

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.


2



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
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OFFSET

1,1


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 with: 9, 7, ?, nil, nil, 2


LINKS

Table of n, a(n) for n=1..34.
Eric S. Rowland, Known Families of Integer Solutions of x^3 + y^3 + z^3 = n
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), 841851.
D.R. HeathBrown, 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
AndreasStephan 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


FORMULA

Author conjectures that explicit formula or recurrence does not exist.


EXAMPLE

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 solution 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, this the third term is 2.


PROG

(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..(z1)).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"


CROSSREFS

Cf. A050788, A159935
Sequence in context: A155792 A247226 A197834 * A091558 A244659 A002205
Adjacent sequences: A173512 A173513 A173514 * A173516 A173517 A173518


KEYWORD

nonn


AUTHOR

Andy Martin, Feb 20 2010


STATUS

approved



