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

1,2

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.

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.)

Example

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.

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}]

Crossrefs

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

Sequence in context: A029130 A081611 A326539 * A081228 A270826 A363069

Adjacent sequences: A297992 A297993 A297994 * A297996 A297997 A297998

Keyword

nonn

Author

Zhi-Wei Sun, Jan 10 2018

Status

approved