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 17 results. Next

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]

A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 01 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 4, we can write 2*n as p + 2^x + 5^y, where p is an odd prime, and x and y are positive integers.
This has been verified for n up to 10^10.
See also A303934 and A304081 for further refinements, and A303932 and A304034 for similar conjectures.

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime.
a(5616) = 2 since 2*5616 = 9059 + 2^11 + 5^3 = 10979 + 2^7 + 5^3 with 9059 and 10979 both prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[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,1,80}];Print[tab]

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]

A304331 Number of integers k > 1 such that n - F(k) is a positive squarefree number, where F(k) denotes the k-th Fibonacci number A000045(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, every n = 2,3,... can be written as the sum of a positive Fibonacci number and a positive squarefree number.
This has been verified for n up to 10^10.
See also A304333 for a similar conjecture involving Lucas numbers.

Examples

			a(2) = 1 with 2 - F(2) = 1 squarefree.
a(53) = 2 with 53 - F(3) = 3*17 and 53 - F(9) = 19 both squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=Fibonacci[n];
    tab={};Do[r=0;k=2;Label[bb];If[f[k]>=n,Goto[aa]];If[SquareFreeQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]

A304333 Number of positive integers k such that n - L(k) is a positive squarefree number, where L(k) denotes the k-th Lucas number A000204(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for n up to 5*10^9.
See also A304331 for a similar conjecture involving Fibonacci numbers.
For all n, a(n) <= A130241(n). - Antti Karttunen, May 13 2018

Examples

			a(2) = 1 with 2 - L(1) = 1 squarefree.
a(3) = 1 with 3 - L(1) = 2 squarefree.
a(67) = 2 with 67 - L(1) = 2*3*11 and 67 - L(7) = 2*19 both squarefree.
		

Crossrefs

Programs

  • Maple
    a := proc(n) local count, lucas, newcas;
    count := 0; lucas := 1; newcas := 2;
    while lucas < n do
        if numtheory:-issqrfree(n - lucas) then count := count + 1 fi;
        lucas, newcas := lucas + newcas, lucas;
    od;
    count end:
    seq(a(n), n=1..90); # Peter Luschny, May 15 2018
  • Mathematica
    f[n_]:=f[n]=LucasL[n];
    tab={};Do[r=0;k=1;Label[bb];If[f[k]>=n,Goto[aa]];If[SquareFreeQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]
  • PARI
    A304333(n) = { my(u1=1,u2=3,old_u1,c=0); if(n<=2,n-1,while(u1Antti Karttunen, May 13 2018

A304522 Number of ordered ways to write n as the sum of a Fibonacci number and a positive odd squarefree number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 13 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 27, 83, 31509.
This conjecture implies that any integer n > 1 not equal to 83 can be written as the sum of a positive Fibonacci number and a positive odd squarefree number, which has been verified for n up to 10^10. Note that 83 = 0 + 83 = 1 + 2*41, where 0 and 1 are Fibonacci numbers, and 83 and 2*41 are squarefree.
The author would like to offer 1000 US dollars as the prize for the first complete solution to his conjecture that any positive integer is the sum of a Fibonacci number and a positive odd squarefree number.
See also A304331, A304333 and A304523 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0 + 1 with 0 a Fibonacci number and 1 odd and squarefree.
a(2) = 1 since 2 = 1 + 1 with 1 = A000045(1) = A000045(2) a Fibonacci number and 1 odd and squarefree.
a(27) = 1 since 27 = 8 + 19 with 8 = A000045(6) a Fibonacci number and 19 odd and squarefree.
a(83) = 1 since 83 = 0 + 83 with 0 = A000045(0) a Fibonacci number and 83 odd and squarefree.
a(31509) = 1 since 31509 = 10946 + 20563 with 10946 = A000045(21) a Fibonacci number and 20563 odd and squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=Fibonacci[n];
    QQ[n_]:=QQ[n]=n>0&&Mod[n,2]==1&&SquareFreeQ[n];
    tab={};Do[r=0;k=0;Label[bb];If[f[k]>=n,Goto[aa]];If[QQ[n-f[k]],r=r+1];k=k+1+Boole[k==1];Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]

A304523 Number of ordered ways to write n as the sum of a Lucas number (A000032) and a positive odd squarefree number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 13 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 11, 13, 27, 29, 67, 139, 193, 247, 851.
It has been verified that a(n) > 0 for all n = 2..5*10^9.
See also A304331, A304333 and A304522 for similar conjectures.

Examples

			a(3) = 1 since 3 = A000032(0) + 1 with 1 odd and squarefree.
a(27) = 1 since 27 = A000032(3) + 23 with 23 odd and squarefree.
a(29) = 1 since 29 = A000032(6) + 11 with 11 odd and squarefree.
a(67) = 1 since 67 = A000032(0) + 5*13 with 5*13 odd and squarefree.
a(247) = 1 since 247 = A000032(6) + 229 with 229 odd and squarefree.
a(851) = 1 since 851 = A000032(0) + 3*283 with 3*283 odd and squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=LucasL[n];
    QQ[n_]:=QQ[n]=n>0&&Mod[n,2]==1&&SquareFreeQ[n];
    tab={};Do[r=0;k=0;Label[bb];If[k>0&&f[k]>=n,Goto[aa]];If[QQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];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]
Showing 1-10 of 17 results. Next