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 1-10 of 24 results. Next

A303233 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c + 2^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 4, 6, 7, 7, 7, 9, 7, 8, 9, 9, 8, 12, 11, 11, 11, 11, 11, 14, 11, 13, 12, 11, 10, 14, 11, 12, 17, 15, 12, 16, 14, 15, 17, 19, 15, 16, 13, 15, 17, 17, 16, 20, 16, 14, 17, 17, 14, 22, 17, 14, 14, 17, 15, 19
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 20 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two triangular numbers and two powers of 2.
a(n) > 0 for all n = 2..10^9. See A303234 for numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers. See also A303363 for a stronger conjecture.
In contrast, Crocker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.

Examples

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

References

  • R. C. Crocker, On the sum of two squares and two powers of k, Colloq. Math. 112(2008), 235-267.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[4(n-2^k-2^j)+1],Do[If[SQ[8(n-2^k-2^j-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-2^k-2^j)+1]-1)/2}]],{k,0,Log[2,n]-1},{j,k,Log[2,n-2^k]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303363 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c + 2^d, where a,b,c,d are nonnegative integers with a <= b, c <= d and 2|c.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 2, 4, 6, 3, 5, 6, 4, 6, 7, 4, 4, 9, 6, 6, 8, 4, 9, 9, 5, 7, 7, 5, 7, 9, 4, 8, 13, 7, 6, 11, 7, 10, 13, 8, 9, 10, 7, 9, 11, 7, 9, 15, 8, 8, 14, 6, 9, 16, 6, 8, 11, 11, 10, 12, 8, 7, 15, 10, 8, 11, 9, 14, 15, 9
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 22 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is stronger than the author's conjecture in A303233. I have verified a(n) > 0 for all n = 2..10^9.
In contrast, Corcker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.

Examples

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

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[4(n-4^j-2^k)+1],Do[If[SQ[8(n-4^j-2^k-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-4^j-2^k)+1]-1)/2}]],{j,0,Log[4,n/2]},{k,2j,Log[2,n-4^j]}];tab=Append[tab,r],{n,1,70}];Print[tab]

A303338 Number of ways to write n as x^2 + 2*y^2 + 3*2^z + 4^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 3, 3, 2, 4, 3, 2, 6, 2, 4, 8, 2, 4, 7, 3, 4, 8, 5, 5, 10, 6, 4, 10, 8, 5, 12, 7, 3, 12, 4, 5, 12, 5, 5, 14, 7, 4, 12, 7, 6, 12, 6, 6, 10, 7, 7, 12, 7, 6, 14, 6, 8, 16, 4, 8, 18, 5, 6, 16, 5, 9, 13, 7, 7, 14
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 22 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
This is stronger than the author's previous conjecture in A302983. It has been verified that a(n) > 0 for all n = 4..10^9.
Jiao-Min Lin (a student at Nanjing University) has found a counterexample to the conjecture: a(12558941213) = 0. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(4) = 1 with 4 = 0^2 + 2*0^2 + 3*2^0 + 4^0.
a(5) = 1 with 5 = 1^2 + 2*0^2 + 3*2^0 + 4^0.
a(6) = 1 with 6 = 0^2 + 2*1^2 + 3*2^0 + 4^0.
a(9) = 2 with 9 = 0^2 + 2*1^2 + 3*2^0 + 4^1 = 0^2 + 2*1^2 + 3*2^1 + 4^0.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[MemberQ[{5,7},Mod[Part[Part[f[n],i],1],8]]&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[QQ[n-3*2^k-4^j],Do[If[SQ[n-3*2^k-4^j-2x^2],r=r+1],{x,0,Sqrt[(n-3*2^k-4^j)/2]}]],{k,0,Log[2,n/3]},{j,0,If[3*2^k==n,-1,Log[4,n-3*2^k]]}];tab=Append[tab,r],{n,1,70}];Print[tab]

A302982 Number of ways to write n as x^2 + 5*y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
Clearly, a(4*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 4..2*10^8.
See also A302983 and A302984 for similar conjectures.

Examples

			a(4) = 1 with 4 = 0^2 + 5*0^2 + 2^0 + 3*2^0.
a(5) = 2 with 5 =  1^2 + 5*0^2 + 2^0 + 3*2^0 = 0^2 + 5*0^2 + 2^1 + 3*2^0.
a(6) = 1 with 6 = 1^2 + 3*0^2 + 2^1 + 3*2^0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[n-3*2^k-2^j-5x^2],r=r+1],{k,0,Log[2,n/3]},{j,0,If[n==3*2^k,-1,Log[2,n-3*2^k]]},{x,0,Sqrt[(n-3*2^k-2^j)/5]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A302984 Number of ways to write n as x^2 + 2*y^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 3, 4, 5, 5, 8, 5, 5, 7, 4, 6, 7, 9, 9, 10, 10, 7, 9, 8, 10, 15, 10, 9, 10, 8, 6, 10, 10, 11, 14, 14, 8, 12, 13, 13, 20, 15, 12, 16, 10, 15, 12, 10, 15, 17, 16, 12, 16, 14, 14, 21
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
Clearly, a(2*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 6...10^9.
See also A302982 and A302983 for similar conjectures.

Examples

			a(6) = 1 with 6 = 0^2 + 2*0^2 + 2^0 + 5*2^0.
a(7) = 2 with 7 = 1^2 + 2*0^2 + 2^0 + 5*2^0 = 0^2 + 2*0^2 + 2^1 + 5*2^0.
a(8) = 2 with 8 = 0^2 + 2*1^2 + 2^0 + 5*2^0 = 1^2 + 2*0^2 + 2^1 + 5*2^0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[MemberQ[{5,7},Mod[Part[Part[f[n],i],1],8]]&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[n-5*2^k-2^j],Do[If[SQ[n-5*2^k-2^j-2x^2],r=r+1],{x,0,Sqrt[(n-5*2^k-2^j)/2]}]],{k,0,Log[2,n/5]},{j,0,Log[2,Max[1,n-5*2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303234 Numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 22, 23, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 42, 44, 46, 47, 49, 52, 53, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 71, 74, 77, 79, 80
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 20 2018

Keywords

Comments

Conjecture: Any integer n > 1 can be written as the sum of two terms of the current sequence.
This is equivalent to the author's conjecture in A303233.

Examples

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

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[Do[If[SQ[8(n-2^k)+1],tab=Append[tab,n];Goto[aa]],{k,0,Log[2,n]}];Label[aa],{n,1,80}];Print[tab]

A303389 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
See also A303401, A303432 and A303540 for similar conjectures.

Examples

			a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-5^j-5^k)+1]-1)/2}]],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303393 Numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 4, 5, 6, 7, 8, 11, 15, 16, 20, 22, 25, 26, 28, 29, 31, 33, 35, 37, 40, 41, 46, 50, 53, 56, 60, 61, 67, 70, 71, 79, 80, 83, 91, 92, 96, 103, 106, 110, 116, 121, 125, 126, 128, 130, 131, 135, 137, 140, 141, 145, 146, 153, 154, 158, 161, 170, 172, 176
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

The author's conjecture in A303389 has the following equivalent version: Each integer n > 1 can be expressed as the sum of two terms of the current sequence.
This has been verified for all n = 2..2*10^8.

Examples

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

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
    tab={};Do[Do[If[TQ[m-5^k],tab=Append[tab,m];Goto[aa]],{k,0,Log[5,m]}];Label[aa],{m,1,176}];Print[tab]

A303399 Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is equivalent to the author's conjecture in A303389. It has been verified that a(n) > 0 for all n = 2..6*10^9.
Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m + 1 can be written as the sum of two squares.

Examples

			a(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2.
a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2.
a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],r=r+1],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301376 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that x^2-(3*y)^2 = 4^k for some k = 0,1,2,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 19 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0. Moreover, any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and y even such that x^2 - (3*y)^2 = 4^k for some k = 0,1,2,....
We have verifed this for all n = 1..10^7.
Compare this conjecture with the conjectures in A299537.
As 3*A001353(n)^2 + 1 = A001075(n)^2, the conjecture in A300441 implies that any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x^2 - 3*y^2 = 4^k for some k = 0,1,2,....
See also A301391 for a similar conjecture.

Examples

			a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1^2 - (3*0)^2 = 4^0.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (3*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (3*0)^2 = 4^1.
a(31) = 3 since 31^2 = 10^2 + 2^2 + 4^2 + 29^2 with 10^2 - (3*2)^2 = 4^3, and 31^2 = 20^2 + 4^2 + 4^2 + 23^2 = 20^2 + 4^2 + 16^2 + 17^2 with 20^2 - (3*4)^2 = 4^4.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[4^k+9y^2]&&QQ[n^2-4^k-10y^2],Do[If[SQ[n^2-(4^k+10y^2)-z^2],r=r+1],{z,0,Sqrt[(n^2-4^k-10y^2)/2]}]],{k,0,Log[2,n]},{y,0,Sqrt[(n^2-4^k)/10]}];tab=Append[tab,r],{n,1,80}];Print[tab]
Showing 1-10 of 24 results. Next