| Displaying 1-10 of 10 results found.
page
1
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x + 3*y + 5*z a positive square, where x,y,z,w are nonnegative integers such that 3*x or y or z is a square.
+10
11
1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 4, 2, 1, 5, 5, 2, 1, 2, 2, 2, 1, 4, 5, 2, 1, 4, 7, 1, 2, 5, 3, 2, 1, 3, 5, 4, 2, 8, 5, 1, 3, 5, 6, 1, 2, 4, 8, 5, 4, 2, 4, 4, 2, 6, 5, 5, 2, 1, 6, 4, 1, 8, 9, 6, 2, 3, 4, 1, 2, 6, 8, 5, 4, 5, 8, 2, 1
COMMENTS
Conjecture: a(n) > 0 for all n = 1,2,3,....
This is stronger than the author's 1-3-5 conjecture in A271518. See also A300751 for a similar conjecture stronger than the 1-3-5 conjecture. In 2020, A. Machiavelo, R. Reis and N. Tsopanidis verified a(n) > 0 for n up to 1.05*10^11. - Zhi-Wei Sun, Oct 06 2020
EXAMPLE
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 1 = 1^2 and 3 + 3*1 + 5*6 = 6^2.
a(248) = 1 since 248 = 10^2 + 2^2 + 0^2 + 12^2 with 0 = 0^2 and 10 + 3*2 + 5*0 = 4^2.
a(263) = 1 since 263 = 13^2 + 2^2 + 9^2 + 3^2 with 9 = 3^2 and 13 + 3*2 + 5*9 = 8^2.
a(808) = 1 since 808 = 12^2 + 14^2 + 18^2 + 12^2 with 3*12 = 6^2 and 12 + 3*14 + 5*18 = 12^2.
a(1288) = 1 since 1288 = 12^2 + 18^2 + 26^2 + 12^2 with 3*12 = 6^2 and 12 + 3*18 + 5*26 = 14^2.
a(3544) = 1 since 3544 = 14^2 + 34^2 + 16^2 + 44^2 with 16 = 4^2 and 14 + 3*34 + 5*16 = 14^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[(SQ[3(m^2-3y-5z)]||SQ[y]||SQ[z])&&SQ[n-(m^2-3y-5z)^2-y^2-z^2], r=r+1], {m, 1, (35n)^(1/4)}, {y, 0, Min[m^2/3, Sqrt[n]]}, {z, 0, Min[(m^2-3y)/5, Sqrt[n-y^2]]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers for which x or y or z is a square and (12*x)^2 + (15*y)^2 + (20*z)^2 is also a square.
+10
8
1, 3, 1, 1, 6, 1, 1, 3, 2, 8, 2, 2, 7, 2, 2, 1, 8, 6, 2, 8, 1, 3, 1, 1, 9, 8, 4, 3, 7, 3, 3, 3, 6, 9, 4, 4, 7, 5, 1, 8, 8, 4, 3, 3, 11, 2, 1, 1, 4, 11, 3, 8, 8, 4, 4, 2, 3, 8, 4, 2, 8, 3, 4, 1, 15, 9, 3, 9, 3, 5, 2, 6, 10, 11, 5, 3, 5, 6, 2, 6
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 1, 3, 4, 6, 7, 21, 23, 24, 39, 47, 86, 95, 344, 651, 764.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as the sum of a fourth power and three squares.
See also A300792 for two similar conjectures.
EXAMPLE
a(6) = 1 since 6 = 0^2 + 1^2 + 1^2 + 2^2 with 0 = 0^2 and (12*0)^2 + (15*1)^2 + (20*1)^2 = 25^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (15*2)^2 + (20*1)^2 = 38^2.
a(21) = 1 since 21 = 4^2 + 0^2 + 1^2 + 2^2 with 4 = 2^2 and (12*4)^2 + (15*0)^2 + (20*1)^2 = 52^2.
a(39) = 1 since 39 = 5^2 + 2^2 + 1^2 + 3^2 with 1 = 1^2 and (12*5)^2 + (15*2)^2 + (20*1)^2 = 70^2.
a(344) = 1 since 344 = 0^2 + 10^2 + 10^2 + 12^2 with 0 = 0^2 and (12*0)^2 + (15*10)^2 + (20*10)^2 = 250^2.
a(764) = 1 since 764 = 7^2 + 3^2 + 25^2 + 9^2 with 25 = 5^2 and (12*7)^2 + (15*3)^2 + (20*25)^2 = 509^2.
a(8312) = 2 since 8312 = 42^2 + 36^2 + 34^2 + 64^2 with 36 = 6^2 and (12*42)^2 + (15*36)^2 + (20*34)^2 = 1004^2, and 8312 = 66^2 + 16^2 + 44^2 + 42^2 with 16 = 4^2 and (12*66)^2 + (15*16)^2 + (20*44)^2 = 1208^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[(SQ[x]||SQ[y]||SQ[z])&&SQ[(12x)^2+(15y)^2+(20z)^2]&&SQ[n-x^2-y^2-z^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300792.
Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x or y or z is a square and 9*x^2 + 16*y^2 + 24*z^2 is also a square.
+10
8
1, 2, 1, 2, 4, 1, 2, 2, 2, 6, 2, 3, 5, 1, 4, 1, 5, 7, 4, 5, 1, 5, 2, 1, 9, 6, 5, 3, 4, 7, 2, 2, 6, 7, 3, 5, 7, 4, 4, 6, 6, 4, 5, 3, 9, 4, 2, 1, 4, 11, 5, 9, 5, 6, 4, 1, 9, 7, 3, 6, 5, 4, 4, 2, 14, 4, 6, 5, 2, 8, 2, 7, 9, 5, 5, 4, 3, 8, 1, 4
COMMENTS
Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 1, 3, 6, 14, 21, 24, 56, 79, 119, 143, 248, 301, 383, 591, 728, 959, 1223, 1751, 2311, 6119.
Conjecture 2: Any positive integer n can be written as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x or y or z is a square and 36*x^2 + 40*y^2 + 45*z^2 is also a square.
See also A300791 for a similar conjecture.
EXAMPLE
a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 1 = 1^2 and 9*1^2 + 16*1^2 + 24*0^2 = 5^2.
a(14) = 1 since 14 = 1^2 + 0^2 + 3^2 + 2^2 with 1 = 1^2 and 9*1^2 + 16*0^2 + 24*3^2 = 15^2.
a(728) = 1 since 728 = 10^2 + 0^2 + 12^2 + 22^2 with 0 = 0^2 and 9*10^2 + 16*0^2 + 24*12^2 = 66^2.
a(959) = 1 since 959 = 25^2 + 18^2 + 3^2 + 1^2 with 25 = 5^2 and 9*25^2 + 16*18^2 + 24*3^2 = 105^2.
a(1751) = 1 since 1751 = 19^2 + 25^2 + 18^2 + 21^2 with 25 = 5^2 and 9*19^2 + 16*25^2 + 24*18^2 = 145^2.
a(2311) = 1 since 2311 = 1^2 + 41^2 + 23^2 + 10^2 with 1 = 1^2 and 9*1^2 + 16*41^2 + 24*23^2 = 199^2.
a(6119) = 1 since 6119 = 1^2 + 5^2 + 3^2 + 78^2 with 1 = 1^2 and 9*1^2 + 16*5^2 + 24*3^2 = 25^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[(SQ[x]||SQ[y]||SQ[z])&&SQ[9x^2+16y^2+24z^2]&&SQ[n-x^2-y^2-z^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or 2*y or z is a square and (12*x)^2 + (21*y)^2 + (28*z)^2 is also a square.
+10
6
1, 4, 4, 2, 5, 6, 2, 2, 4, 5, 7, 1, 3, 7, 2, 3, 5, 7, 7, 2, 6, 1, 2, 2, 2, 11, 7, 3, 3, 8, 5, 1, 4, 5, 9, 4, 6, 8, 6, 4, 7, 9, 3, 3, 2, 9, 2, 1, 3, 6, 16, 5, 9, 7, 6, 5, 1, 5, 9, 4, 4, 7, 5, 5, 5, 17, 6, 4, 7, 3, 6, 3, 6, 11, 11, 4, 3, 1, 8, 2, 6
COMMENTS
Conjecture 1: a(n) > 0 for all n = 0,1,2,....
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with w a positive integer and x,y,z nonnegative integers such that x or y or z is a square and 144*x^2 + 505*y^2 + 720*z^2 is also a square.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x (or 2*x) is a square.
In 2016, the author conjectured in A271510 that each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y >= z such that (3*x)^2 + (4*y)^2 + (12*z)^2 is a square.
EXAMPLE
a(11) = 1 since 11 = 0^2 + 1^2 + 1^2 + 3^2 with 0 = 0^2 and (12*0)^2 + (21*1)^2 + (28*1)^2 = 35^2.
a(56) = 1 since 56 = 4^2 + 6^2 + 2^2 + 0^2 with 4 = 2^2 and (12*4)^2 + (21*6)^2 + (28*2)^2 = 146^2.
a(77) = 1 since 77 = 4^2 + 0^2 + 5^2 + 6^2 with 4 = 2^2 and (12*4)^2 + (21*0)^2 + (28*5)^2 = 148^2.
a(184) = 1 since 184 = 12^2 + 2^2 + 0^2 + 6^2 with 0 = 0^2 and (12*12)^2 + (21*2)^2 + (28*0)^2 = 150^2.
a(599) = 1 since 599 = 21^2 + 11^2 + 1^2 + 6^2 with 1 = 1^2 and (12*21)^2 + (21*11)^2 + (28*1)^2 = 343^2.
a(7836) = 1 since 7836 = 38^2 + 18^2 + 68^2 + 38^2 with 2*18 = 6^2 and (12*38)^2 + (21*18)^2 + (28*68)^2 = 1994^2.
a(15096) = 1 since 15096 = 16^2 + 6^2 + 52^2 + 110^2 with 16 = 4^2 and (12*16)^2 + (21*6)^2 + (28*52)^2 = 1474^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[(SQ[x]||SQ[2y]||SQ[z])&&SQ[(12x)^2+(21y)^2+(28z)^2]&&SQ[n-x^2-y^2-z^2], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; tab=Append[tab, r], {n, 0, 80}]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square , where w is a positive integer and x,y,z are nonnegative integers such that z or 2*z or 3*z is a square.
+10
5
1, 3, 1, 2, 6, 1, 2, 3, 2, 7, 2, 2, 7, 1, 5, 2, 7, 9, 3, 6, 2, 3, 4, 1, 9, 7, 3, 4, 5, 7, 2, 3, 5, 6, 3, 4, 7, 3, 8, 6, 7, 6, 4, 3, 7, 3, 2, 2, 4, 13, 5, 8, 6, 5, 3, 1, 8, 8, 3, 6, 5, 1, 11, 2, 16, 5, 4, 8, 1, 8, 2, 7, 11, 7, 5, 4, 2, 13, 2, 6
COMMENTS
Conjecture 1: a(n) > 0 for all n > 0. Moreover, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where w is a positive integer and x,y,z are nonnegative integers for which one of z, z/2, z/3 is a square and z/2 (or z/3) is a power of 4 (including 1).
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where x,y,z,w are nonnegative integers for which x or z is a square or y = 2^k for some k = 0,1,2,....
Conjecture 3. Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c) = 1. If each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that (a*x)^2 + (b*y)^2 + (c*z)^2 is a square, then (a,b,c) must be among the three triples (3,4,12), (12,15,20) and (12,21,28).
We have verified Conjectures 1 and 2 for n up to 5*10^5 and 10^6 respectively.
EXAMPLE
a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 = 0^2 and (3*1)^2 + (4*1)^2 + (12*0)^2 = 5^2.
a(14) = 1 since 14 = 3^2 + 0^2 + 1^2 + 2^2 with 1 = 1^2 and (3*3)^2 + (4*0)^2 + (12*1)^2 = 15^2.
a(69) = 1 since 69 = 0^2 + 8^2 + 2^2 + 1^2 with 2*2 = 2^2 and (3*0)^2 + (4*8)^2 + (12*2)^2 = 40^2.
a(671) = 1 since 671 = 18^2 + 17^2 + 3^2 + 7^2 with 3*3 = 3^2 and (3*18)^2 + (4*17)^2 + (12*3)^2 = 94^2.
a(1175) = 1 since 1175 = 30^2 + 5^2 + 9^2 + 13^2 with 9 = 3^2 and (3*30)^2 + (4*5)^2 + (12*9)^2 = 142^2.
a(12151) = 1 since 12151 = 50^2 + 71^2 + 49^2 + 47^2 with 49 = 7^2 and (3*50)^2 + (4*71)^2 + (12*49)^2 = 670^2.
a(16204) = 1 since 16204 = 90^2 + 90^2 + 0^2 + 2^2 with 0 = 0^2 and (3*90)^2 + (4*90)^2 + (12*0)^2 = 450^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[z]||SQ[2z]||SQ[3z], Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(3x)^2+(4y)^2+(12z)^2], r=r+1], {x, 0, Sqrt[n-1-z^2]}, {y, 0, Sqrt[n-1-x^2-z^2]}]], {z, 0, Sqrt[n-1]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 23*y^2 = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....
+10
4
1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 2, 2, 2, 4, 2, 1, 3, 5, 3, 5, 5, 4, 3, 3, 3, 5, 3, 3, 4, 6, 2, 1, 4, 3, 4, 5, 3, 4, 2, 3, 6, 5, 2, 4, 6, 4, 2, 2, 3, 6, 3, 3, 4, 6, 4, 4, 5, 4, 4, 5, 2, 4, 4, 1, 7, 8, 2, 7, 8, 5, 3, 5, 5, 7, 4, 5, 6, 5, 2, 5
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k (k = 0,1,2,...), 3, 7, 115, 151, 219, 267, 1151, 1367.
We have verified a(n) > 0 for all n = 1..10^8.
EXAMPLE
a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 23*0 = 1^3.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 23*0 = 2^2*1^3.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2^2 + 23*1^2 = 3^3.
a(48) = 2 since 48 = 4^2 + 0^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 with 4^2 + 23*0^2 = 2*2^3 and 6^2 + 23*2^2 = 2*4^3.
a(115) = 1 since 115 = 3^2 + 3^2 + 4^2 + 9^2 with 3^2 + 23*3^2 = 6^3.
a(267) = 1 since 267 = 3^2 + 1^2 + 1^2 + 16^2 with 3^2 + 23*1^2 = 2^2*2^3.
a(1151) = 1 since 1151 = 7^2 + 3^2 + 2^2 + 33^2 with 7^2 + 23*3^2 = 2^2*4^3.
a(1367) = 1 since 1367 = 17^2 + 5^2 + 18^2 + 27^2 with 17^2 + 23*5^2 = 2^2*6^3.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
tab={}; Do[r=0; Do[If[QQ[x^2+23y^2], Do[If[SQ[n-x^2-y^2-z^2], r=r+1], {z, 0, Sqrt[(n-x^2-y^2)/2]}]], {y, 0, Sqrt[n/2]}, {x, y, Sqrt[n-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301304, A301314.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 7*y^2 = 2^k*m for some k = 0,1,2 and m = 1,2,3,....
+10
4
1, 2, 2, 2, 2, 3, 1, 2, 3, 4, 3, 4, 2, 3, 2, 2, 4, 5, 3, 4, 4, 2, 1, 3, 2, 6, 5, 4, 3, 4, 2, 2, 6, 5, 4, 6, 3, 3, 3, 4, 6, 8, 2, 5, 5, 3, 2, 4, 4, 5, 6, 5, 5, 4, 3, 3, 6, 5, 2, 6, 3, 4, 3, 2, 6, 10, 3, 5, 8, 1, 2, 5, 5, 6, 5, 6, 5, 3, 1, 4
COMMENTS
Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 10^8.
EXAMPLE
a(79) = 1 since 79 = 5^2 + 1^2 + 2^2 + 7^2 with 5^2 + 7*1^2 = 2^2*2^3.
a(323) = 1 since 323 = 3^2 + 1^2 + 12^2 + 13^2 with 3^2 + 7*1^2 = 2*2^3.
a(646) = 1 since 646 = 22^2 + 11^2 + 4^2 + 5^2 with 22^2 + 7*11^2 = 11^3.
a(815) = 1 since 815 = 9^2 + 5^2 + 15^2 + 22^2 with 9^2 + 7*5^2 = 2^2*4^3.
a(1111) = 1 since 1111 = 1^2 + 1^2 + 22^2 + 25^2 with 1^2 + 7*1^2 = 2^3.
a(2822) = 1 since 2822 = 2^2 + 0^2 + 3^2 + 53^2 with 2^2 + 7*0^2 = 2^2*1^3.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
tab={}; Do[r=0; Do[If[QQ[x^2+7y^2], Do[If[SQ[n-x^2-y^2-z^2], r=r+1], {z, 0, Sqrt[(n-x^2-y^2)/2]}]], {y, 0, Sqrt[n/2]}, {x, y, Sqrt[n-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301303, A301314.
Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x + 3*y + 9*z = 2^k*m^3 for some k,m = 0,1,2,....
+10
4
1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 2, 1, 4, 4, 1, 2, 1, 3, 1, 1, 4, 2, 4, 1, 2, 3, 1, 1, 4, 2, 4, 3, 2, 5, 4, 3, 3, 7, 3, 2, 3, 1, 3, 1, 3, 6, 7, 2, 4, 7, 3, 1, 5, 2, 6, 3, 2, 7, 7, 1, 4, 9, 3, 4, 2, 5, 5, 4, 5, 4, 6, 1, 5, 5, 1, 2
COMMENTS
Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2, where x is a positive integer and y,z,w are nonnegative integers such that 2*x + 7*y = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....
We have verified a(n) > 0 for all n = 1..10^7.
EXAMPLE
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 + 3*2 + 9*1 = 2*2^3.
a(19) = 1 since 19 = 4^2 + 1^2 + 1^2 + 1^2 with 4 + 3*1 + 9*1 = 2*2^3.
a(46) = 1 since 46 = 0^2 + 6^2 + 1^2 + 3^2 with 0 + 3*6 + 9*1 = 3^3.
a(79) = 1 since 79 = 2^2 + 7^2 + 1^2 + 5^2 with 2 + 3*7 + 9*1 = 2^2*2^3.
a(125) = 1 since 125 = 2^2 + 0^2 + 0^2 + 11^2 with 2 + 3*0 + 9*0 = 2*1^3.
a(736) = 1 since 736 = 0^2 + 24^2 + 4^2 + 12^2 with 0 + 3*24 + 9*4 = 2^2*3^3.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
QQ[n_]:=CQ[n]||CQ[n/2]||CQ[n/4];
tab={}; Do[r=0; Do[If[QQ[x+3y+9z]&&SQ[n-x^2-y^2-z^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301303, A301304.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x - 3*y a square, where x,y,z,w are nonnegative integers with z <= w such that 10*x or y is a square.
+10
1
1, 2, 3, 2, 2, 2, 2, 1, 1, 2, 4, 3, 2, 1, 2, 2, 2, 3, 5, 3, 4, 2, 1, 1, 1, 4, 6, 5, 2, 3, 3, 1, 3, 4, 5, 4, 5, 3, 3, 2, 2, 6, 6, 2, 1, 4, 2, 2, 2, 2, 9, 6, 6, 3, 4, 3, 1, 4, 3, 4, 4, 4, 3, 3, 2, 6, 9, 4, 5, 4, 4, 1, 2, 4, 7, 9, 2, 3, 3, 1, 2
COMMENTS
Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 7, 8, 13, 22, 23, 24, 31, 44, 56, 71, 79, 88, 109, 120, 152, 184, 472, 1912, 6008, 9080.
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with 3*x - y twice a square, where x,y,z,w are nonnegative integers such that 5*x or y is a square.
By the author's 2017 JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.
a(n) > 0 for all n = 0..10^8. Also, Conjecture 2 holds for all n = 0..10^8. - Zhi-Wei Sun, Oct 05 2020
EXAMPLE
a(22) = 1 since 22 = 1^2 + 1^2 + 2^2 + 4^2 with 1 = 1^2 and 4*1 - 3*1 = 1^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 1 = 1^2 and 4*3 - 3*1 = 3^2.
a(109) = 1 since 109 = 0^2 + 0^2 + 3^2 + 10^2 with 0 = 0^2 and 4*0 - 3*0 = 0^2.
a(184) = 1 since 184 = 10^2 + 8^2 + 2^2 + 4^2 with 10*10 = 10^2 and 4*10 - 3*8 = 4^2.
a(6008) = 1 since 6008 = 12^2 + 16^2 + 42^2 + 62^2 with 16 = 4^2 and 4*12 - 3*16 = 0^2.
a(9080) = 1 since 9080 = 10^2 + 12^2 + 0^2 + 94^2 with 10*10 = 10^2 and 4*10 - 3*12 = 2^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[Mod[m^2+3y, 4]==0&&(SQ[10(m^2+3y)/4]||SQ[y]), Do[If[SQ[n-((m^2+3y)/4)^2-y^2-z^2], r=r+1], {z, 0, Sqrt[Max[0, (n-((m^2+3y)/4)^2-y^2)/2]]}]], {m, 0, 2n^(1/4)}, {y, 0, 4/5*Sqrt[n-m^4/16]}]; tab=Append[tab, r], {n, 0, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271518, A281976, A282542, A300666, A300667, A300219, A300708, A300712, A300751, A300752.
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that x*(y+3*z) is a cube or half a cube and y <= z <= w if x = 0.
+10
1
1, 2, 2, 2, 3, 2, 2, 2, 3, 5, 2, 3, 4, 1, 1, 1, 5, 7, 3, 3, 5, 3, 1, 4, 6, 7, 4, 3, 6, 1, 4, 2, 6, 7, 1, 5, 4, 4, 2, 5, 5, 4, 5, 2, 8, 4, 2, 1, 4, 7, 5, 7, 6, 4, 3, 3, 4, 7, 2, 1, 5, 3, 2, 2, 7, 10, 6, 4, 6, 3, 4, 5, 9, 8, 5, 4, 2, 6, 2, 3
COMMENTS
Conjecture: a(n) > 0 for all n > 0.
We have verified this for all n = 1..3*10^6.
EXAMPLE
a(14) = 1 since 14 = 0^2 + 1^2 + 2^2 + 3^2 with 0*(1+3*2) = 0^3.
a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with 2*(1+3*1) = 2^3.
a(16) = 1 since 16 = 0^2 + 0^2 + 0^2 + 4^2 with 0*(0+3*0) = 0^3.
a(60) = 1 since 60 = 4^2 + 2^2 + 2^2 + 6^2 with 4*(2+3*2) = 4^3/2.
a(92) = 1 since 92 = 6^2 + 6^2 + 4^2 + 2^2 with 6*(6+3*4) = 6^3/2.
a(240) = 1 since 240 = 2^2 + 14^2 + 6^2 + 2^2 with 2*(14+3*6) = 4^3.
a(807) = 1 since 807 = 1^2 + 21^2 + 2^2 + 19^2 with 1*(21+3*2) = 3^3.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
QQ[n_]:=QQ[n]=CQ[n]||CQ[2n];
tab={}; Do[r=0; Do[If[QQ[x(y+3z)]&&SQ[n-x^2-y^2-z^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[If[x==0, n/3, n-1-x^2]]}, {z, If[x==0, y, 0], Sqrt[If[x==0, (n-x^2-y^2)/2, n-1-x^2-y^2]]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301303, A301304, A301314.
Search completed in 0.009 seconds
|