A304435 - OEIS

%I #12 Jun 22 2020 01:54:59

%S 3,5,6,8,10,13,17,19,20,24,25,26,28,29,34,36,37,40,41,45,50,52,53,54,

%T 58,61,62,65,68,73,74,75,80,81,82,85,88,89,90,96,97,98,100,101,104,

%U 106,109,113,116,117,122,125,130,136,137,145,146,148,149,150,153,157,160,164,168,169,170,173,176,178,180,181,185,192,193,194,197

%N Numbers n such that n^5 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

%C 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.

%e 3^5 = 3^3 + 6^3; 5^5 = 10^2 + 55^2, 6^5 = 2^5 + 88^2, ...

%p LIM:= 200^5: P:={seq(seq(x^k, k=3..floor(log[x](LIM))), x=2..floor(LIM^(1/3)))}:

%p is_A304435:= proc(n) local n5, Pp;  n5:= n^5;  if remove(t -> subs(t, x)<=1 or subs(t, y)<=1 or subs(t, x-y)=0, [isolve(x^2+y^2=n^5)]) <> [] then return true fi;  Pp:= map(t ->n5-t, P minus {n5, n5/2});  (Pp intersect P <> {}) or (select(issqr, Pp) <> {})

%p end proc: # adapted from code by _Robert Israel_ for A304434

%t M = 200; LIM = M^5;

%t P = Flatten @ Table[Table[x^k, {k, 3, Floor[Log[x, LIM]]}], {x, 2,Floor[ LIM^(1/3)]}];

%t filterQ[n_] := Module[{n5 = n^5, Pp, x, y}, If[Solve[x > 1 && y > 1 && x != y && x^2 + y^2 == n5, {x, y}, Integers] != {}, Return[True]]; Pp = n5 - (P ~Complement~ {n5, n5/2}); (Pp ~Intersection~ P) != {} || Select[Pp, IntegerQ[Sqrt[#]]&] != {}];

%t Select[Range[2, M], filterQ] (* _Jean-François Alcover_, Jun 21 2020, after Maple *)

%o (PARI) N=200;L=N^5;P=List(); for(x=2,sqrtnint(L,3),for(k=3,logint(L,x),listput(P,x^k)));P=Set(P);

%o is_A304435(n)={for(i=1,#s=sum2sqr(n=n^5),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))} \\ For sum2sqr() see A133388.

%Y Cf. A304433, A304434, A001597 (perfect powers).

%K nonn

%O 1,1

%A _M. F. Hasler_, May 25 2018