A352627 - OEIS

%I #12 Mar 26 2022 21:19:32

%S 1,2,2,2,3,2,3,3,3,4,4,1,4,3,1,3,3,4,5,4,3,1,5,3,5,6,3,4,6,1,2,3,3,8,

%T 5,3,4,4,4,3,5,3,6,4,3,2,1,2,4,6,4,5,5,1,5,5,2,7,5,2,6,2,1,3,3,5,4,7,

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

%N Number of ways to write n as a^2 + 2*b^2 + c^4 + 4*d^4 + c^2*d^2, where a,b,c,d are nonnegative integers.

%C Conjecture: a(n) > 0 for all n = 0,1,2,.... In other words, each nonnegative integer can be written as a^2 + 2*b^2 + c^4 + 4*d^4 + c^2*d^2 with a,b,c,d integers.

%C It seems that a(n) = 1 only for n = 0, 11, 14, 21, 29, 46, 53, 62, 149, 174, 221, 239, 254, 1039, 1709, 2239.

%C See also A352628, A352629 and A352632 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A352627/b352627.txt">Table of n, a(n) for n = 0..10000</a>

%e a(11) = 1 with 11 = 3^2 + 2*1^2 + 0^4 + 4*0^4 + 0^2*0^2.

%e a(14) = 1 with 14 = 0^2 + 2*2^2 + 1^4 + 4*1^4 + 1^2*1^2.

%e a(221) = 1 with 221 = 12^2 + 2*2^2 + 1^4 + 4*2^4 + 1^2*2^2.

%e a(239) = 1 with 239 = 15^2 + 2*2^2 + 1^4 + 4*1^4 + 1^2*1^2.

%e a(254) = 1 with 254 = 1^2 + 2*6^2 + 3^4 + 4*2^4 + 3^2*2^2.

%e a(1039) = 1 with 1039 = 31^2 + 2*6^2 + 1^4 + 4*1^4 + 1^2*1^2.

%e a(1709) = 1 with 1709 = 9^2 + 2*26^2 + 4^4 + 4*1^4 + 4^2*1^2.

%e a(2239) = 1 with 2239 = 41^2 + 2*6^2 + 3^4 + 4*3^4 + 3^2*3^2.

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

%t tab={};Do[r=0;Do[If[SQ[n-4d^4-c^4-c^2*d^2-2b^2],r=r+1],{d,0,(n/4)^(1/4)},{c,0,Sqrt[(Sqrt[4n-15*d^4]-d^2)/2]},{b,0,Sqrt[(n-4d^4-c^4-c^2*d^2)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]

%Y Cf. A000290, A000583, A352628, A352629, A352632.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Mar 24 2022