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.

A209320 Number of ways to write 2n = p+q with p and q both prime, p+1 and q-1 both practical.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 19 2013

Keywords

Comments

Conjecture: a(n)>0 for all n>2.
As p+q=(p+1)+(q-1), this unifies Goldbach's conjecture and its analog involving practical numbers.
The conjecture has been verified for n up to 10^7.

Examples

			a(8) = 2 since 2*8 = 3+13 = 11+5 with 3, 5, 11, 13 all prime and 3+1, 13-1, 11+1, 5-1 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[PrimeQ[2n-Prime[k]]==True&&pr[Prime[k]+1]==True&&pr[2n-Prime[k]-1]==True,1,0],{k,1,PrimePi[2n-2]}]
    Do[Print[n," ",a[n]],{n,1,100}]

A210445 Least positive integer k with k*n practical.

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 4, 1, 2, 2, 6, 1, 6, 2, 2, 1, 12, 1, 12, 1, 2, 3, 12, 1, 4, 3, 2, 1, 12, 1, 16, 1, 2, 6, 4, 1, 18, 6, 2, 1, 20, 1, 20, 2, 2, 6, 24, 1, 4, 2, 4, 2, 24, 1, 4, 1, 4, 6, 24, 1, 24, 8, 2, 1, 4, 1, 30, 3, 4, 2, 30, 1, 30, 9, 2, 3, 4, 1, 36, 1, 2, 10, 36, 1, 4, 10, 4, 1, 36, 1, 4, 3, 6, 12, 4, 1, 42, 2, 2, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 20 2013

Keywords

Comments

Conjecture: a(n) < n for all n>1, and a(n) < n/2 for all n>47.
Large values are obtained for prime n: The corresponding subsequence is a(p(n)) = (1, 2, 4, 4, 6, 6, 12, 12, 12, 12, 16, 18, 20, 20, 24, 24, 24, 24, ...), while for composite indices, a(c(n)) = (1, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 3, 1, 4, 3, 2, 1, 1, 1, 2, ...). - M. F. Hasler, Jan 21 2013

Examples

			a(10)=2 since 2*10=20 is practical but 1*10=10 is not.
		

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)
    Do[Do[If[pr[k*n]==True,Print[n," ",k];Goto[aa]],{k,1,n}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]
  • PARI
    A210445(n)={for(k=1,n,is_A005153(k*n)&&return(k))} \\ (Would return 0 if a(n)>n.) - M. F. Hasler, Jan 20 2013

Formula

a(n) = 1 iff n is in A005153, therefore a(n) > 1 for all odd n>1. - M. F. Hasler, Jan 21 2013

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}]

A219314 Composition of the inverse binomial transform of Fibonacci numbers and the Catalan transform of Fibonacci numbers.

Original entry on oeis.org

0, 1, 0, 3, 3, 13, 26, 77, 192, 529, 1412, 3873, 10603, 29315, 81318, 226763, 634627, 1782637, 5022840, 14193457, 40211105, 114191159, 324981030, 926720807, 2647513282, 7576475383, 21716189676, 62336237007, 179182653117, 515717424109, 1486119467026
Offset: 0

Author

Arkadiusz Wesolowski, Nov 17 2012

Keywords

Crossrefs

Formula

G.f.: ((1+2*x)*sqrt(1-2*x-3*x^2) - 1 + x + 2*x^2)/(2*(1+x)*(1-2*x-4*x^2)).
Asymptotics: a(n) ~ 3^(n+2)*5/(8*sqrt(3*Pi*n^3)). - Fung Lam, Apr 07 2014
Conjecture: n*a(n) -2*n*a(n-1) +11*(-n+2)*a(n-2) +4*(2*n-5)*a(n-3) +8*(5*n-17)*a(n-4) +24*(n-4)*a(n-5)=0. - R. J. Mathar, Jun 14 2016
Conjecture: n*(5*n-7)*a(n) -4*(5*n^2-12*n+6)*a(n-1) -(15*n^2-11*n-30) *a(n-2) +2*(35*n^2-119*n+66)*a(n-3) +12*(n-3)*(5*n2)*a(n-4)=0. - R. J. Mathar, Jun 14 2016
Showing 1-5 of 5 results.