cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A346643 Number of ways to write n as w^2 + 2*x^2 + 3*y^4 + 4*z^4, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 3, 3, 4, 2, 3, 4, 3, 1, 3, 4, 1, 3, 3, 2, 3, 4, 2, 2, 4, 2, 3, 2, 2, 2, 3, 2, 2, 3, 1, 5, 3, 2, 3, 4, 3, 1, 3, 2, 2, 1, 2, 4, 2, 3, 5, 4, 3, 7, 4, 3, 7, 5, 2, 4, 6, 1, 2, 6, 2, 6, 5, 5, 4, 8, 5, 5, 7, 2, 8, 8, 2, 2, 7, 4, 6, 5, 4, 7, 8, 7, 1, 7, 6, 3, 5, 4, 3, 2, 2, 5, 4, 3, 7, 8
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 24 2022

Keywords

Comments

1-2-3-4 Conjecture: a(n) > 0 except for n = 158.
This has been verified for n up to 10^8.
It seems that a(n) = 1 only for n = 0, 1, 2, 14, 17, 35, 42, 46, 62, 87, 119, 122, 168, 189, 206, 234, 237, 302, 317, 398, 545, 1037, 1437, 4254.
See also A347865 and A350857 for similar conjectures.

Examples

			a(46) = 1 with 46 = 5^2 + 2*3^2 + 3*1^4 + 4*0^4.
a(119) = 1 with 119 = 7^2 + 2*3^2 + 3*2^4 + 4*1^4.
a(398) = 1 with 398 = 13^2 + 2*9^2 + 3*1^4 + 4*2^4.
a(545) = 1 with 545 = 19^2 + 2*6^2 + 3*2^4 + 4*2^4.
a(1037) = 1 with 1037 = 31^2 + 2*6^2 + 3*0^4 + 4*1^4.
a(1437) = 1 with 1437 = 9^2 + 2*26^2 + 3*0^4 + 4*1^4.
a(4254) = 1 with 4254 = 45^2 + 2*31^2 + 3*3^4 + 4*2^4.
		

Crossrefs

Programs

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

A347865 Number of ways to write n as w^2 + 2*x^2 + y^4 + 3*z^4, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 3, 3, 2, 4, 3, 2, 5, 3, 1, 2, 3, 3, 4, 6, 5, 4, 6, 3, 2, 6, 2, 5, 7, 1, 3, 3, 2, 4, 5, 4, 6, 7, 4, 3, 3, 4, 2, 4, 4, 2, 3, 2, 4, 6, 5, 7, 10, 4, 7, 7, 1, 9, 6, 3, 7, 3, 2, 2, 4, 5, 7, 11, 6, 4, 9, 3, 5, 11, 2, 7, 10, 2, 2, 2, 4, 8, 12, 7, 9, 10, 7, 6, 5, 7, 6, 7, 8, 5, 1, 2, 4, 10, 7, 11, 15
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 24 2022

Keywords

Comments

Conjecture 1: a(n) > 0 except for n = 744.
This has been verified for n up to 10^8.
It seems that a(n) = 1 only for n = 0, 14, 29, 56, 94, 110, 158, 159, 224, 239, 296, 464, 589, 1214, 1454, 1709.
Conjecture 2: For any positive odd integer a, all sufficiently large integers can be written as a*w^4 + 2*x^4 + (2*y)^2 + z^2 with w,x,y,z integers. If M(a) denotes the largest integer not of the form a*w^4 + 2*x^4 + (2*y)^2 + z^2 (with w,x,y,z integers), then M(1) = 255, M(3) = 303, M(5) = 497, M(7) = 3182, M(9) = 4748, M(11) = 5662, M(13) = 5982, M(15) = 10526, M(17) = 4028 and M(19) = 11934.
Conjecture 3: Let E(a,b,c) be the set of nonnegative integers not of the form w^2 + a*x^2 + b*y^4 + c*z^4 with w,x,y,z integers. Then E(1,2,4) = {135, 190, 510}, E(1,2,5) = {35, 254, 334}, E(2,1,4) = {190, 270, 590} and E(2,3,7) = {94, 490, 983} and E(3,1,2) = {56, 168, 378}.
See also A346643 and A350857 for similar conjectures.

Examples

			a(14) = 1 with 14 = 3^2 + 2*1^2 + 0^4 + 3*1^4.
a(158) = 1 with 158 = 11^2 + 2*3^2 + 2^4 + 3*1^4.
a(589) = 1 with 589 = 14^2 + 2*14^2 + 1^4 + 3*0^4.
a(1214) = 1 with 1214 = 27^2 + 2*11^2 + 0^4 + 3*3^4.
a(1454) = 1 with 1454 = 27^2 + 2*19^2 + 0^4 + 3*1^4.
a(1709) = 1 with 1709 = 29^2 + 2*0^2 + 5^4 + 3*3^4.
		

Crossrefs

Programs

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

A214891 Numbers that are not the sum of two squares and two fourth powers.

Original entry on oeis.org

23, 44, 71, 79, 184, 368, 519, 599, 704, 1136, 1264, 2944, 4024, 5888, 8304, 9584, 11264, 18176, 20224, 47104, 64384, 94208, 132864, 153344, 180224, 290816, 323584, 753664, 1030144, 1507328, 2125824, 2453504, 2883584, 4653056, 5177344, 12058624, 16482304
Offset: 1

Views

Author

Joerg Arndt, Jul 29 2012

Keywords

Comments

From XU Pingya, Feb 07 2018: (Start)
When n is a term, 16n is also. This can be proved as follows:
(1) If w is odd, then 16n - w^4 == 7 (mod 8), and it follows from Legendre's three-square theorem that the equation x^2 + y^2 + z^4 + w^4 = 16n has no solution (it is the same when x, y or z are odd numbers).
(2) If x, y, z and w are even numbers (x = 2a, y = 2b, z = 2c, w = 2d) such that x^2 + y^2 + z^4 + w^4 = 16n, then a^2 + b^2 = 4(n - c^4 - d^4). So there are integers u and v satisfying u^2 + v^2 = n - c^4 - d^4. i.e. u^2 + v^2 + c^4 + d^4 = n, which is a contradiction.
(End)
Conjecture: The set {a(n): n > 0} coincides with {16^k*m: k = 0, 1, 2, ... and m = 23, 44, 71, 79, 184, 519, 599, 4024}. - Zhi-Wei Sun, Jan 27 2022

Crossrefs

Programs

  • PARI
    N=10^6;  x='x+O('x^N);
    S(e)=sum(j=0, ceil(N^(1/e)), x^(j^e));
    v=Vec( S(4)^2 * S(2)^2 );
    for(n=1,#v,if(!v[n],print1(n-1,", ")));

Extensions

a(29)-a(37) from Donovan Johnson, Jul 29 2012

A350860 Number of ways to write n as w^4 + (x^4 + y^2 + z^2)/81, where w,x,y,z are nonnegative integers with y <= z.

Original entry on oeis.org

1, 6, 8, 5, 4, 7, 8, 3, 2, 5, 10, 10, 3, 4, 6, 2, 6, 14, 12, 10, 8, 19, 18, 4, 4, 11, 23, 15, 2, 7, 8, 3, 8, 13, 19, 18, 15, 21, 13, 4, 4, 17, 24, 10, 3, 6, 13, 7, 5, 10, 21, 23, 14, 15, 13, 3, 5, 17, 15, 12, 4, 13, 21, 4, 4, 13, 36, 25, 14, 20, 14, 3, 6, 13, 19, 18, 5, 14, 11, 3, 7, 32, 45, 19, 17, 22, 21, 8, 4, 17, 31
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 19 2022

Keywords

Comments

Conjecture 1: a(n) > 0 for any nonnegative integer n. Also, every n = 0,1,2,... can be written as 4*w^4 + (x^4 + y^2 + z^2)/81 with w,x,y,z integers.
This implies that each nonnegative rational number can be written as w^4 + x^4 + y^2 + z^2 (or 4*w^4 + x^4 + y^2 + z^2) with w,x,y,z rational numbers.
Conjecture 2: For any positive integer c, there is a positive integer m such that every n = 0,1,2,... can be written as w^4 + (c^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive integer c each nonnegative rational number can be written as w^4 + c^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.
Conjecture 3: For any positive odd number d, there is a positive integer m such that every n = 0,1,2,... can be written as 4*w^4 + (d^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive odd number d each nonnegative rational number can be written as 4*w^4 + d^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.

Examples

			a(8) = 2 with 8 = 0^4 + (0^4 + 18^2 + 18^2)/81 = 0^4 + (4^4 + 14^2 + 14^2)/81.
a(15) = 2 with 15 = 1^4 + (3^4 + 18^2 + 27^2)/81 = 1^4 + (5^4 + 5^2 + 22^2)/81.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[81(n-w^4)-x^4-y^2],r=r+1],{w,0,n^(1/4)},{x,0,3(n-w^4)^(1/4)},{y,0,Sqrt[(81(n-w^4)-x^4)/2]}];tab=Append[tab,r],{n,0,90}];Print[tab]

A348890 Number of ways to write n as 5*w^4 + x^4 + y^2 + z^2, where w,x,y,z are nonnegative integers with y <= z.

Original entry on oeis.org

1, 2, 2, 1, 1, 3, 3, 2, 2, 3, 4, 2, 0, 2, 3, 2, 3, 3, 4, 2, 2, 4, 3, 3, 2, 5, 6, 1, 0, 3, 4, 4, 3, 2, 4, 2, 2, 4, 3, 2, 2, 6, 4, 1, 0, 3, 5, 2, 1, 1, 6, 3, 2, 4, 2, 4, 3, 3, 4, 2, 0, 3, 2, 1, 2, 4, 6, 1, 2, 3, 4, 4, 1, 4, 5, 1, 0, 2, 2, 3, 4, 7, 6, 3, 2, 7, 9, 3, 4, 6, 9, 6, 0, 2, 5, 4, 5, 6, 7, 4, 4
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 28 2022

Keywords

Comments

Conjecture: a(n) = 0 only for n == 12 (mod 16).
This has been verified for n up to 10^8.
Now we show that a(n) = 0 whenever n == 12 (mod 16). If 16*q + 12 = 5*w^4 + x^4 + y^2 + z^2 with q,w,x,y,z integers, then the equality modulo 8 yields that w,x,y,z are all even, hence 4*q + 3 == 20*(w/2)^4 + 4*(x/2)^4 + (y/2)^2 + (z/2)^2 and thus (y/2)^2 + (z/2)^2 == 3 (mod 4) which is impossible.
It seems that a(n) = 1 only for n = 0, 3, 4, 27, 43, 48, 49, 63, 67, 72, 75, 192, 215, 303, 1092.

Examples

			a(192) = 1 with 192 = 5*1^4 + 3^4 + 5^2 + 9^2.
a(215) = 1 with 215 = 5*1^4 + 2^4 + 5^2 + 13^2.
a(303) = 1 with 303 = 5*1^4 + 0^4 + 3^2 + 17^2.
a(1092) = 1 with 1092 = 5*0^4 + 2^4 + 20^2 + 26^2.
		

Crossrefs

Programs

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

A351206 Least positive integer m such that n = x^4 + (y^4 + z^4 + 7*w^2)/m^4 for some nonnegative integers x,y,z,w with y <= z.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 4, 3, 1, 1, 1, 6, 2, 2, 10, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 4, 2, 4, 10, 1, 1, 2, 4, 2, 1, 1, 1, 3, 1, 3, 2, 2, 2, 2, 2, 1, 3, 4, 2, 2, 1, 1, 3, 1, 1, 1, 1, 2, 4, 2, 2, 3, 8, 10, 2, 2, 1, 10, 5, 1, 1, 1, 1, 1, 4, 2, 2, 8, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 4, 4
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 04 2022

Keywords

Comments

Conjecture: a(n) exists for any nonnegative integer n.
This implies that each nonnegative rational number can be written as 7*w^2 + x^4 + y^4 + z^4 with w,x,y,z rational numbers.

Examples

			a(6) = 2 with 6 = 1^4 + (1^4 + 2^4 + 7*3^2)/2^4.
a(19) = 6 with 19 = 0^4 + (1^4 + 4^4 + 7*59^2)/6^4.
a(22) = 10 with 22 = 2^4 + (2^4 + 13^4 + 7*67^2)/10^4.
a(5797) = 20 with 5797 = 0^4 + (81^4 + 164^4 + 7*4797^2)/20^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[m=1; Label[bb]; k=m^4; Do[If[SQ[(k*(n-x^4)-y^4-z^4)/7], tab=Append[tab,m]; Goto[aa]],  {x, 0, n^(1/4)}, {y, 0, (k*(n-x^4)/2)^(1/4)},{z,y,(k*(n-x^4)-y^4)^(1/4)}]; m=m+1; Goto[bb]; Label[aa], {n,0,100}];Print[tab]
Showing 1-6 of 6 results.