A051395 ?= A051386

[all at abouthugo.de]

SeqFans,

when looking for duplicates in OEIS I found:

%S A051395
2,9,16,28,35,54,65,72,91,126,128,133,134,152,182,183,189,201,217,219,
%T A051395 224,243,250,273,278,280,309,341,344,351,370
%N A051395 Numbers whose 7th power is expressible as the sum of positive
cubes.
                                          should read ...... 2 positive
...
%e A051395 201^7 = 148137^3 + 215472^3
%K A051395 nonn
%O A051395 0,1
%A A051395 Jud McCranie (judmccr(AT)bellsouth.net)

%I A051386
%S A051386
2,9,16,28,35,54,65,72,91,126,128,133,134,152,182,183,189,201,217,219,
%T A051386
224,243,250,273,278,280,309,341,344,351,370,399,407,422,432,453,468,
%U A051386
497,513,520,539,559,576,579,637,651,658,686,728,730,737,756,793,854
%N A051386 Numbers whose 4th power is the sum of two positive cubes.
%C A051386 When x is the sum of 2 positive cubes (A003325) there is a
trivial solution.
%e A051386 134^4 = 469^3 + 603^3
%Y A051386 Cf. A003325, A051387.
%K A051386 nonn
%O A051386 1,1
%A A051386 Jud McCranie (judmccr(AT)bellsouth.net)

Obviously not much difference.

The first terms of both sequences:

      2^4=           2^3   +     2^3   < 1/1
      2^7=           4^3   +     4^3   < 
                                    
      9^4=          18^3   +     9^3   < 2/1
      9^7=         162^3   +    81^3   <
                                    
     16^4=          32^3   +    32^3   < 1/1
     16^7=         512^3   +   512^3   <
                                    
     28^4=          84^3   +    28^3   < 3/1
     28^7=        2352^3   +   784^3   <
                                    
     35^4=         105^3   +    70^3   < 3/2
     35^7=        3675^3   +  2450^3   <
                                    
     54^4=         162^3   +   162^3   < 1/1
     54^7=        8748^3   +  8748^3   <
                                    
     65^4=         260^3   +    65^3   < 4/1
     65^7=       16900^3   +  4225^3   <
                                    
     72^4=         288^3   +   144^3   < 2/1
     72^7=       20736^3   + 10368^3   <
                                    
     91^4=         364^3   +   273^3   < 4/3
     91^7=       33124^3   + 24843^3   <
                                    
    126^4=         630^3   +   126^3   < 5/1
    126^7=       79380^3   + 15876^3   <
                                    
    128^4=         512^3   +   512^3   < 1/1
    128^7=       65536^3   + 65536^3   <

If we can write a^4 = p^3 + q^3 p>=q then it seems that
we can also always write a^7 = u^3 + v^3 with u/v=p/q

Can we express v=f(p,q) and thus merge both sequences into one?

Please excuse my ignorance if you see that it's obvious ;-)

Hugo