A304436 -id:A304436 - cp's OEIS frontend

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

M. F. Hasler, May 12 2018

nonn

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.

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

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

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

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

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 *)

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.

A298591 Numbers which are the sum of two distinct perfect powers x^k + y^m with x, y, k, m >= 2.

Original entry on oeis.org

12, 13, 17, 20, 24, 25, 29, 31, 33, 34, 35, 36, 40, 41, 43, 44, 45, 48, 52, 53, 57, 58, 59, 61, 63, 65, 68, 72, 73, 74, 76, 80, 81, 85, 89, 90, 91, 96, 97, 100, 104, 106, 108, 109, 113, 116, 117, 125, 127, 129, 130, 132, 133, 134, 136, 137, 141, 144, 145, 146, 148, 149, 150

Offset: 1

M. F. Hasler, May 26 2018

nonn

The number of terms between 2^(n-1) and 2^n-1 is, for n = 1, 2, 3, ...: 0, 0, 0, 2, 6, 17, 24, 69, 129*, 215, 425, 891, 1571, 2994, 5655*, 10535, 20132, 38840, 73510, 140730, 268438*, 514262, ... (For terms with * the next larger power of 2 is in the sequence, so it would be, e.g., ..., 130, 214, ... if we count from 2^n+1 to 2^(n+1).) At 2^22 this corresponds to a density of about 25%, decreasing by about 1% at each power of 2.

			12 = 2^2 + 2^3, 13 = 2^2 + 3^2, 17 = 2^3 + 3^2, ...

Robert Israel, Table of n, a(n) for n = 1..10000
Andrew Lohr, Several Topics in Experimental Mathematics, arXiv:1805.00076 [math.CO], 2018.

Cf. A076467, A111231, A304433, A304434, A304435, A304436.

N:= 1000: # for all terms <= N
PP:= {seq(seq(x^k,k=2..floor(log[x](N))),x=2..floor(sqrt(N)))}:
sort(convert(select(`<=`,{seq(seq(PP[i]+PP[j],i=1..j-1),j=2..nops(PP))},N),list)); # Robert Israel, May 27 2018

max = 150; Table[If[x^k == y^m, Nothing, x^k + y^m], {x, 2, Sqrt[max-4]}, {y, x, Sqrt[max-4]}, {k, 2, Log[2, max-4]}, {m, 2, Log[2, max-4]}] // Flatten // Select[#, # <= max &]& // Union (* Jean-François Alcover, Sep 18 2018 *)

is(n,A=A076467,s=sum2sqr(n))={for(i=1,#s, vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=2,#A, n>A[i]||return; ispower(n-A[i]) && A[i]*2!=n && return(1))} \\ A076467 must be computed up to limit n. See A133388 for sum2sqr.

cp's OEIS Frontend

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).

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