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A271237
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Number of ordered ways to write n as u^3 + 2*v^3 + 3*x^3 + 4*y^3 + 5*z^3, where u, v, x, y and z are nonnegative integers.
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5
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1, 1, 1, 2, 2, 3, 3, 3, 4, 3, 4, 3, 3, 3, 2, 3, 2, 3, 1, 2, 3, 2, 2, 1, 4, 3, 2, 3, 3, 5, 3, 4, 6, 4, 5, 4, 6, 4, 4, 3, 5, 5, 3, 6, 3, 6, 4, 4, 6, 3, 5, 4, 4, 4, 3, 4, 5, 7, 4, 6, 4, 5, 6, 4, 10, 2, 6, 8, 3, 7, 4, 8, 6, 5, 5, 4, 5, 2, 6, 1, 5, 3, 3, 8, 5, 7, 6, 6, 9, 6, 7, 6, 6, 5, 5, 6, 4, 6, 6, 8, 1
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OFFSET
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0,4
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COMMENTS
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Conjecture: We have {u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever (a,b,c,d) is among the following 32 quadruples: (1,2,2,3), (1,2,2,4), (1,2,3,4), (1,2,4,5), (1,2,4,6), (1,2,4,9), (1,2,4,10), (1,2,4,11), (1,2,4,18), (1,3,4,6), (1,3,4,9), (1,3,4,10), (2,2,4,5), (2,2,6,9), (2,3,4,5), (2,3,4,6), (2,3,4,7), (2,3,4,8), (2,3,4,9), (2,3,4,10), (2,3,4,12), (2,3,4,15), (2,3,4,18), (2,3,5,6), (2,3,6,12), (2,3,6,15), (2,4,5,6), (2,4,5,8), (2,4,5,9), (2,4,5,10), (2,4,6,7), (2,4,7,10).
In particular, this implies that a(n) > 0 for all n = 0,1,2,... We guess that a(n) = 1 only for n = 0, 1, 2, 18, 23, 79, 100.
If {m*u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} with 1 <= m <= a <= b <= c <= d, then m = 1, and we can show that (a,b,c,d) must be among the 32 quadruples listed in the conjecture (cf. Theorem 1.2 of the linked 2017 paper).
Conjecture verified for all the 32 quadruples up to 10^11. - Mauro Fiorentini, Jul 09 2023
It is known that there are exactly 54 quadruples (a,b,c,d) with 1 <= a <= b <= c <= d such that {a*w^2+b*x^2+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A271099 and A271169 for conjectures refining Waring's problem.
We also conjecture that if P(u,v,x,y,z) is one of the four polynomials u^6+v^3+2*x^3+4*y^3+5*z^3 and a*u^6+v^3+2*x^3+3*y^3+4*z^3 (a = 5,8,12) then any natural number can be written as P(u,v,x,y,z) with u,v,x,y,z nonnegative integers. - Zhi-Wei Sun, Apr 06 2016
Conjecture verified for all the 4 polynomials up to 10^11. - Mauro Fiorentini, Jul 09 2023
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REFERENCES
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S. Ramanujan, On the expression of a number in the form a*x^2 + b*y^2 + c*z^2 + d*w^2, Proc. Cambridge Philos. Soc. 19(1917), 11-21.
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 0^3 + 2*1^3 + 3*0^3 + 4*0^3 + 5*0^3.
a(18) = 1 since 18 = 2^3 + 2*1^3 + 3*1^3 + 4*0^3 + 5*1^3.
a(23) = 1 since 23 = 0^3 + 2*2^3 + 3*1^3 + 4*1^3 + 5*0^3.
a(79) = 1 since 79 = 1^3 + 2*3^3 + 3*2^3 + 4*0^3 + 5*0^3.
a(100) = 1 since 100 = 2^3 + 2*1^3 + 3*3^3 + 4*1^3 + 5*1^3.
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MATHEMATICA
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CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0; Do[If[CQ[n-5z^3-4y^3-3x^3-2v^3], r=r+1], {z, 0, (n/5)^(1/3)}, {y, 0, ((n-5z^3)/4)^(1/3)}, {x, 0, ((n-5z^3-4y^3)/3)^(1/3)}, {v, 0, ((n-5z^3-4y^3-3x^3)/2)^(1/3)}]; Print[n, " ", r]; Continue, {n, 0, 100}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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