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.

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A303934 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m squarefree, where k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 03 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for all n = 2..10^10.
Note that a(n) <= A303821(n).

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime and 2^0 + 5^0 squarefree.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime and 2^1 + 5^0 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304031 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m a product of at most three distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

a(n) > 0 for all 1 < n <= 10^10 with the only exception n = 3114603841, and 2*3114603841 = 6219442049 + 2^3 + 5^10 with 6219442049 prime and 2^3 + 5^10 = 3*17*419*457 squarefree.
Note that a(n) <= A303934(n) <= A303821(n).

Examples

			a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 = 2^1 + 5^0  prime.
a(7) = 2 since 2*7 = 7 + 2^1 + 5^1 with 7 = 2^1 + 5^1 prime, and 2*7 = 11 + 2^1 + 5^0 with 11 and 2^1 + 5^0 both prime.
a(42908) = 2 since 2*42908 = 85751 + 2^6 + 5^0 with 85751 prime and 2^6 + 5^0 = 5*13, and 2*42908 = 69431 + 2^14 + 5^0 with 69431 prime and 2^14 + 5^0 = 5*29*113.
		

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
    tab={};Do[r=0;Do[If[qq[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303949 Number of ways to write 2*n+1 as p + 2*(2^k+5^m) with p prime and 2^k+5^m a product of at most three distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 05 2018

Keywords

Comments

4787449 is the first value of n > 2 with a(n) = 0, and 2*4787449+1 = 9574899 has the unique representation as p + 2*(2^k+5^m): 9050609 + 2*(2^18+5^0) with 9050609 prime and 2^18+5^0 = 5*13*37*109.
See also A303934 and A304081 for related conjectures.

Examples

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

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n+1-2(2^k+5^m)],r=r+1],{k,0,Log[2,n]},{m,0,Log[5,n+1/2-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304032 Number of ways to write 2*n as p + 2^k + 3^m with p prime and 2^k + 3^m a product of at most two distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 4, 4, 4, 6, 6, 5, 8, 9, 4, 6, 7, 4, 9, 10, 6, 9, 10, 6, 11, 14, 7, 9, 11, 5, 10, 9, 6, 12, 10, 3, 11, 15, 7, 12, 16, 7, 9, 14, 9, 12, 14, 8, 12, 16, 5, 12, 18, 10, 12, 16, 9, 12, 19, 10, 13, 17, 6, 10, 15, 6, 10, 16, 10, 12, 15, 10, 17, 20, 8, 14, 15, 8, 11, 18, 9, 12
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

The even number 58958 cannot be written as p + 2^k + 3^m with p and 2^k + 3^m both prime.
Clearly, a(n) <= A303702(n). We note that a(n) > 0 for all n = 2..5*10^8.
See also A304034 for a related conjecture.

Examples

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

References

  • J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes, Sci. Sinica 16(1973), 157-176.

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=2;
    tab={};Do[r=0;Do[If[qq[2^k+3^m]&&PrimeQ[2n-2^k-3^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[3,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A304122 Squarefree numbers of the form 2^k + 5^m, where k is a positive integer and m is a nonnegative integer.

Original entry on oeis.org

3, 5, 7, 13, 17, 21, 29, 33, 37, 41, 57, 65, 69, 89, 127, 129, 133, 141, 157, 253, 257, 281, 381, 517, 537, 627, 629, 633, 641, 689, 753, 881, 1049, 1137, 1149, 1649, 2049, 2053, 2073, 2173, 3127, 3129, 3133, 3157, 3189, 3253, 3637, 4097, 4101, 4121
Offset: 1

Views

Author

Zhi-Wei Sun, May 07 2018

Keywords

Comments

The conjecture in A304081 has the following equivalent version: Any even number greater than 4 can be written as the sum of a prime and a term of the current sequence, and also any odd number greater than 8 can be written as the sum of a prime and twice a term of the current sequence.

Examples

			a(1) = 3 since 3 = 2^1 + 5^0 is squarefree.
a(6) = 21 since 21 = 2^4 + 5^1 = 3*7 is squarefree.
		

Crossrefs

Programs

  • Mathematica
    V={}; Do[If[SquareFreeQ[2^k+5^m],V=Append[V,2^k+5^m]],{k,1,12},{m,0,5}];
    LL:=LL=Sort[DeleteDuplicates[V]];
    a[n_]:=a[n]=LL[[n]];
    Table[a[n],{n,1,50}]

A308547 Number of ways to write n as a^2 + 2*b^2 + 2^c*3^d, where a,b,c,d are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 06 2019

Keywords

Comments

As 3*(a^2 + 2*b^2 + 2^c*3^d) = (a+2*b)^2 + 2*(a-b)^2 + 2^c*3^(d+1), we have a(3*n) > 0 if a(n) > 0.
The first positive integer n with a(n) = 0 is 139571911. We also have a(142991573) = 0.

Examples

			a(1) = 1 with 1 = 0^2 + 2*0^2 + 2^0*3^0.
a(2) = 2 with 2 = 0^2 + 2*0^2 + 2^1*3^0 = 1^2 + 2*0^2 + 2^0*3^0.
a(1117) = 2 with 1117 = 10^2 + 2*12^2 + 2^0*3^6 = 19^2 + 2*18^2 + 2^2*3^3.
a(78373) = 1 with 78373 = 271^2 + 2*48^2 + 2^2*3^4.
a(448159) = 1 with 448159 = 610^2 + 2*195^2 + 2^0*3^2.
a(82816213) = 2 with 82816213 = 4353^2 + 2*5651^2 + 2^1*3^0 = 3681^2 + 2*5885^2 + 2^1*3^0.
a(90685253) = 2 with 90685253 = 7007^2 + 2*4560^2 + 2^2*3^0 = 607^2 + 2*6720^2 + 2^2*3^0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[n-3^k*2^m-2x^2],r=r+1],{k,0,Log[3,n]},{m,0,Log[2,n/3^k]},{x,0,Sqrt[(n-3^k*2^m)/2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303429 Number of ordered pairs (k, m) of nonnegative integers such that n - 3^k - 5^m can be written as the sum of two squares.

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

Views

Author

Zhi-Wei Sun, Apr 28 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is equivalent to the author's conjecture in A303656.
It has been verified that a(n) > 0 for all n = 2..10^9.

Crossrefs

Programs

  • Maple
    a(5) = 1 with 5 - 3^1 - 5^0 = 0^2 + 1^2.
    a(25) = 1 with 25 - 3^1 - 5^1 = 1^2 + 4^2.
  • 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],r=r+1],{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]

A303997 Number of ways to write 2*n as p + 3^k + binomial(2*m,m), where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

502743678 is the first value of n > 1 with a(n) = 0.

Examples

			a(2) = 1 since 2*2 = 2 + 3^0 + binomial(2*0,0) with 2 prime.
a(3) = 2 since 2*3 = 3 + 3^0 + binomial(2*1,1) = 2 + 3^1 + binomial(2*0,0) with 3 and 2 both prime.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>=2n,Goto[aa]];Do[If[PrimeQ[2n-c[k]-3^m],r=r+1],{m,0,Log[3,2n-c[k]]}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]

A303998 Number of ways to write 2*n+1 as p + 2^k + binomial(2*m,m), where p is a prime, and k and m are positive integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2.
This has been verified for n up to 10^9.

Examples

			a(3) = 1 since 2*3+1 = 3 + 2^1 + binomial(2*1,1) with 3 prime.
a(4) = 2 since 2*4+1 = 3 + 2^2 + binomial(2*1,1) = 5 + 2^1 + binomial(2*1,1) with 3 and 5 both prime.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    tab={};Do[r=0;k=1;Label[bb];If[c[k]>2n,Goto[aa]];Do[If[PrimeQ[2n+1-c[k]-2^m],r=r+1],{m,1,Log[2,2n+1-c[k]]}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A308403 Number of ways to write n as 6^i + 3^j + A008347(k), where i, j and k > 0 are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 25 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 2. In other words, each n = 3,4,... can be written as 6^i + 3^j + prime(k) - prime(k-1) + ... + (-1)^(k-1)*prime(1), where i, j and k > 0 are nonnegative integers.
Conjecture 2: If {a,b} is among {2,m} (m = 3..14), {3,4}, {3,5}, then any integer n > 2 can be written as a^i + b^j + A008347(k) with i, j and k > 0 nonnegative integers.
Using Qing-Hu Hou's program, we have verified Conjectures 1 and 2 for n up to 10^9 and 10^7 respectively. - Zhi-Wei Sun, May 28 2019
Conjecture 1 verified up to 10^10. Conjecture 2 holds up to 10^10 for all cases except {2, 12} since 4551086841 cannot be written as 2^i + 12^j + A008347(k). - Giovanni Resta, May 28 2019

Examples

			a(3) = 1 with 3 - (6^0 + 3^0) = 1 = A008347(2).
a(4) = 1 with 4 - (6^0 + 3^0) = 2 = A008347(1).
a(24) = 1 with 24 - (6^0 + 3^0) = 22 = A008347(13).
a(234) = 1 with 234 - (6^1 + 3^3) = 201 = A008347(90).
a(1134) = 1 with 1134 - (6^2 + 3^0) = 1097 = A008347(322).
a(4330) = 1 with 4330 - (6^3 + 3^0) = 4113 = A008347(1016).
a(5619) = 1 with 5619 - (6^1 + 3^3) = 5586 = A008347(1379).
a(6128) = 1 with 6128 - (6^0 + 3^0) = 6126 = A008347(1499).
a(16161) = 1 with 16161 - (6^3 + 3^0) = 15944 = A008347(3445).
a(133544) = 1 with 133544 - (6^0 + 3^8) = 126982 = A008347(22579).
		

Crossrefs

Programs

  • Mathematica
    Pow[n_]:=Pow[n]=n>0&&IntegerQ[Log[3,n]];
    s[0]=0;s[n_]:=s[n]=Prime[n]-s[n-1];
    tab={};Do[r=0;Do[If[s[k]>=n,Goto[bb]];Do[If[Pow[n-s[k]-6^m],r=r+1],{m,0,Log[6,n-s[k]]}];Label[bb],{k,1,2n-1}];tab=Append[tab,r],{n,1,100}];Print[tab]
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