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Integers n such that n^3 is the average of a nonzero square and a nonzero fourth power.

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`%I #25 Jul 18 2016 20:46:53
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`%S 1,5,16,25,26,40,41,50,80,81,125,250,256,365,386,400,405,416,425,450,
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`%T 457,477,625,626,640,656,800,841,845,1000,1125,1153,1210,1225,1280,
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`%U 1296,1681,1825,2000,2025,2057,2106,2197,2312,2401,3042,3125,3240,3250,3321,3362,3400,3625
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`%N Integers n such that n^3 is the average of a nonzero square and a nonzero fourth power.
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`%C Numbers n such that 2*n^3 = x^2 + y^4 where x and y are nonzero integers, is soluble.
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`%C Square terms of this sequence are 1, 16, 25, 81, 256, 400, 625, 841, 1225, 1296, 1681, 2025, 2401, ...
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`%C From _David A. Corneth_, Jun 06 2016 (Start):
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`%C A000351, the powers of 5, is a subsequence.
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`%C If n is a term, then n * k^4 is a term; as 2*n^3 = x^4 + y^2, 2 * (n * k^4)^3 = (k^3 * x)^4 + (k^6 * y)^2. (End)
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`%H Chai Wah Wu, <a href="/A274012/b274012.txt">Table of n, a(n) for n = 1..10000</a>
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`%e 5 is a term because 5^3 = (13^2 + 3^4) / 2.
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`%o (PARI) is(n) = for(x=1, (2*n) ^ 0.75, if(issquare(2*n^3 - x^4)&&2*n^3-x^4>0, return(1)); 0) \\ _David A. Corneth_, Jun 06 2016
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`%Y Cf. A266212.
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`%K nonn
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`%O 1,2
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`%A _Altug Alkan_, Jun 06 2016
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