OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 9, 10, 11, 13, 14, 15, 16, 23, 24, 38, 39, 61, 76, 77, 104, 118, 188, 214, 229.
(ii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^4+x^3+2*y^3+5*z^3, u^5+2*v^4+x^3+2*y^3+3*z^3, u^5+3*v^4+x^3+2*y^3+3*z^3, 2*u^5+v^4+x^3+y^3+4*z^3, 2*u^5+v^4+x^3+2*y^3+4*z^3, 3*u^5+v^4+x^3+2*y^3+4*z^3, 5*u^5+v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+3*z^3, u^4+2*v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+6*z^3, u^4+2*v^4+x^3+3*y^3+4*z^3, u^4+2*v^4+x^3+4*y^3+5*z^3, u^4+2*v^4+x^3+4*y^3+6*z^3, u^4+2*v^4+x^3+4*y^3+10*z^3, u^4+3*v^4+x^3+2*y^3+3*z^3, u^4+3*v^4+x^3+2*y^3+4*z^3, u^4+3*v^4+x^3+2*y^3+6*z^3, u^4+4*v^4+x^3+y^3+2*z^3, u^4+4*v^4+x^3+2*y^3+3*z^3, u^4+4*v^4+x^3+2*y^3+4*z^3, u^4+5*v^4+x^3+2*y^3+4*z^3, u^4+6*v^4+x^3+2*y^3+3*z^3, u^4+7*v^4+x^3+2*y^3+3*z^3, u^4+9*v^4+x^3+2*y^3+4*z^3, 2*u^4+4*v^4+x^3+2*y^3+3*z^3,2*u^4+6*v^4+x^3+2*y^3+4*z^3, 3*u^4+6*v^4+x^3+2*y^3+4*z^3.
(iii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^3+x^3+2*y^3+4*z^3, u^5+v^3+2*x^3+3*y^3+c*z^3 (c = 6,9), a*u^5+v^3+2*x^3+4*y^3+5*z^3 (a = 1,2,6), b*u^5+v^3+2*x^3+4*y^3+6*z^3 (b = 2,3), u^5+v^3+2*x^3+4*y^3+d*z^3 (d = 9,13).
The listed polynomials in part (ii), together with u^5+v^4+x^3+2*y^3+3*z^3, should essentially exhaust all those polynomials P(u,v,x,y,z) = s*u^k+t*v^j+a*x^3+b*y^3+c*z^3 with s,t,a,b,c positive integers and k >= j > 3, such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
There are also finitely many (but quite a lot) polynomials P(u,v,x,y,z) of the form m*u^4+a*v^3+b*x^3+c*y^3+d*z^3 with a,b,c,d and m positive integers such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
Conjectures (i), (ii) and (iii) verified for n up to 10^11 for all polynomials. - Mauro Fiorentini, Sep 20 2023
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(16) = 1 since 16 = 0^5 + 2^4 +0^3 + 2*0^3 + 3*0^3.
a(104) = 1 since 104 = 0^5 + 2^4 + 4^3 + 2*0^3 + 3*2^3.
a(188) = 1 since 188 = 2^5 + 1^4 + 3^3 + 2*4^3 + 3*0^3.
a(229) = 1 since 229 = 1^5 + 3^4 + 4^3 + 2*1^3 + 3*3^3.
MATHEMATICA
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0; Do[If[CQ[n-u^5-v^4-3z^3-2y^3], r=r+1], {u, 0, (n-1)^(1/5)}, {v, 1, (n-u^5)^(1/4)}, {z, 0, ((n-u^5-v^4)/3)^(1/3)}, {y, 0, ((n-u^5-v^4-3z^3)/2)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 07 2016
STATUS
approved