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.

A308641 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers with c*(3c+1)/2 <= d*(3d+1)/2.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 3, 2, 2, 3, 1, 3, 2, 2, 4, 3, 6, 2, 3, 4, 1, 5, 3, 4, 4, 4, 8, 3, 5, 6, 5, 4, 3, 4, 2, 4, 6, 5, 4, 5, 6, 5, 4, 5, 4, 3, 4, 5, 1, 5, 6, 6, 4, 2, 7, 4, 5, 4, 4, 4, 5, 6, 3, 5, 7, 7, 5, 3, 5, 5, 5, 7, 6, 3, 7, 6, 8, 5, 5, 7, 5, 7, 4, 2, 8, 6, 6, 3, 3, 7, 2, 9, 4, 7, 6, 5, 7, 2, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Jun 13 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: If f(x) is one of the polynomials x^2, x*(5x+1), x*(5x+1)/2, x*(7x+1)/2, x*(7x+5)/2, then any positive integers n can be written as (2^a*5^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 3: Any positive integers n can be written as (2^a*7^b)^2 + c*(7c+5)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
See also A308640 for similar conjectures.

Examples

			a(12) = 1 with 12 = (2^1*5^0)^2 + (-1)*(3*(-1)+1)/2 + 2*(3*2+1)/2.
a(22) = 1 with 22 = (2^2*5^0)^2 + (-1)*(3*(-1)+1)/2 + (-2)*(3*(-2)+1)/2.
a(50) = 1 with 50 = (2^2*5^0)^2 + (-3)*(3*(-3)+1)/2 + (-4)*(3*(-4)+1)/2.
a(330) = 1 with 330 = (2^2*5^0)^2 + (-8)*(3*(-8)+1)/2 + 12*(3*12+1)/2.
a(8650) = 1 with 8650 = (2^5*5^0)^2 + 8*(3*8+1)/2 + (-71)*(3*(-71)+1)/2.
a(29440) = 1 with 29440 = (2*5)^2 + (-80)*(3*(-80)+1)/2 + (-115)*(3*(-115)+1)/2.
a(48459) = 1 with 48459 = (2^7*5^0)^2 + 20*(3*20+1)/2 + (-145)*(3*(-145)+1)/2.
a(153035) = 1 with 153035 = (2*5^2)^2 + 35*(3*35+1)/2 + (-315)*(3*(-315)+1)/2.
a(164043) = 1 with 164043 = (2^2*5^2)^2 + (-46)*(3*(-46)+1)/2 + 317*(3*317+1)/2.

Links

Crossrefs

Cf. A000079, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640.

Programs

  • Mathematica
    PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-4^a*25^b-c(3c+1)/2],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{c,-Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6],(Sqrt[12(n-4^a*25^b)+1]-1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308644 Number of ways to write n as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+1)/2, where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 13 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Also, any positive integer n can be written as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+3)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 2: Let r be 1 or 3. Then, any positive integer n can be written as (3^a*4^b)^2 + c*(3c+1)/2 + d*(7d+r)/2, where a and b are nonnegative integers, and c and d are integers.
We have verified Conjectures 1-2 for all n = 1..10^6.
See also A308640, A308641 and A308656 for similar conjectures.

Examples

			a(152) = 1 with 152 = (3^0*5^0)^2 + (-4)*(3*(-4)+1)/2 + 6*(7*6+1)/2.
a(129894) = 1 with 129894 = (3^0*5^1)^2 + 154*(3*154+1)/2 + 164*(7*164+1)/2.
a(200963) = 1 with 200963 = (3^1*5^0)^2 + 364*(3*364+1)/2 + 24*(7*24+1)/2.
a(371278) = 1 with 371278 = (3^3*5^1)^2 + (-382)*(3*(-382)+1)/2 + (-196)*(7*(-196)+1)/2.
a(534699) = 1 with 534699 = (3^2*5^2)^2 + 543*(3*543+1)/2 + (-109)*(3*(-109)+1)/2.

Links

Crossrefs

Cf. A000244, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308656.

Programs

  • Mathematica
    PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-9^a*25^b-x(7x+1)/2],r=r+1],{a,0,Log[9,n]},{b,0,Log[25,n/9^a]},{x,-Floor[(Sqrt[56(n-9^a*25^b)+1]+1)/14],(Sqrt[56(n-9^a*25^b)+1]-1)/14}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308656 Number of ways to write n as (2^a*9^b)^2 + c*(2c+1) + d*(3d+1), where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 14 2019

Keywords

Comments

Note that {x*(2x+1): x is an integer} = {n*(n+1)/2: n = 0,1,2,...}.
Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: If f(x) is one of the polynomials x*(4x+1), x*(5x+2), x*(5x+4), x*(7x+3)/2 and x(7x+5)/2, then any positive integer n can be written as (2^a*9^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 3: Let r be 1 or 2. Then any positive integer n can be written as (2^a*7^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.
Conjecture 4: If g(x) is one of the polynomials x*(x+1), x*(4x+3), x*(7x+1)/2, x*(7x+3)/2 and x*(7x+5)/2, then any positive integer n can be written as (2^a*7^b)^2 + g(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
We have verified a(n) > 0 for all n = 1..10^8, and Conjectures 2-4 for all n = 1..10^6.
See also A308640, A308641, and A308644 for similar conjectures.
Jiao-Min Lin (a student at Nanjing University) has found a counterexample to Conjecture 1: a(2109982225) = 0. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(13) = 1 with 13 = (2^0*9^0)^2 + 2*(2*2+1) + (-1)*(3*(-1)+1).
a(3515) = 1 with 3515 = (2^0*9^1)^2 + 0*(2*0+1) + (-34)*(3*(-34)+1).
a(124076) = 1 with 124076 = (2^3*9^1)^2 + 206*(2*206+1) + 106*(3*106+1).
a(141518) = 1 with 141518 = (2^1*9^2)^2 + (-188)*(2*(-188)+1) + 122*(3*122+1).
a(345402) = 1 with 345402 = (2^7*9^0)^2 + 18*(2*18+1) + (-331)*(3*(-331)+1).

Links

Crossrefs

A000079, A000217, A000420, A001019, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644.

Programs

  • Mathematica
    PQ[n_]:=PQ[n]=IntegerQ[Sqrt[12n+1]];
tab={};Do[r=0;Do[If[PQ[n-81^a*4^b-x(2x+1)],r=r+1],{a,0,Log[81,n]},{b,0,Log[4,n/81^a]},{x,-Floor[(Sqrt[8(n-81^a*4^b)+1]+1)/4],(Sqrt[8(n-81^a*4^b)+1]-1)/4}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308661 Number of ways to write 12*n+5 as (2^a*5^b)^2 + c^2 + d^2, where a,b,c,d are nonnegative integers with a > 0 and c <= d.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 15 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,.... Equivalently, for each nonnegative integer n we can write 3*n+1 as a*(a+1)/2 + b*(b+1)/2 + (2^c*5^d)^2 with a,b,c,d nonnegative integers.
Conjecture 2: For each n = 0,1,2,... we can write 24*n+10 as a^2 + b^2 + (2^c*3^d)^2 with a,b,c,d nonnegative integers and d > 0.
We have verified Conjectures 1 and 2 for n up to 2*10^8 and 10^8 respectively.
By the Gauss-Legendre theorem on sums of three squares, for each n = 0,1,... we can write 4*n+1 (or 4*n+2, or 8*n+3) as the sum of three squares.
Conjecture 1 holds for n < 8.33*10^9. - Giovanni Resta, Jun 19 2019

Examples

			a(0) = 1 with 12*0 + 5 = (2^1*5^0)^2 + 0^2 + 1^2.
a(4) = 2 with 12*4 + 5 = 53 = (2^1*5^0)^2 + 0^2 + 7^2 = (2^2*5^0)^2 + 1^2 + 6^2.
a(441019) = 2 with 12*441019 + 5 = 5292233 = (2^1*5^2)^2 + 513^2 + 2242^2 = (2^3*5^1)^2 + 757^2 + 2172^2.

Links

Crossrefs

Cf. A000079, A000217, A000244, A000290, A000351, A000378, A001481, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308662.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[12n+5-4^a*25^b-x^2],r=r+1],{a,1,Log[4,12n+5]},{b,0,Log[25,(12n+5)/4^a]},{x,0,Sqrt[(12n+5-4^a*25^b)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A308662 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 15 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Let r be 1 or 2. Then, any positive integer n can be written as (2^a*5^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.
We have verified Conjectures 1 and 2 for all n = 1..10^8.

Examples

			a(3) = 1 with 3 = (2^0*5^0)^2 + (-1)*(3*(-1)+1) + 0*(3*0+2).
a(7) = 1 with 7 = (2^1*5^0)^2 + (-1)*(3*(-1)+1) + (-1)*(3*(-1)+2).
a(10) = 1 with 10 = (2^0*5^0)^2 + 1*(3*1+1) + 1*(3*1+2).
a(52) = 1 with 52 = (2^0*5^0)^2 + 3*(3*3+1) + (-3)*(3*(-3)+2).
a(98) = 1 with 98 = (2^0*5^1)^2 + 4*(3*4+1) + (-3)*(3*(-3)+2).
a(14596) = 1 with 14596 = (2^3*5^0)^2 + (-36)*(3*(-36)+1) + (-60)*(3*(-60)+2).
a(22163) = 1 with 22163 = (2^3*5^0)^2 + 66*(3*66+1) + (-55)*(3*(-55)+2).
a(150689) = 1 with 150689 = (2^6*5^1)^2 + 117*(3*117+1) + (-49)*(3*(-49)+2).

Links

Crossrefs

Cf. A000079, A000351, A001082, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661.

Programs

  • Mathematica
    OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]];
tab={};Do[r=0;Do[If[OctQ[n-4^a*25^b-x(3x+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{x,-Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6],(Sqrt[12(n-4^a*25^b)+1]-1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308734 Number of ordered ways to write n as (2^a*3^b)^2 + (2^c*5^d)^2 + x^2 + y^2, where a,b,c,d,x,y are nonnegative integers with x <= y.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 3, 1, 3, 5, 2, 3, 4, 4, 5, 1, 4, 8, 4, 4, 8, 8, 4, 3, 8, 7, 7, 6, 5, 13, 6, 1, 10, 11, 7, 7, 10, 9, 9, 5, 7, 18, 7, 5, 14, 11, 6, 3, 10, 11, 9, 8, 7, 15, 9, 4, 14, 12, 5, 10, 9, 10, 11, 1, 11, 19, 10, 6, 17, 21, 6, 8, 14, 12, 13, 7, 14, 21, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Jun 21 2019

Keywords

Comments

Four-square Conjecture: a(n) > 0 for all n > 1.
This is much stronger than Lagrange's four-square theorem. We have verified a(n) > 0 for all n = 2..10^9.
Note that 16265031 cannot be written as (2^a*3^b)^2 + (2^c*3^d)^2 + x^2 + y^2 with a,b,c,d,x,y nonnegative integers.
a(n) > 0 for 1 < n <= 10^10. - Giovanni Resta, Jun 28 2019
I promise to offer 2500 US dollars as the prize for the first correct proof of the Four-square Conjecture. - Zhi-Wei Sun, Jul 09 2019
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 1.6*10^11. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(2^(2k+1)) = 1 with 2^(2k+1) = (2^k*3^0)^2 + (2^k*5^0)^2 + 0^2 + 0^2.
a(2^(2k+2)) = 1 with 2^(2k+2) = (2^k*3^0)^2 + (2^k*5^0)^2 + (2^k)^2 + (2^k)^2.
a(3) = 1 with 3 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
a(5) = 2 with 5 = (2^0*3^0)^2 + (2^1*5^0)^2 + 0^2 + 0^2 = (2^1*3^0)^2 + (2^0*5^0)^2 + 0^2 + 0^2.
a(11) = 2 with 11 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 3^2 = (2^0*3^1)^2 + (2^0*5^0)^2 + 0^2 + 1^2.

Links

Crossrefs

Cf. A000079, A000118, A000290, A000244, A000351, A271518, A281976, A303656, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661, A308662.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-4^a*9^b-4^c*25^d-x^2],r=r+1],{a,0,Log[4,n]},{b,0,Ceiling[Log[9,n/4^a]]-1},
{c,0,Log[4,n-4^a*9^b]},{d,0,Log[25,(n-4^a*9^b)/4^c]},{x,0,Sqrt[(n-4^a*9^b-4^c*25^d)/2]}];tab=Append[tab,r],{n,1,80}];Print[tab]
Showing 1-6 of 6 results.

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