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 11-18 of 18 results.

A210480 Number of primes p

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 23 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>3.
This is stronger than Goldbach's conjecture and the author's conjecture that any odd number greater than one is the sum of a prime and a practical number. Also, it implies that there are infinitely many primes p with p-1 and p+1 both practical.
The author has verified this new conjecture for n up to 10^7.

Examples

			a(1846)=1 since 1846=1289+557 with 1289 and 557 both prime, and 1288 and 1290 both practical.
a(15675)=1 since 15675=919+14756 with 919 prime, and 918, 920, 14756 all practical.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    a[n_]:=a[n]=Sum[If[pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True&&(PrimeQ[n-Prime[k]]==True||pr[n-Prime[k]]==True),1,0],{k,1,PrimePi[n-1]}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210533 Number of ways to write 2n = x+y (x,y>0) with x-1 and x+1 both prime, and x and x^3+y^3 both practical.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 28 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>2. Moreover, for each m=2,3,4,... any sufficiently large even integer can be written as x+y (x,y>0) with x-1 and x+1 both prime, and x and x^m+y^m both practical.

Examples

			a(17)=1 since 2*17=12+22 with 11 and 13 both prime, and 12 and 12^3+22^3=12376 both practical.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    a[n_]:=a[n]=Sum[If[PrimeQ[2k-1]==True&&PrimeQ[2k+1]==True&&pr[2k]==True&&pr[(2k)^3+(2n-2k)^3]==True,1,0],{k,1,n-1}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210681 Number of ways to write 2n = p+q+r (p<=q) with p, q, r-1, r+1 all prime and p-1, p+1, q-1, q+1, r all practical.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 29 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 4.
This conjecture involves two kinds of sandwiches introduced by the author, and it is much stronger than the Goldbach conjecture for odd numbers. We have verified the conjecture for n up to 10^7.
Zhi-Wei Sun also made the following conjectures:
(1) Any even number greater than 10 can be written as the sum of four elements in the set
S = {prime p: p-1 and p+1 are both practical}.
Also, every n=3,4,5,... can be represented as the sum of a prime in S and two triangular numbers.
(2) Each integer n>7 can be written as p + q + x^2 (or p + q + x(x+1)/2), where p is a prime with p-1 and p+1 both practical, and q is a practical number with q-1 and q+1 both prime.
(3) Every n=3,4,... can be written as the sum of three elements in the set
T = {x: 6x is practical with 6x-1 and 6x+1 both prime}.
(4) Any integer n>6 can be represented as the sum of two elements of the set S and one element of the set T.
(5) Each odd number greater than 11 can be written in the form 2p+q+r, where p and q belong to S, and r is a practical number with r-1 and r+1 both prime.

Examples

			a(5)=1 since 2*5=3+3+4 with 3 and 5 both prime, and 2 and 4 both practical.
a(6)=2 since 2*6=3+3+6=3+5+4 with 3,5,7 all prime and 2,4,6 all practical.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    pp[k_]:=pp[k]=pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True
    pq[n_]:=pq[n]=PrimeQ[n-1]==True&&PrimeQ[n+1]==True&&pr[n]==True
    a[n_]:=a[n]=Sum[If[pp[j]==True&&pp[k]==True&&pq[2n-Prime[j]-Prime[k]]==True,1,0],{j,1,PrimePi[n-1]},{k,j,PrimePi[2n-Prime[j]]}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210444 a(n) = |{0

Original entry on oeis.org

0, 0, 1, 2, 0, 4, 1, 0, 2, 2, 0, 4, 0, 1, 4, 2, 0, 6, 1, 3, 2, 2, 0, 5, 2, 1, 3, 1, 2, 11, 0, 1, 4, 1, 2, 6, 0, 2, 4, 3, 1, 9, 2, 3, 4, 2, 0, 7, 1, 4, 4, 5, 0, 8, 4, 1, 3, 3, 0, 15, 0, 3, 4, 4, 4, 13, 2, 4, 2, 5, 2, 10, 0, 2, 11, 2, 3, 12, 0, 6, 6, 2, 2, 13, 3, 5, 7, 5, 1, 16, 4, 4, 6, 3, 2, 11, 0, 8, 6, 7
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 20 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>911.
This implies that for each n=2,3,4,... there is a positive integer k
The conjecture has been verified for n up to 10^6.

Examples

			a(7) = 1 since 6*7 = 42 is practical, and 41 and 43 are twin primes.
		

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    a[n_]:=a[n]=Sum[If[PrimeQ[k*n-1]==True&&PrimeQ[k*n+1]==True&&pr[k*n]==True,1,0],{k,1,n-1}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210452 Number of integers k

Original entry on oeis.org

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

Author

Zhi-Wei Sun, Jan 20 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>4.
This implies the twin prime conjecture since k*p is not practical for any prime p>sigma(k)+1.
Zhi-Wei Sun also made the following conjectures:
(1) For each integer n>197, there is a practical number k
(2) For every n=9,10,... there is a practical number k
(3) For any integer n>26863, the interval [1,n] contains five consecutive integers m-2, m-1, m, m+1, m+2 with m-1 and m+1 both prime, and m-2, m, m+2, m*n all practical.

Examples

			a(11)=1 since 5 and 7 are twin primes, and 6 and 6*11 are both practical.
		

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    a[n_]:=a[n]=Sum[If[PrimeQ[k-1]==True&&PrimeQ[k+1]==True&&pr[k]==True&&pr[k*n]==True,1,0],{k,1,n-1}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210722 Number of ways to write n = (2-(n mod 2))p+q+2^k with p, q-1, q+1 all prime, and p-1, p+1, q all practical.

Original entry on oeis.org

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

Author

Zhi-Wei Sun, Jan 29 2013

Keywords

Comments

Conjecture: a(n)>0 except for n = 1,...,8, 10, 520, 689, 740.
Zhi-Wei Sun also guessed that any integer n>6 different from 407 can be written as p+q+F_k, where p is a prime with p-1 and p+1 practical, q is a practical number with q-1 and q+1 prime, and F_k (k>=0) is a Fibonacci number.

Examples

			a(1832)=1 since 1832=2*881+6+2^6 with 5, 7, 881 all prime and 6, 880, 882 all practical.
a(11969)=1 since 11969=127+11778+2^6 with 127, 11777, 11779 all prime and 126, 128, 11778 all practical.
		

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    pp[k_]:=pp[k]=pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True
    pq[n_]:=pq[n]=PrimeQ[n-1]==True&&PrimeQ[n+1]==True&&pr[n]==True
    a[n_]:=a[n]=Sum[If[pp[j]==True&&pq[n-2^k-(2-Mod[n,2])Prime[j]]==True,1,0],{k,0,Log[2,n]},{j,1,PrimePi[(n-2^k)/(2-Mod[n,2])]}]
    Do[Print[n," ",a[n]],{n,1,100}]

A211165 Number of ways to write n as the sum of a prime p with p-1 and p+1 both practical, a prime q with q+2 also prime, and a Fibonacci number.

Original entry on oeis.org

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

Author

Zhi-Wei Sun, Jan 30 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>5.
This has been verified for n up to 300000.
Note that for n=406 we cannot represent n in the given way with q+1 practical.

Examples

			a(6)=a(7)=1 since 6=3+3+0 and 7=3+3+1 with 3 and 5 both prime, 2 and 4 both practical, 0 and 1 Fibonacci numbers.
		

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
    pp[k_]:=pp[k]=pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True
    q[n_]:=q[n]=PrimeQ[n]==True&&PrimeQ[n+2]==True
    a[n_]:=a[n]=Sum[If[k!=2&&Fibonacci[k]
    				

A213325 Number of ways to write n = q + sum_{k=1}^m(-1)^{m-k}p_k, where p_k is the k-th prime, and q is a practical number with q-4 and q+4 also practical.

Original entry on oeis.org

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

Author

Zhi-Wei Sun, Mar 03 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>8.
The author has verified this for n up to 5*10^6.

Examples

			a(11)=1 since 11=8+(7-5+3-2) with 4, 8, 12 all practical.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n]
    Pow[n_,i_]:=Pow[n,i]=Part[Part[f[n],i],1]^(Part[Part[f[n],i],2])
    Con[n_]:=Con[n]=Sum[If[Part[Part[f[n],s+1],1]<=DivisorSigma[1,Product[Pow[n,i],{i,1,s}]]+1,0,1],{s,1,Length[f[n]]-1}]
    pr[n_]:=pr[n]=n>0&&(n<3||Mod[n,2]+Con[n]==0)
    q[n_]:=q[n]=pr[n-4]==True&&pr[n]==True&&pr[n+4]==True
    s[0_]:=0
    s[n_]:=s[n]=Prime[n]-s[n-1]
    a[n_]:=a[n]=Sum[If[n-s[m]>0&&q[n-s[m]],1,0],{m,1,n}]
    Table[a[n],{n,1,100}]
Previous Showing 11-18 of 18 results.