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 21-24 of 24 results.

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}]

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]
Previous Showing 21-24 of 24 results.