|
%I #11 Mar 26 2022 21:19:46
%S 1,1,1,2,1,2,2,1,2,1,1,2,1,1,2,2,3,2,2,2,1,3,1,2,3,2,3,3,1,3,3,2,3,2,
%T 1,4,3,4,2,2,3,4,3,3,3,3,2,5,4,2,3,4,2,3,1,4,5,4,1,4,3,2,5,4,3,4,4,4,
%U 2,2,1,6,4,3,2,2,2,4,2,4,5,6,3,5,2,3,4,5,3,5,3,4,6,1,3,6,5,4,5,3
%N Number of ways to write n as (a+b)*(a+2*b) + c^4 + 2*d^4, where a is a positive integer, and b,c,d are nonnegative integers.
%C Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as u*v + x^4 + 2*y^4, where u,v,x,y are nonnegative integers with 0 < u <= v < 2*u.
%C It seems that a(n) = 1 only for n = 1, 2, 3, 5, 8, 10, 11, 13, 14, 21, 23, 29, 35, 55, 59, 71, 94, 127, 134, 245, 329, 395, 404.
%C See also A352627, A352628 and A352629 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A352632/b352632.txt">Table of n, a(n) for n = 1..10000</a>
%e a(11) = 1 with 11 = (3+0)*(3+2*0) + 0^4 + 2*1^4.
%e a(23)= 1 with 23 = (3+1)*(3+2*1) + 1^4 + 2*1^4.
%e a(134) = 1 with 134 = (10+1)*(10+2*1) + 0^4 + 2*1^4.
%e a(245) = 1 with 245 = (10+1)*(10+2*1) + 3^4 + 2*2^4.
%e a(329) = 1 with 329 = (6+6)*(6+2*6) + 3^4 + 2*2^4.
%e a(395) = 1 with 395 = (2+11)*(2+2*11) + 3^4 + 2*1^4.
%e a(404) = 1 with 404 = (15+2)*(15+2*2) + 3^4 + 2*0^4.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t tab={};Do[r=0;Do[If[SQ[4(n-2d^4-c^4)+b^2],r=r+1],{d,0,((n-1)/2)^(1/4)},{c,0,(n-1-2d^4)^(1/4)},{b,0,Sqrt[(n-1-2d^4-c^4)/2]}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000290, A000583, A352627, A352628, A352629.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, Mar 25 2022
|
