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 1-5 of 5 results.

A229969 Number of ways to write n = x + y + z with 0 < x <= y <= z such that all the six numbers 2*x-1, 2*y-1, 2*z-1, 2*x*y-1, 2*x*z-1, 2*y*z-1 are prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 04 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5. Moreover, any integer n > 6 can be written as x + y + z with x among 3, 4, 6, 10, 15 such that 2*y-1, 2*z-1, 2*x*y-1, 2*x*z-1, 2*y*z-1 are prime.
We have verified this conjecture for n up to 10^6. As (2*x-1)+(2*y-1)+(2*z-1) = 2*(x+y+z)-3, it implies Goldbach's weak conjecture which has been proved.
Zhi-Wei Sun also had some similar conjectures including the following (i)-(iii):
(i) Any integer n > 6 can be written as x + y + z (x, y, z > 0) with 2*x-1, 2*y-1, 2*z-1 and 2*x*y*z-1 all prime and x among 2, 3, 4. Also, each integer n > 2 can be written as x + y + z (x, y, z > 0) with 2*x+1, 2*y+1, 2*z+1 and 2*x*y*z+1 all prime and x among 1, 2, 3.
(ii) Each integer n > 4 can be written as x + y + z with x = 3 or 6 such that 2*y+1, 2*x*y*z-1 and 2*x*y*z+1 are prime.
(iii) Every integer n > 5 can be written as x + y + z (x, y, z > 0) with x*y-1, x*z-1, y*z-1 all prime and x among 2, 6, 10. Also, any integer n > 2 not equal to 16 can be written as x + y + z (x, y, z > 0) with x*y+1, x*z+1, y*z+1 all prime and x among 1, 2, 6.
See also A229974 for a similar conjecture involving three pairs of twin primes.

Examples

			a(10) = 2 since 10 = 2+2+6 = 3+3+4 with 2*2-1, 2*6-1, 2*2*2-1, 2*2*6 -1, 2*3-1, 2*4-1, 2*3*3-1, 2*3*4-1 all prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[2i-1]&&PrimeQ[2j-1]&&PrimeQ[2(n-i-j)-1]&&PrimeQ[2i*j-1]&&PrimeQ[2i(n-i-j)-1]&&PrimeQ[2j(n-i-j)-1],1,0],{i,1,n/3},{j,i,(n-i)/2}]
    Table[a[n],{n,1,100}]

A230230 Number of ways to write 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 12 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
This is stronger than Goldbach's conjecture for even numbers. If 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime, then 6*n is the sum of the two primes 3*p - 10 and 3*q + 10.
Conjecture verified for 2*n up to 10^9. - Mauro Fiorentini, Jul 08 2023

Examples

			a(5) = 1 since 2*5 = 7 + 3 with 3*7 - 10 = 11 and 3*3 + 10 = 19 both prime.
a(16) = 1 since 2*16 = 13 + 19 with 3*13 - 10 = 29 and 3*19 + 10 = 67 both prime.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=n>3&&PrimeQ[3n-10]
    SQ[n_]:=PrimeQ[n]&&PrimeQ[3n+10]
    a[n_]:=Sum[If[PQ[Prime[i]]&&SQ[2n-Prime[i]],1,0],{i,1,PrimePi[2n-2]}]
    Table[a[n],{n,1,100}]

A230514 Number of ways to write n = a + b + c (0 < a <= b <= c) such that all the three numbers a*(a+1)-1, b*(b+1)-1, c*(c+1)-1 are Sophie Germain primes.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 21 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 22 2023
This implies that there are infinitely many Sophie Germain primes of the form x^2 + x - 1.
See also A230516 for a similar conjecture.

Examples

			a(10) = 2 since 10 = 2 + 2 + 6 = 2 + 3 + 5, and 2*3 - 1 = 5, 6*7 - 1 = 41, 3*4 - 1 = 11, 5*6 - 1 = 29 are all Sophie Germain primes.
a(39) = 1 since 39 = 9 + 15 + 15, and both 9*10 - 1 = 89 and 15*16 - 1 = 239 are Sophie Germain primes.
		

Crossrefs

Programs

  • Mathematica
    pp[n_]:=PrimeQ[n(n+1)-1]&&PrimeQ[2n(n+1)-1]
    a[n_]:=Sum[If[pp[i]&&pp[j]&&pp[n-i-j],1,0],{i,1,n/3},{j,i,(n-i)/2}]
    Table[a[n],{n,1,100}]

A227920 Number of ways to write n = x + y + z with y and z distinct and greater than x such that 6*x-1, 6*y-1, 6*x*y-1 are Sophie Germain primes and {6*x-1, 6*x+1}, {6*z-1, 6*z+1}, {6*x*z-1, 6*x*z+1} are twin prime pairs.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 08 2013

Keywords

Comments

By part (i) of the conjecture in the comments in A227923, for any integer n > 5 not equal to 14 we have a(n) > 0, because there are distinct positive integers x, y, z with x = 1 such that 6*x-1, 6*y-1, 6*x*y-1 are Sophie Germain primes and {6*x-1, 6*x+1}, {6*z-1, 6*z+1}, {6*x*z-1, 6*x*z+1} are twin prime pairs.
Conjecture: Any integer n > 2 can be written as x + y + z (x, y, z > 0) such that 6*x-1, 6*y-1, 6x*y-1, 6*z-1 are Sophie Germain primes, and {6*x-1, 6*x+1}, and {6*y-1, 6*y+1} are twin prime pairs.

Examples

			a(14) = 1 since 14 = 2 + 7 + 5, and 6*2-1 = 11, 6*7-1 = 41, 6*2*7-1 = 83 are Sophie Germain primes, and {6*2-1, 6*2+1} ={11, 13}, {6*5-1, 6*5+1} = {29, 31}, {6*2*5-1, 6*2*5+1} = {59, 61} are twin prime pairs.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=PrimeQ[6n-1]&&PrimeQ[12n-1]
    TQ[n_]:=PrimeQ[6n-1]&&PrimeQ[6n+1]
    RQ[n_]:=TQ[n]&&PrimeQ[12n-1]
    a[n_]:=Sum[If[RQ[i]&&SQ[j]&&SQ[i*j]&&TQ[n-i-j]&&TQ[i(n-i-j)]&&Abs[n-i-2j]>0,1,0],{i,1,n/3-1},{j,i+1,n-1-2i}]
    Table[a[n],{n,1,100}]

A230516 Number of ways to write n = a + b + c with 0 < a <= b <= c such that {a^2+a-1, a^2+a+1}, {b^2+b-1, b^2+b+1}, {c^2+c-1, c^2+c+1} are twin prime pairs.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 22 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 22 2023
This implies that there are infinitely many twin prime pairs of the form {x^2 + x - 1, x^2 + x + 1}.
See also A230514 for a similar conjecture.

Examples

			a(8) = 1 since 8 = 2 + 3 + 3, and {2*3 - 1, 2*3 + 1} = {5, 7} and {3*4 - 1, 3*4 + 1} = {11, 13} are twin prime pairs.
a(39) = 1 since 39 = 3 + 15 + 21, and {3*4 - 1, 3*4 + 1} = {11, 13}, {15*16 - 1, 15*16 + 1} = {239, 241}, {21*22 - 1, 21*22 + 1} = {461, 463} are twin prime pairs.
		

Crossrefs

Programs

  • Mathematica
    pp[n_]:=PrimeQ[n(n+1)-1]&&PrimeQ[n(n+1)+1]
    a[n_]:=Sum[If[pp[i]&&pp[j]&&pp[n-i-j],1,0],{i,1,n/3},{j,i,(n-i)/2}]
    Table[a[n],{n,1,100}]
Showing 1-5 of 5 results.