A349778 - OEIS

%I #13 Dec 01 2021 03:57:07

%S 4,4,8,4,8,4,4,4,4,6,8,4,8,4,2,2,7,10,12,8,9,3,4,2,5,11,10,8,8,3,1,3,

%T 7,10,11,5,12,7,7,4,5,8,8,7,8,8,2,3,4,9,11,8,18,5,11,8,4,8,11,8,7,6,3,

%U 8,7,12,12,12,11,4,7,5,10,9,11,7,11,4,3,6,11,13,17,9,10,6,5,7,7,13,13,12,5,6,3,3,5,14,12,10,18

%N Number of ways to write n as x^2 + y^k + 2*z^m, where x,y,z are nonnegative integers with x >= y, and k and m belong to the set {2,3}.

%C Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 30, 120, 142.

%C We have verified this for all n <= 10^6.

%C Conjecture 2: Let S = {x^k: k = 2,3 and x = 0,1,2,...}, and let a be 3 or 4 or 5. Then any nonnegative integer can be written as x + 2*y + a*z, where x,y,z are elements of the set S.

%C Conjecture 3: Let T = {x^k: k = 2,3,4,... and x = 0,1,2,...}. If (b,c) is among the ordered pairs (1,2), (2,4), (2,5) and (3,2), then each n = 0,1,... can be written as x + b*y + c*z, where x and y are elements of T, and z is a square.

%H Zhi-Wei Sun, <a href="/A349778/b349778.txt">Table of n, a(n) for n = 0..10000</a>

%H Zhi-Wei Sun, <a href="http://hitpress.hit.edu.cn/2021/1015/c12593a261001/page.htm">New Conjectures in Number Theory and Combinatorics</a> (in Chinese), Harbin Institute of Technology Press, 2021.

%e a(3) = 4. In fact, 3 = 1^2 + 0^2 + 2*1^2 = 1^2 + 0^2 + 2*1^3 = 1^2 + 0^3 + 2*1^2 = 1^2 + 0^3 + 2*1^3 with 1 >= 0.

%e a(30) = 1 with 30 = 2^2 + 2^3 + 2*3^2 and 2 >= 2.

%e a(120) = 1 with 120 = 10^2 + 2^2 + 2*2^3 and 10 >= 2.

%e a(142) = 1 with 142 = 6^2 + 2^3 + 2*7^2 and 6 >= 2.

%t tab={};Do[r=0;Do[If[IntegerQ[((n-x^2-y^k)/2)^(1/m)],r=r+1],{x,0,Sqrt[n]},{k,2,3},{y,0,Min[x,(n-x^2)^(1/k)]},{m,2,3}];tab=Append[tab,r],{n,0,100}];Print[tab]

%Y Cf. A000290, A000578, A001597, A002760, A275168, A275169, A349787.

%K nonn

%O 0,1

%A _Zhi-Wei Sun_, Nov 29 2021