OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 23, 1135, 6415, 6471.
(ii) If P(x,y) is one of the polynomials 3*x^4 + y^3 and x^6 + 3*y^2, then any positive integer n can be written as P(x,y) + z^2 + 2^k with x,y,z and k nonnegative integers.
We have verified that a(n) > 0 for all n = 2..2*10^7, and that part (ii) of the conjecture holds for all n = 1..10^7.
We also find finitely many polynomials of the form a*x^m + b*y^2 (including x^4 + y^2 and 10*x^5 + y^2) with a and b positive integers and m <= 5, for which it seems that any positive integer can be written as P(x,y) + z^2 + 2^k with x,y,z,k nonnegative integers.
See also A280153 for a similar conjecture involving powers of 4 or 8.
Qing-Hu Hou at Tianjin Univ. has verified that a(n) > 0 for all n = 2..10^9. In 2017, the author announced to offer US $234 as the prize for the first correct solution to his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Dec 30 2017
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.
EXAMPLE
a(2) = 1 since 2 = 0^4 + 0^3 + 0^2 + 2^1.
a(23) = 1 since 23 = 2^4 + 1^3 + 2^2 + 2^1.
a(1135) = 1 since 1135 = 0^4 + 7^3 + 28^2 + 2^3.
a(6415) = 1 since 6415 = 1^4 + 13^3 + 11^2 + 2^12.
a(6471) = 1 since 6471 = 1^4 + 13^3 + 57^2 + 2^10.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
In[2]:= Do[r=0; Do[If[SQ[n-2^k-x^4-y^3], r=r+1], {k, 1, Log[2, n]}, {x, 0, (n-2^k)^(1/4)}, {y, 0, (n-2^k-x^4)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 01 2017
STATUS
approved