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 A262824 Number of ordered ways to write n as w^2 + x^3 + 2*y^3 + 3*z^3, where w, x, y and z are nonnegative integers. 8
 1, 2, 2, 3, 4, 3, 3, 3, 2, 3, 3, 3, 4, 2, 3, 2, 2, 5, 2, 4, 5, 3, 2, 1, 4, 5, 5, 6, 8, 5, 4, 5, 3, 7, 3, 4, 8, 1, 4, 3, 4, 7, 4, 5, 4, 3, 3, 3, 3, 6, 5, 3, 9, 3, 4, 7, 3, 7, 3, 5, 4, 2, 6, 5, 4, 6, 8, 7, 8, 5, 5, 5, 1, 6, 4, 3, 7, 2, 5, 5, 5, 8, 8, 10, 9, 6, 3, 7, 6, 8, 9, 9, 8, 5, 6, 4, 3, 6, 7, 4, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Conjecture: (i) For any m = 3, 4, 5, 6 and n >= 0, there are nonnegative integers w, x, y, z such that n = w^2 + x^3 + 2*y^3 + m*z^3. (ii) For P(w,x,y,z) = w^2 + x^3 + 2*y^3 + z^4, w^2 + x^3 + 2*y^3 + 3*z^4, w^2 + x^3 + 2*y^3 + 6*z^4, 2*w^2 + x^3 + 4*y^3 + z^4, we have {P(w,x,y,z): w,x,y,z = 0,1,2,...} ={0,1,2,...}. See also A262827 and A262857 for similar conjectures. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 EXAMPLE a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3. a(8) = 2 since 8 = 2^2 + 1^3 + 2*0^3 + 3*1^3 = 0^2 + 2^3 + 2*0^3 + 3*0^3. a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3. a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3. a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3. MATHEMATICA SQ[n_]:=IntegerQ[Sqrt[n]] Do[r=0; Do[If[SQ[n-x^3-2y^3-3z^3], r=r+1], {x, 0, n^(1/3)}, {y, 0, ((n-x^3)/2)^(1/3)}, {z, 0, ((n-x^3-2y^3)/3)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 100}] CROSSREFS Cf. A000290, A000578, A262813, A262815, A262816, A262827, A262857. Sequence in context: A285732 A144909 A117114 * A242899 A229989 A126688 Adjacent sequences:  A262821 A262822 A262823 * A262825 A262826 A262827 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 03 2015 STATUS approved

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Last modified August 20 20:36 EDT 2018. Contains 313927 sequences. (Running on oeis4.)