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A280356 Number of ways to write n as x^4 + y^3 + z^2 + 2^k, where x,y,z are nonnegative integers and k is a positive integer. 5

%I #26 Feb 20 2018 04:35:56

%S 0,1,3,4,4,4,3,3,5,5,4,5,6,5,2,3,7,8,7,7,8,5,1,4,9,8,5,7,8,6,3,8,14,

%T 11,7,8,7,4,4,8,13,9,4,8,8,5,4,8,11,5,5,8,8,6,4,6,9,6,6,10,6,2,3,4,10,

%U 10,9,13,12,7,2,7,11,9,7,9,6,2,3,7

%N Number of ways to write n as x^4 + y^3 + z^2 + 2^k, where x,y,z are nonnegative integers and k is a positive integer.

%C Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 23, 1135, 6415, 6471.

%C (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.

%C 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.

%C 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.

%C See also A280153 for a similar conjecture involving powers of 4 or 8.

%C 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

%H Zhi-Wei Sun, <a href="/A280356/b280356.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/179b.pdf">New conjectures on representations of integers (I)</a>, Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

%e a(2) = 1 since 2 = 0^4 + 0^3 + 0^2 + 2^1.

%e a(23) = 1 since 23 = 2^4 + 1^3 + 2^2 + 2^1.

%e a(1135) = 1 since 1135 = 0^4 + 7^3 + 28^2 + 2^3.

%e a(6415) = 1 since 6415 = 1^4 + 13^3 + 11^2 + 2^12.

%e a(6471) = 1 since 6471 = 1^4 + 13^3 + 57^2 + 2^10.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t 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}]

%Y Cf. A000079, A000290, A000578, A000583, A262827, A262956, A270533, A270559.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Jan 01 2017

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