来自在线整数百科全书的问候语!http://oeis.org/ Search: id:a267121 Showing 1-1 of 1 %I A267121 %S A267121 1,3,2,1,6,7,1,3,7,7,6,2,6,12,1,1,12,10,7,6,13,7,2,7,8,19,8,1,18,12,2, %T A267121 3,14,15,13,7,7,18,1,7,25,14,6,6,19,13,2,2,14,22,12,6,18,27,4,12,13,9, %U A267121 19,1,18,25,5,1,24,26,6,12,26,14,2,10,16,31,16,7,24,13,4,6 %N A267121 Number of ordered ways to write n as x^2 +y ^ ^ 2+z ^ 2+w ^ 2与x*y*z(x+**y+4*Z+10×w)为正方形,其中x为正整数,y,z,w为非负整数。(%1)A267121猜想:(i)a(n)>0,n=0,a(n)=1仅为n=4 ^ k*m(k=0,1,2,…)m=1, 7, 15,39, 119, 127,159, 239, 359,391, 527, 543,863, 5791).% c~a267121(ii)任何正整数都可以写成x* 2+y^ 2+z ^ 2+w ^ 2,其中2×x*y*(x+2y+z +2W)(或2×x*y*(x+6y+z +2w),或者x*y*(x+11y+z +2w))为正方形,其中x,y,z,w是非负整数,具有z>0(或w>0)。D*Z)是一个正方形,假设(A,B,C,D)是以下四个(1,1,2,3),(1,1,4,5),(1,1,6,9),(1,2,6,34),(1,3,6,M)(M,12, 21, 27,36),(1,3,9,18),(1,3,9,36),(1,3,18,27),(1,3,24117),(1,3,90,99),(1,6,6,18),(1,6,6,30),(1,8,16,24),(1,8,16,24)。x^ 2+y^ 2+z ^ 2与w,x,y,z非负整数,使得w*(a*W+b*x+c*y++)拉格朗日四方定理的改进,参见ARXIV:1604.06723 .%%H A267121支伟隼,n,a(n)n=1…10000的表%H A267121支伟隼,拉格朗日四方定理的改进,ARXIV:1604.06723 [数学,GM ],2016。%%H A267121支伟隼,细化拉格朗日四方定理,一个数字理论列表,4月26日,2016. %e A267121 a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2 > 0 and 2*0*0*(2+9*0+11*0+10*0) = 0^2. %e A267121 a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2 > 0 and 2*1*1*(2+9*1+11*1+10*1) = 8^2. %e A267121 a(15) = 1 since 15 = 2^2 + 1^2 + 3^2 + 1^2 with 2 > 0 and 2*1*3*(2+9*1+11*3+10*1) = 18^2. %e A267121 a(39) = 1 since 39 = 1^2 + 1^2 + 1^2 + 6^2 with 1 > 0 and 1*1*1*(1+9*1+11*1+10*6) = 9^2. %e A267121 a(119) = 1 since 119 = 1^2 + 1^2 + 9^2 + 6^2 with 1 > 0 and 1*1*9*(1+9*1+11*9+10*6) = 39^2. %e A267121 a(127) = 1 since 127 = 5^2 + 1^2 + 1^2 + 10^2 with 5 > 0 and 5*1*1*(5+9*1+11*1+10*10) = 25^2. %e A267121 a(159) = 1 since 159 = 3^2 + 1^2 + 7^2 + 10^2 with 3 > 0 and α=α^ +α^ +α^ +α+ ^,α*和* * * * *(α+ * + * + * * + + * *)=α^。3×1×7 *(3+9×1+11×7+10×10)=63 ^ 2。α=α^ +α^ +α^ +α+ ^,α*和* * * * *(α+ * + * + * * + + * *)=α^。19×5×1 *(19+9×5+11×1+10×2)=95 ^ 2。9*9*5*(9+9*9+11*5+10*26) = 405^2. %e A267121 a(5791) = 1 since 5791 = 57^2 + 38^2 + 33^2 + 3^2 with 57 > 0 and 57*38*33*(57+9*38+11*33+10*3) = 7524^2. %t A267121 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]] %t A267121 Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x*y*z(x+9y+11z+10*Sqrt[n-x^2-y^2-z^2])],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];CF.A000 0118、A266076、A2623、A26857、A26850、A269400、A27 1510、A27 1513、A27 1518、A27 1608、A27 1765、A27 1714、A27 1721、A27 1724、A27 1775、A27 1778、A27 1824、A27 204、A27 23 32、A27 23 51、K %A267121、NN 2%A267121、1 2 2 % A267121打印[n,],“r];继续,{n,1,80}[y%A267121在OEIS最终用户许可协议下,可以获得01 2016×2016的内容:HTTP:/OEIS.Org/许可证