cp's OEIS Frontend

A302174 Smallest solution x of x^n + y^(n+1) = z^(n+2), x, y, z >= 1.

%I A302174 #50 May 05 2018 04:10:28
%S A302174 1,2,27,256,472392,262144,13759414272,4294967296,4057816381784064
%N A302174 Smallest solution x of x^n + y^(n+1) = z^(n+2), x, y, z >= 1.
%C A302174 From _M. F. Hasler_, Apr 13 2018: (Start)
%C A302174 Proofs for the upper limits given in the formula section:
%C A302174 For odd n = 2k-1, x = 2^(2*k^2) yields a solution with x^(2k-1) = y^(2k) = 1/2*z^(2k+1), y = 2^(k(2k-1)) and z = 2^(2k(k-1)+1), because 2*k^2*(2k-1) + 1 = (2k+1)*(2k(k-1)+1).
%C A302174 For even n = 2k, x = 2^(k*(2k+1))*3^(2k+2) yields a solution, with y = 2^(2*k^2)*3^(2k+1) and z = 2^(2*k^2-k+1)*3^(2k), because for the exponents of 3, (2k+1)^2 = (2k+2)2k + 1 and the factor 1+3 = 2^2 adds 2 to the (identical) exponent of 2 in x^(2k) and y^(2k+1), to factor as 2*k^2(2k+1) + 2 = (2k+2)(k(2k-1)+1). (End)
%F A302174 For odd n, a(n) <= 2^((n+1)^2/2); for even n, a(n) <= 2^(n*(n+1)/2)*3^(n+2).
%F A302174 We may conjecture that, for n > 4, a(n) is given by these upper limits.
%e A302174 1^0 + 3^1 = 2^2, therefore a(0) = 1.
%e A302174 2^1 + 5^2 = 3^3, so a(1) = 2. (No solution can have x = 1 because z^3 - 1 = (z - 1)(z^2 + z + 1) cannot be a square: if a = z - 1, then z^2 + z + 1 = a^2 + 3a + 3 is congruent to 3 modulo any factor of a, and a = 3b yields z^3 - 1 = 9*b*(3*b^2 + 3b + 1), the last factor being congruent to 1 modulo any factor of b, and cannot be a square.)
%e A302174 27^2 + 18^3 = 9^4, so a(2) = 27.
%e A302174 256^3 + 64^4 = 32^5, so a(3) = 256.
%e A302174 472392^4 + 52488^5 = 8748^6, so a(4) = 472392.
%Y A302174 Cf. A001105 (2*k^2), A060757 (4^k^2 = 2^(2k^2)), A000244 (3^k).
%Y A302174 Conjectured to be a subsequence of A003586 (2^i*3^j).
%Y A302174 Cf. A300564, A300565, A300566, A300567, A300568 (z^4 = x^2 + y^3, ..., z^8 = x^6 + y^7).
%K A302174 nonn,more
%O A302174 0,2
%A A302174 _Jacques Tramu_, Apr 07 2018
%E A302174 Extended to a(0) = 1 by _M. F. Hasler_, Apr 13 2018