A349957 -id:A349957

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A349661

Number of ways to write n as x^4 + y^2 + (z^2 + 4^w)/2 with x,y,z,w nonnegative integers.

+10
4

1, 3, 3, 2, 4, 5, 2, 2, 4, 5, 6, 4, 3, 6, 4, 1, 6, 7, 4, 6, 8, 5, 1, 4, 6, 10, 10, 3, 6, 10, 2, 3, 8, 6, 10, 10, 5, 6, 4, 5, 12, 14, 6, 5, 9, 6, 2, 3, 6, 12, 14, 7, 5, 8, 2, 7, 14, 6, 9, 9, 5, 9, 4, 2, 10, 15, 7, 7, 8, 7, 3, 5, 5, 7, 14, 5, 9, 9, 1, 4, 11, 8, 11, 13, 7, 13, 7, 2, 11, 17, 12, 8, 5, 6, 7, 5, 7, 11, 12, 8

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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

This has been verified for n up to 10^6.

As (x^2 + y^2)/2 = ((x+y)/2)^2 + ((x-y)/2)^2, the conjecture gives a new refinement of Lagrange's four-square theorem.

See also A350012 for a similar conjecture.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.

Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15(2019), 1863-1893. See also arXiv:1701.05868 [math.NT].

Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

EXAMPLE

a(1) = 1 with 1 = 0^4 + 0^2 + (1^2 + 4^0)/2.

a(23) = 1 with 23 = 1^4 + 3^2 + (5^2 + 4^0)/2.

a(79) = 1 with 79 = 1^4 + 2^2 + (12^2 + 4^1)/2.

a(1199) = 1 with 1199 = 5^4 + 18^2 + (22^2 + 4^2)/2.

a(3679) = 1 with 3679 = 5^4 + 2^2 + (78^2 + 4^2)/2.

a(6079) = 1 with 6079 = 3^4 + 42^2 + (92^2 + 4^1)/2.

a(33439) = 1 with 33439 = 1^4 + 175^2 + (75^2 + 4^0)/2.

a(50399) = 1 with 50399 = 13^4 + 135^2 + (85^2 + 4^0)/2.

a(207439) = 1 with 207439 = 1^4 + 142^2 + (612^2 + 4^1)/2.

MATHEMATICA

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

tab={}; Do[r=0; Do[If[SQ[2(n-x^4-y^2)-4^z], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 0, Sqrt[n-1-x^4]}, {z, 0, Log[4, 2(n-x^4-y^2)]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

CROSSREFS

Cf. A000118, A000290, A000302, A000583, A349957, A349992, A350012.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 08 2021

STATUS

approved

A349945

Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/5 with a,b,c,d nonnegative integers.

+10
3

1, 3, 4, 3, 3, 6, 7, 3, 1, 4, 5, 4, 2, 3, 8, 5, 3, 9, 10, 6, 7, 11, 10, 3, 2, 6, 8, 9, 3, 9, 16, 5, 4, 11, 9, 7, 9, 9, 12, 7, 2, 8, 11, 7, 2, 11, 14, 4, 3, 10, 10, 9, 8, 9, 21, 9, 3, 9, 5, 7, 4, 10, 17, 8, 3, 15, 15, 9, 9, 16, 20, 5, 3, 5, 7, 11, 3, 11, 18, 4, 6, 22, 18, 11, 14, 15, 19, 10, 2, 9, 16, 10, 3, 9, 16, 11, 7, 19, 16, 13, 12

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 2^(4*k+3) (k = 0,1,2,...).

This has been verified for all n <= 10^5.

Conjecture 2: Each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + w^2)/5 with x,y,z,w nonnegative integers, provided that (a,b,c) is among the four triples (1,2,4), (2,1,1), (6,1,1), (6,1,6).

See also A349942 for a similar conjecture.

Via a computer search, we have found many tuples (a,b,c,d,m) of positive integers (such as (1,1,4,2,3), (4,1,1,2,3) and (1,1,19,1,4900)) for which we guess that each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + d*w^2)/m with x,y,z,w nonnegative integers.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.

Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

EXAMPLE

a(8) = 1 with 8 = 0^4 + 2^2 + (2^4 + 2^2)/5.

MATHEMATICA

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

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

CROSSREFS

Cf. A000290, A000583, A349942, A349956, A349957.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 06 2021

STATUS

approved

A349992

Number of ways to write n as x^4 + y^2 + (z^2 + 2*4^w)/3, where x, y, z are nonnegative integers, and w is 0 or 1.

+10
3

1, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 2, 4, 8, 7, 7, 6, 5, 6, 6, 6, 8, 7, 8, 6, 1, 4, 2, 6, 8, 6, 7, 5, 7, 6, 6, 6, 7, 7, 8, 7, 3, 5, 3, 4, 6, 6, 6, 7, 5, 3, 5, 4, 9, 8, 9, 8, 2, 4, 1, 2, 9, 8, 10, 8, 4, 6, 4, 9, 6, 6, 6, 4, 2, 2, 1, 2, 10, 10, 13, 8, 9, 7, 9, 9, 7, 10, 6, 10, 4, 3, 4, 3, 11, 10, 9

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 30, 64, 80, 302, 350, 472, 480, 847, 3497, 13582, 25630, 38064.

This has been verified for n up to 10^6.

Conjecture 2: If (a,b,c,m) is one of the ordered tuples (1,1,11,12), (1,1,11,60), (1,1,14,15), (1,1,23,24), (1,1,23,32), (1,1,23,48), (1,2,23,96), (2,1,11,60), (2,1,23,24), (2,1,23,48), (4,1,23,48), then each n = 1 2,3,... can be written as a*x^4 + b*y^2 + (z^2 + c*4^w)/m, where x,y,z are nonnegative integers, and w is 0 or 1.

We have verified Conjecture 2 for n up to 2*10^5.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

EXAMPLE

a(30) = 1 with 30 = 1^4 + 5^2 + (2^2 + 2*4)/3.

a(480) = 1 with 480 = 1^4 + 14^2 + (29^2 + 2*4)/3.

a(847) = 1 with 847 = 0^4 + 29^2 + (4^2 + 2*4^0)/3.

a(3497) = 1 with 3497 = 4^4 + 48^2 + (53^2 + 2*4^0)/3.

a(13582) = 1 with 13582 = 9^4 + 28^2 + (53^2 + 2*4^0)/3.

a(25630) = 1 with 25630 = 5^4 + 158^2 + (11^2 + 2*4^0)/3.

a(38064) = 1 with 38064 = 3^4 + 157^2 + (200^2 + 2*4^0)/3.

MATHEMATICA

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

tab={}; Do[r=0; Do[If[SQ[3(n-x^4-y^2)-2*4^z], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 0, Sqrt[n-1-x^4]}, {z, 0, 1}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

CROSSREFS

Cf. A000290, A000583, A349942, A349943, A349957.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 08 2021

STATUS

approved

A350012

Number of ways to write n as 4*x^4 + y^2 + (z^2 + 4^w)/2 with x,y,z,w nonnegative integers.

+10
3

1, 2, 1, 1, 4, 4, 1, 3, 5, 5, 3, 3, 4, 7, 3, 2, 6, 5, 2, 4, 6, 2, 2, 5, 4, 6, 2, 2, 6, 7, 2, 2, 6, 5, 5, 4, 3, 7, 5, 5, 8, 6, 2, 6, 9, 4, 2, 4, 5, 8, 3, 3, 5, 8, 3, 6, 5, 3, 6, 4, 6, 5, 6, 1, 10, 9, 2, 6, 11, 8, 1, 7, 5, 11, 6, 4, 7, 10, 3, 6, 10, 4, 8, 8, 6, 8, 6, 5, 11, 13, 5, 1, 11, 8, 3, 4, 4, 9, 7, 6

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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

This is a new refinement of Lagrange's four-square theorem since (x^2 + y^2)/2 = ((x+y)/2)^2 + ((x-y)/2)^2. We have verified the conjecture for n up to 10^6.

See also A349661 for a similar conjecture.

We also have some other conjectures of such a type.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.

Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15(2019), 1863-1893. See also arXiv:1701.05868 [math.NT].

Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

EXAMPLE

a(1) = 4*0^4 + 0^2 + (1^2 + 4^0)/2.

a(3) = 1 with 3 = 4*0^4 + 1^2 + (0^2 + 4)/2.

a(4) = 1 with 4 = 4*0^4 + 0^2 + (2^2 + 4)/2.

a(7) = 1 with 7 = 4*1^4 + 1^2 + (0^2 + 4)/2.

a(71) = 1 with 71 = 4*1^4 + 3^2 + (10^2 + 4^2)/2.

a(92) = 1 with 92 = 4*1^4 + 6^2 + (10^2 + 4)/2.

a(167) = 1 with 167 = 4*1^4 + 9^2 + (10^2 + 4^3)/2.

a(271) = 1 with 271 = 4*1^4 + 11^2 + (6^2 + 4^4)/2.

a(316) = 1 with 316 = 4*1^4 + 4^2 + (24^2 + 4^2)/2.

a(4796) = 1 with 4796 = 4*5^4 + 36^2 + (44^2 + 4^3)/2.

a(14716) = 1 with 14716 = 4*5^4 + 4^2 + (156^2 + 4^3)/2.

a(24316) = 1 with 24316 = 4*3^4 + 84^2 + (184^2 + 4^2)/2.

MATHEMATICA

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

tab={}; Do[r=0; Do[If[SQ[2(n-4x^4-y^2)-4^z], r=r+1], {x, 0, ((n-1)/4)^(1/4)}, {y, 0, Sqrt[n-1-4x^4]}, {z, 0, Log[4, 2(n-4x^4-y^2)]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

CROSSREFS

Cf. A000118, A000290, A000302, A000583, A349661, A349957, A349992.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 08 2021

STATUS

approved

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