A297995 Number of ways to write n as 4*u + v^2 + x^3 + y^4 + 2*z^8, where u is 0 or 1, v is a positive integer and x,y,z are nonnegative integers.

1, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 6, 5, 5, 5, 7, 6, 6, 5, 4, 5, 5, 6, 3, 5, 4, 4, 5, 5, 5, 5, 6, 5, 5, 5, 4, 4, 4, 2, 2, 2, 3, 4, 6, 6, 5, 7, 6, 7, 5, 6, 4, 4, 4, 3, 4, 3, 5, 4

Offset: 1

Author

Zhi-Wei Sun, Jan 10 2018

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 398, 496, 498, 500, 1507, 6419, 7843, 7983, 8688, 8947, 9175, 9251, 12923, 12976, 48381.

(ii) For any positive integers a,b,c,d,e and integers g,h,i,j,k greater than one, if each n = 0,1,2,... can be written as a*u^g + b*v^h + c*x^i + d*y^j + e*z^k with u,v,x,y,z nonnegative integers, then 1/g + 1/h + 1/i + 1/j + 1/k is greater than 1/2 + 1/3 + 1/4 + 1/8 = 29/24.

Examples

			a(500) = 1 since 500 = 4*0 + 22^2 + 0^3 + 2^4 + 2*0^8.
a(1507) = 1 since 1507 = 4*1 + 13^2 + 11^3 + 1^4 + 2*1^8.
a(9251) = 1 since 9251 = 4*0 + 91^2 + 7^3 + 5^4 + 2*1^8.
a(12923) = 1 since 12923 = 4*1 + 54^2 + 1^3 + 10^4 + 2*1^8.
a(12976) = 1 since 12976 = 4*0 + 92^2 + 15^3 + 5^4 + 2*2^8.
a(48381) = 1 since 48381 = 4*1 + 140^2 + 6^3 + 13^4 + 2*0^8.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Remark 3.5.)

Crossrefs

Cf. A000290, A000578, A000583, A262827, A280356.

Programs

Mathematica

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-4u-2z^8-y^4-x^3],r=r+1],{u,0,Boole[n>3]},{z,0,((n-4u)/2)^(1/8)},{y,0,(n-4u-2z^8)^(1/4)},{x,0,(n-4u-2z^8-y^4)^(1/3)}];Print[n," ",r],{n,1,80}]