OFFSET
1,3
COMMENTS
Is there a solution k such that k^k = (x^3 + y^3) / 2 where x > y > 0?
The answer to the above question is yes: 76^76 = (523974089123227128080087214816032969930445946880^3 + 314384453473936276848052328889619781958267568128^3)/2. Other examples include 112^112 and 172^172. - Chai Wah Wu, Jan 18 2016
From David A. Corneth, Feb 05 2026: (Start)
152 = 5^3 + 3^3 so 76 = (5^3 + 3^3)/2 and so 76^76 = 76 * 76^75 = 76*(76^25)^3 = (5^3 + 3^3)/2 * (76^25)^3 = ((5*(76^25))^3 + (3*(76^25))^3) / 2.
Similarly ((6*(112^37))^3 + (2*(112^37))^3)/2 = 112^112 and ((7*172^57)^3 + (172^57)^3)/2 = 172^172.
Other terms k in first 228 terms that are not multiples of 3 and are not cubes are 364 = (8^3 + 6^3)/2, 427 = (9^3 + 5^3)/2, 532 = (10^3 + 4^3)/2 and so k^k can be written in a similar way like above for those cases.
Does "Number k such that k is a multiple of 3 or 2*k is the sum of two positive cubes and k is one more than a multiple of 3." define the sequence? (End)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..228
FORMULA
For n > 6, 1.5*n < a(n) < 3*n. - Charles R Greathouse IV, Feb 05 2026
EXAMPLE
1 is a term because 1^1 = 1 = (1^3 + 1^3) / 2.
3 is a term because 3^3 = 27 = (3^3 + 3^3) / 2.
8 is a term because 8^8 = 2^24 = (256^3 + 256^3) / 2.
MATHEMATICA
Select[Range@ 24, Resolve[Exists[{x, y}, And[Reduce[#^# == (x^3 + y^3)/2, {x, y}, Integers], x > 0, y > 0]]] &] (* Michael De Vlieger, Jan 15 2016 *)
PROG
(PARI) T=thueinit('z^3+1);
is(n) = #select(v->min(v[1], v[2])>0, thue(T, n))>0;
for(n=0, 28, if(is(2*n^n), print1(n, ", ")));
(PARI) T=thueinit('z^3+1, 1);
is(n) = bittest(15261560708789483359, n%63) && (n%3==0 || ispower(n, 3) || #select(v->min(v[1], v[2])>0, thue(T, 2*n^n))>0) \\ Charles R Greathouse IV, Feb 05 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Altug Alkan, Jan 14 2016
EXTENSIONS
a(13) from Michael De Vlieger, Jan 15 2016
a(14)-a(60) from Chai Wah Wu, Jan 18 2016
STATUS
approved