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]

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]

A303656 Number of ways to write n as a^2 + b^2 + 3^c + 5^d, where a,b,c,d are nonnegative integers with a <= b.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 27 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 squares, a power of 3 and a power of 5.
It has been verified that a(n) > 0 for all n = 2..2*10^10.
It seems that any integer n > 1 also can be written as the sum of two squares, a power of 2 and a power of 3.
The author would like to offer 3500 US dollars as the prize for the first proof of his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Jun 05 2018
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 2.4*10^11. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(2) = 1 with 2 = 0^2 + 0^2 + 3^0 + 5^0.
a(5) = 1 with 5 = 0^2 + 1^2 + 3^1 + 5^0.
a(25) = 1 with 25 = 1^2 + 4^2 + 3^1 + 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[n-3^k-5^m],Do[If[SQ[n-3^k-5^m-x^2],r=r+1],{x,0,Sqrt[(n-3^k-5^m)/2]}]],{k,0,Log[3,n]},{m,0,If[n==3^k,-1,Log[5,n-3^k]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303637 Number of ways to write n as x^2 + y^2 + 2^z + 5*2^w, where x,y,z,w are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
This has been verified for n up to 5*10^9. Note that 321256731 cannot be written as x^2 + (2*y)^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.
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.
570143 cannot be written as x^2 + y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers, and 2284095 cannot be written as x^2 + y^2 + 2^z + 7*2^w with x,y,z,w nonnegative integers.
Jiao-Min Lin (a student at Nanjing University) found a counterexample to the conjecture: a(18836421387) = 0. - Zhi-Wei Sun, Jul 21 2022

Examples

			a(6) = 1 with 6 = 0^2 + 0^2 + 2^0 + 5*2^0.
a(8) = 2 with 8 = 1^2 + 1^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^1 + 5*2^0.
a(9) = 2 with 9 = 1^2 + 1^2 + 2^1 + 5*2^0 = 0^2 + 0^2 + 2^2 + 5*2^0.
a(10) = 2 with 10 = 0^2 + 2^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^2 + 5*2^0.
		

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[n-5*2^k-2^m],Do[If[SQ[n-5*2^k-2^m-x^2],r=r+1],{x,0,Sqrt[(n-5*2^k-2^m)/2]}]],{k,0,Log[2,n/5]},{m,0,If[n/5==2^k,-1,Log[2,n-5*2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303639 Number of ways to write n as a^2 + b^2 + binomial(2*c+1,c) + binomial(2*d+1,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is similar to the author's conjecture in A303540.
It has been verified that a(n) > 0 for all n = 2..6*10^8.

Examples

			a(9) = 1 with 9 = 1^2 + 2^2 + binomial(2*0+1,0) + binomial(2*1+1,1).
a(2530) = 1 with 2530 = 0^2 + 49^2 + binomial(2*1+1,1) + binomial(2*4+1,4).
a(3258) = 1 with 3258 = 22^2 + 52^2 + binomial(2*3+1,3) + binomial(2*3+1,3).
a(5300) = 1 with 5300 = 10^2 + 59^2 + binomial(2*1+1,1) + binomial(2*6+1,6).
a(13453) = 1 with 13453 = 51^2 + 104^2 + binomial(2*0+1,0) + binomial(2*3+1,3).
a(20964) = 1 with 20964 = 13^2 + 138^2 + binomial(2*3+1,3) + binomial(2*6+1,6).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    c[n_]:=c[n]=Binomial[2n+1,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;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))*5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))*5^m squarefree.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 7.
This has been verified for n up to 2*10^10.
See also A303821, A303934, A303949, A304031 and A304122 for related information, and A304034 for a similar conjecture.
The author would like to offer 2500 US dollars as the prize to the first proof of the conjecture, and 250 US dollars as the prize to the first explicit counterexample. - Zhi-Wei Sun, May 08 2018

Examples

			a(6) = 1 since 6 = 3 + 2^1 + 5^0 with 3 an odd prime and 2^1 + 5^0 = 3 squarefree.
a(15) = 1 since 15 = 5 + 2^3 + 2*5^0 with 5 an odd prime and 2^3 + 2*5^0 = 2*5 squarefree.
a(35) = 1 since 35 = 29 + 2^2 + 2*5^0 with 29 an odd prime and 2^2 + 2*5^0 = 2*3 squarefree.
a(91) = 1 since 91 = 17 + 2^6 + 2*5^1 with 17 an odd prime and 2^6 + 2*5^1 = 2*37 squarefree.
a(9574899) = 1 since 9574899 = 9050609 + 2^19 + 2*5^0 with 9050609 an odd prime and 2^19 + 2*5^0 = 2*5*13*37*109 squarefree.
a(6447154629) = 2 since 6447154629 = 6447121859 + 2^15 + 2*5^0 with 6447121859 prime and 2^15 + 2*5^0 = 2*5*29*113 squarefree, and 6447154629 = 5958840611 + 2^15 + 2*5^12 with 5958840611 prime and 2^15 + 2*5^12 = 2*17*41*433*809 squarefree.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=n>2&&PrimeQ[n];
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*5^m]&&PQ[n-2^k-(1+Mod[n,2])*5^m],r=r+1],{k,0,Log[2,n]},{m,0,If[2^k==n,-1,Log[5,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))*3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))*3^m squarefree.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 11.
This has been verified for n up to 10^10.
See also A304081 for a similar conjecture.

Examples

			a(8) = 1 since 8 = 3 + 2^1 + 3^1 with 3 prime and 2^1 + 3^1 = 5 squarefree.
a(13) = 1 since 13 = 3 + 2^2 + 2*3^1 with 3 prime and 2^2 + 2*3^1 = 2*5 squarefree.
a(19) = 1 since 19 = 5 + 2^3 + 2*3^1 with 5 prime and 2^3 + 2*3^1 = 2*7 squarefree.
a(23) = 1 since 23 = 13 + 2^2 + 2*3^1 with 13 prime and 2^2 + 2*3 = 2*5 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*3^m]&&PrimeQ[n-2^k-(1+Mod[n,2])*3^m],r=r+1],{k,1,Log[2,n]},{m,1,If[2^k==n,-1,Log[3,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]
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