%I #42 Jan 24 2025 23:49:03

%S 8192,7593750,8605184

%N Numbers z such that z^7 = x^5 + y^6 for some integers x, y >= 1.

%C Also in the sequence: 72900000 = 2^5*3^6*5^5, 51018336 = 2^5*3^13, 6083264512 = 2^14*13^5, 916132832 = 2^5*31^5, 6530347008 = 2^12*3^13, 16307453952 = 2^26*3^5, 233861123808 = 2^5*3^9*13^5, 850305600000 = 2^9*3^12*5^5.

%C Consider a solution (x,y,z) of x^5 + y^6 = z^7. For any m, (x*m^42, y*m^35, z*m^30) is also a solution. Reciprocally, if (x/m^42, y/m^35, z/m^30) is a triple of integers for some m, then this is also a solution. We call primitive a solution for which there is no such m > 1.

%C When S = a^5 + b^12/4 is a square, then z = b^6/2 + sqrt(S) is a solution, with x = a*z and y = b*z. All known solutions are of this form. Sequence A303267 lists the y-values, thus equal to b*z where z = a(n), with corresponding x = a*z = (a(n)^7 - A303267(n)^6)^(1/5).

%H Hayden Chesnut, <a href="/A300567/a300567_3.py.txt">A300567 Python Code</a>

%e a(1) = 8192 = 2^13 is in the sequence because (2^13)^7 = (2^18)^5 + (2^15)^6, using 18*5 = 15*6 = 90 = 13*7 - 1 and 1 + 1 = 2.

%o (PARI) is(z)=for(y=1,sqrtnint(-1+z=z^7,6),ispower(z-y^6,5)&&return(y))

%o /* Code below for illustration only, not guaranteed to give a complete list. */

%o S=[]; N=1e5; forstep(b=1,99,1/6, forstep(a=1,N,1/6, issquare(b^12/4+a^5,&r)&& !frac(z=b^6/2+r)&& S=setunion(S,[z])); print1([b])); S

%o (Python) # See Hayden link. This code is built to identify valid z values based on specific conjectures outlined in the file.

%Y Cf. A300564 (z^4 = x^2 + y^3) and A242183, A300565 (z^5 = x^3 + y^4), A300566 (z^6 = x^4 + y^5), A302174.

%Y See A303267 for the y-values.

%Y Cf. A303375 (numbers of the form a^5 + b^6).

%K nonn,more,bref,hard

%O 1,1

%A _M. F. Hasler_, Apr 16 2018