A352503 - OEIS

%I #27 Mar 31 2022 03:03:03

%S 1,1,1,1,1,2,2,3,2,2,2,2,3,2,2,1,2,2,1,1,1,2,1,1,1,1,2,2,3,2,1,2,4,4,

%T 2,2,1,2,2,2,3,2,3,2,2,2,3,5,3,2,1,2,2,2,3,2,2,1,2,3,4,4,1,4,5,3,6,4,

%U 5,4,5,5,3,5,3,5,1,1,1,3,6,2,3,2,4,4,3,3,2,4,2,2,3,1,3,4,5,2,5,4

%N Number of ways to write n as w^3 + 2*x^3 + 4*y^3 + 5*z^3 + t^6, where w is a positive integer, and x,y,z,t are nonnegative integers.

%C Conjecture: a(n) > 0 for all n > 0.

%C This has been verified for n = 1..10^6.

%C It seems that a(n) = 1 only for n = 1..5, 16, 19, 20, 21, 23, 24, 25, 26, 31, 37, 51, 58, 63, 77, 78, 79, 94, 108, 207, 208, 218, 316, 487, 490, 559.

%H Zhi-Wei Sun, <a href="/A352503/b352503.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. (See Conjecture 3.4(i).)

%e a(20) = 1 with 20 = 2^3 + 2*1^3 + 4*1^3 + 5*1^3 + 1^6.

%e a(79) = 1 with 79 = 2^3 + 2*1^3 + 4*0^3 + 5*1^3 + 2^6.

%e a(316) = 1 with 316 = 1^3 + 2*3^3 + 4*4^3 + 5*1^3 + 0^6.

%e a(487) = 1 with 487 = 5^3 + 2*!^3 + 4*4^3 + 5*2^3 + 2^6.

%e a(490) = 1 with 490 = 2^3 + 2*3^3 + 4*3^3 + 5*4^3 + 0^6.

%e a(559) = 1 with 559 = 8^3 + 2*1^3 + 4*1^3 + 5*2^3 + 1^6.

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];

%t tab={};Do[r=0;Do[If[CQ[n-t^6-2x^3-4y^3-5z^3],r=r+1],{t,0,(n-1)^(1/6)},{x,0,((n-1-t^6)/2)^(1/3)},{y,0,((n-1-t^6-2x^3)/4)^(1/3)},{z,0,((n-1-t^6-2x^3-4y^3)/5)^(1/3)}];tab=Append[tab,r],{n,1,100}];Print[tab]

%Y Cf. A000578, A001014, A271099, A271237.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Mar 28 2022