A304433 Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

5, 7, 8, 10, 12, 13, 14, 17, 20, 25, 26, 28, 29, 32, 33, 34, 37, 40, 41, 45, 48, 50, 52, 53, 56, 57, 58, 61, 63, 65, 68, 71, 72, 73, 74, 78, 80, 82, 85, 89, 90, 97, 98, 100, 101, 104, 105, 106, 109, 112, 113, 114, 116, 117, 122, 125, 126, 128

Offset: 1

Author

M. F. Hasler, May 12 2018

Keywords

• nonn

Comments

Motivated by the search of solutions to a^n + b^(2n+2)/4 = (perfect square), which arises when searching solutions to x^n + y^(n+1) = z^(n+2) of the form x = a*z, y = b*z. It turns out that many solutions are of the form a^n = d (b^(n+1) + d), where d is a perfect power.

Examples

			5^3 = 125 = 4^2 + 11^2; 7^3 = 10^2 + 3^5; 8^3 = 13^2 + 7^3, ...
		

Links

• Robert Israel, Table of n, a(n) for n = 1..5970

Crossrefs

Cf. A001597 (perfect powers), A076467 (cubes and higher powers), A304434, A304435, A304436 (analog for n^4, n^5, n^6).

Programs

Maple

N:= 200: # to get terms <= N
N3:= N^3:
P:= {seq(seq(x^k, k=3..floor(log[x](N3))), x=2..N)}:
filter:= proc(n) local n3, Pp, x, y;
  n3:= n^3;
  if remove(t -> subs(t, x)<=1 or subs(t, y)<=1 or subs(t, x-y)=0, [isolve(x^2+y^2=n3)]) <> [] then return true fi;
  Pp:= map(t ->n3-t, P minus {n3, n3/2});
   (Pp intersect P <> {}) or (select(issqr, Pp) <> {})
end proc:
select(filter, [$2..N]); # Robert Israel, Jun 01 2018

Mathematica

M = 200;
M3 = M^3;
P = Union@ Flatten@ Table[Table[x^k, {k, 3, Floor[Log[x, M3]]}], {x, 2, M}];
filterQ[n_] := Module[{n3, Pp, x, y}, n3 = n^3; If[Solve[x > 1 && y > 1 && x != y && x^2 + y^2 == n3, {x, y}, Integers] != {}, Return[True]]; Pp = n3 - (P ~Complement~ {n3, n3/2}); (Pp ~Intersection~ P) != {} || Select[ Pp, IntegerQ[Sqrt[#]]&] != {}];
Select[Range[2, M], filterQ] (* Jean-François Alcover, Jun 21 2020, after Robert Israel *)

PARI

L=200^3; P=List(); for(x=2, sqrtnint(L,3), for(k=3, logint(L, x), listput(P, x^k))); #P=Set(P) \\ This P = A076467 \ {1} = A111231 \ {0} up to limit L.
is(n,e=3)={for(i=1, #s=sum2sqr(n=n^e), vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=1, #P, n>P[i]||return; ispower(n-P[i])&& P[i]*2 != n && return(1))} \\ The above P must be computed up to L >= n^3. For sum2sqr() see A133388.