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 A266212 Positive integers x such that x^3 = y^4 + z^2 for some positive integers y and z. 9
 8, 13, 20, 25, 40, 125, 128, 193, 200, 208, 225, 313, 320, 328, 400, 500, 605, 640, 648, 1000, 1053, 1156, 1521, 1620, 1625, 1681, 1700, 2000, 2025, 2048, 2125, 2465, 2493, 2873, 2920, 3025, 3088, 3185, 3200, 3240, 3328, 3400, 3600, 3656, 3748, 3816, 4225, 4625, 4913, 5000, 5008, 5120, 5248, 6400, 6728, 6760, 6793, 6845, 7225, 8000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS If x^3 = y^4 + z^2, then (a^(4k)*x)^3 = (a^(3k)*y)^4 + (a^(6k)*z)^2 for all a = 1,2,3,... and k = 0,1,2,... So the sequence has infinitely many terms. Conjecture: For any integer m, there are infinitely many triples (x,y,z) of positive integers with x^4 - y^3 + z^2 = m. This is stronger than the conjecture in A266152. LINKS Zhi-Wei Sun and Chai Wah Wu, Table of n, a(n) for n = 1..698 n = 1..100 from Zhi-Wei Sun Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. EXAMPLE a(1) = 8 since 8^3 = 4^4 + 16^2. a(2) = 13 since 13^3 = 3^4 + 46^2. a(3) = 20 since 20^3 = 4^4 + 88^2. a(8) = 193 since 193^3 = 6^4 + 2681^2. a(12) = 313 since 313^3 = 66^4 + 3419^2. a(20) = 1000 since 1000^3 = 100^4 + 30000^2. MATHEMATICA SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]] n=0; Do[Do[If[SQ[x^3-y^4], n=n+1; Print[n, " ", x]; Goto[aa]], {y, 1, x^(3/4)}]; Label[aa]; Continue, {x, 1, 8000}] CROSSREFS Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266152, A266153. Sequence in context: A219721 A176209 A227453 * A063849 A273980 A101642 Adjacent sequences: A266209 A266210 A266211 * A266213 A266214 A266215 KEYWORD nonn AUTHOR Zhi-Wei Sun, Dec 23 2015 STATUS approved

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Last modified April 22 17:24 EDT 2024. Contains 371905 sequences. (Running on oeis4.)