ?, nil, nil, 2.

Examples

			6^3 + 8^3 = 9^3 - 1: There are no solutions when n = 1 for z < 9, thus the first term is 9.
5^3 + 6^3 = 7^3 - 2: There are no solutions for z < 7, thus the second term is 7.
It is unknown if there is a solution when n = 3.
It is known there are no solutions when n = 4 and 5.
1^3 + 1^3 = 2^3 - 6, thus the third term is 2.
		

Links

• Andreas-Stephan Elsenhans and Joerg Jahnel, List of solutions of x^3 + y^3 + z^3 = n for n < 1000 neither a cube nor twice a cube
• D.R. Heath-Brown, W.M. Lioen and H.J.J. te Riele, on Solving the Diophantine Equation x^3 + y^3 + z^3 = k on a Vector Computer
• Kenji Koyama, Yukio Tsuruoka, and Hiroshi Sekigawa, On Searching For Solutions of the Diophantine Equation x^3 + y^3 + z^3 = n, Math. Comp. 66 (1997), 841-851.
• Eric S. Rowland, Known Families of Integer Solutions of x^3 + y^3 + z^3 = n

Crossrefs

Cf. A050788, A159935.

Programs

Ruby

# x^3 + y^3 = z^3 - n
# Solve for all z less than z_limit, and
# n less than n_limit.
# When n = 7, z = 812918 and faster code and language are needed.
# However, by optimizing this code slightly and running for 2 days
# the author was able to search all z < 164000 and n < 100
#
n_limit = 7 # Configure as desired
z_limit = 20 # Configure as desired
h = {}
(2..z_limit).each{ |z|
  (1..(z-1)).each{ |y|
  (1..(y)).each{ |x|
  n = z*z*z - x*x*x - y*y*y
  if n > 0 && n < n_limit && h[n].nil?
  puts "Found z = #{z} when #{x}^^3 + #{y}^^3 = #{z}^^3 - #{n}"
  h[n] = z
  end
} } } print "\nPartial sequence generated when n < #{n_limit} and z is searched to #{z_limit} is:\n"
h.sort.each{|k,v| print "#{v}, " }
print "\b\b \n"

Formula

Author conjectures that no explicit formula or recurrence exists.