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.

Previous Showing 11-16 of 16 results.

A301391 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y even such that x^2 - (6*y)^2 = 4^k for some k = 0,1,2,....

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 8, 2, 2, 2, 3, 2, 6, 1, 2, 2, 1, 1, 11, 3, 2, 4, 4, 3, 3, 1, 6, 10, 6, 2, 7, 2, 3, 2, 6, 3, 8, 2, 7, 7, 2, 1, 11, 4, 4, 2, 2, 1, 6, 1, 3, 11, 3, 3, 16, 3, 5, 4, 8, 5, 2, 3, 11, 5, 8, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 20 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 11, 13, 19, 2^k*m (k = 0,1,2,... and m = 1, 5, 7, 31).
We have verified a(n) > 0 for all n = 1..10^7.
See also A301376 for a similar conjecture.

Examples

			a(2) = 1 since 2^2 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 - (6*0)^2 = 4^1.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (6*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (6*0)^2 = 4^1.
a(11) = 1 since 11 = 2^2 + 0^2 + 6^2 + 9^2 with 2^2 - (6*0)^2 = 4^1.
a(13) = 1 since 13 = 4^2 + 0^2 + 3^2 + 12^2 with 4^2 - (6*0)^2 = 4^2.
a(19) = 1 since 19 = 1^2 + 0^2 + 6^2 + 18^2 with 1^2 - (6*0)^2 = 4^0.
a(31) = 1 since 31^2 = 20^2 + 2^2 + 14^2 + 19^2 with 20^2 - (6*2)^2 = 4^4.
a(75) = 2 since 75^2 = 68^2 + 10^2 + 1^2 + 30^2 = 68^2 + 10^2 + 15^2 + 26^2 with 68^2 - (6*10)^2 = 4^5.

Links

Crossrefs

Cf. A000118, A000290, A000302, A299537, A299794, A299924, A300219, A300396, A300441, A300510, A301376.

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+144m^2]&&QQ[n^2-4^k-148m^2], Do[If[SQ[n^2-(4^k+148m^2)-z^2],r=r+1],{z,0,Sqrt[(n^2-4^k-148m^2)/2]}]],{k,0,Log[2,n]},{m,0,Sqrt[(n^2-4^k)/148]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A302981 Number of ways to write n as x^2 + 2*y^2 + 2^z + 2^w, where x,y,z,w are nonnegative integers with z <= w.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Clearly, a(2*n) > 0 if a(n) > 0. We note that 52603423 is the first value of n > 1 with a(n) = 0.
See also A302982 and A302983 for related things.

Examples

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

Links

Crossrefs

Cf. A000079, A000290, A002479, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302982, A302983.

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-2^k-2^j],Do[If[SQ[n-2^k-2^j-2x^2],r=r+1],{x,0,Sqrt[(n-2^k-2^j)/2]}]],{k,0,Log[2,n]-1},{j,k,Log[2,Max[1,n-2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A302983 Number of ways to write n as x^2 + 2*y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

0, 0, 0, 1, 2, 2, 4, 5, 4, 5, 6, 4, 8, 8, 7, 12, 8, 6, 9, 9, 6, 13, 13, 8, 13, 12, 8, 13, 14, 11, 15, 17, 8, 14, 11, 11, 16, 17, 11, 17, 19, 8, 17, 19, 10, 19, 18, 12, 15, 17, 12, 20, 17, 13, 20, 18, 16, 24, 18, 15
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
Clearly, a(2*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 4..6*10^9.
See also A302982 and A302984 for similar conjectures.

Examples

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

Links

Crossrefs

Cf. A000079, A000290, A002479, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302984, A302985.

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

A302985 Number of ordered pairs (x, y) of nonnegative integers such that n - 2^x - 3*2^y has the form u^2 + 2*v^2 with u and v integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
This is equivalent to the author's conjecture in A302983. We have verified a(n) > 0 for all n = 4...6*10^9.
See also A302982 and A302984 for similar conjectures.

Examples

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

Links

Crossrefs

Cf. A000079, A000290, A002479, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984.

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]);
tab={};Do[r=0;Do[If[QQ[n-3*2^k-2^j],r=r+1],{k,0,Log[2,n/3]},{j,0,If[3*2^k==n,-1,Log[2,n-3*2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303235 Number of ordered pairs (x, y) with 0 <= x <= y such that n - 2^x - 2^y can be written as the sum of two triangular numbers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 20 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m+1 (or 8*m+2) can be written as the sum of two squares.
We have verified a(n) > 0 for all n = 2..4*10^8. See also the related sequences A303233 and A303234.

Examples

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

Links

Crossrefs

Cf. A000079, A000217, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234.

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-2^k-2^j)+1],r=r+1],{k,0,Log[2,n]-1},{j,k,Log[2,n-2^k]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A300448 Number of the integers 4^k*m with k >= 0 and m = 1, 2, 5 such that n^2 - 4^k*m can be written as the sum of two squares.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 06 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Each positive square n^2 can be written as 4^k*m + x^2 + y^2 with k,x,y nonnegative integers and m among 1, 5 17.
Conjecture 3: For each n = 1,2,3,... we can write n^2 as 4^k*m + x^2 + y^2 with k,x,y nonnegative integers and m among 1, 5, 65.
Conjecture 3 implies that A300441(n) > 0 for all n > 0 since 4*A001353(0)^2 + 1 = 1, 4*A001353(1)^2 + 1 = 5 and 4*A001353(2)^2 + 1 = 65.
We have verified Conjectures 1-3 for all n = 1..2*10^7.

Examples

			a(1) = 1 since 1^2 - 4^0*1 = 0 = 0^2 + 0^2.
a(2) = 2 since 2^2 - 4^0*2 = 2 = 1^2 + 1^2 and 2^2 - 4*1 = 0 = 0^2 + 0^2.
a(4) = 2 since 4^2 - 4^1*2 = 8 = 2^2 + 2^2 and 4^2 - 4^2*1 = 0 = 0^2 + 0^2.

Links

Crossrefs

Cf. A000118, A000302, A001481, A271518, A281976, A299924, A299537, A299794, A300362, A300441, A300396.

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[If[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1,1,0],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
u[1]=1; u[2]=2; u[3]=5;
tab={};Do[r=0;Do[If[QQ[n^2-4^k*u[m]],r=r+1],{m,1,3},{k,0,Log[4,n^2/u[m]]}];tab=Append[tab,r],{n,1,80}];Print[tab]
Previous Showing 11-16 of 16 results.

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