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 1-4 of 4 results.

A238766 Number of ordered ways to write n = k + m (k > 0 and m > 0) such that prime(prime(k)) - prime(k) + 1, prime(prime(2*k+1)) - prime(2*k+1) + 1 and prime(prime(m)) - prime(m) + 1 are all prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 05 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 11, 44, 55, 149, 371.
This suggests that there are infinitely many prime pairs {p, q} with 2*pi(p) + 1 = pi(q) such that prime(p) - p + 1 and prime(q) - q + 1 are both prime.

Examples

			a(3) = 1 since 3 = 1 + 2 with prime(prime(1)) - prime(1) + 1 = prime(2) - 2 + 1 = 2, prime(prime(2*1+1)) - prime(2*1+1) + 1 = prime(5) - 5 + 1 = 7 and prime(prime(2)) - prime(2) + 1 = prime(3) - 3 + 1 = 3 all prime.
a(371) = 1 since 371 = 66 + 305 with prime(prime(66)) - prime(66) + 1 = prime(317) - 317 + 1 = 2099 - 316 = 1783, prime(prime(2*66+1)) - prime(2*66+1) + 1 = prime(751) - 751 + 1 = 5701 - 750 = 4951 and prime(prime(305)) - prime(305) + 1 = prime(2011) - 2011 + 1 = 17483 - 2010 = 15473 all prime.

Links

Crossrefs

Cf. A000040, A234694, A234695, A235189, A236832, A238134, A238756, A238776.

Programs

  • Mathematica
    pq[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]
a[n_]:=Sum[If[pq[k]&&pq[2k+1]&&pq[n-k],1,0],{k,1,n-1}]
Table[a[n],{n,1,80}]

A238878 a(n) = |{0 < k <= n: prime(prime(k)) - prime(k) + 1 and prime(prime(k*n)) - prime(k*n) + 1 are both prime}|.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 06 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 19, 77.
(ii) For any integer n > 0, there is a number k among 1, ..., n such that 2*k + 1 and prime(prime(k^2*n)) - prime(k^2*n) + 1 are both prime.

Examples

			a(5) = 1 since prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 and prime(prime(4*5)) - prime(4*5) + 1 = prime(71) - 71 + 1 = 353 - 70 = 283 are both prime.
a(77) = 1 since prime(prime(3)) - prime(3) + 1 = prime(5) - 5 + 1 = 11 - 4 = 7 and prime(prime(3*77)) - prime(3*77) + 1 = prime(1453) - 1453 + 1 = 12143 - 1452 = 10691 are both prime.

Links

Crossrefs

Cf. A000040, A234695, A237715, A238573, A238576, A238756, A238766, A238776, A238814, A238881.

Programs

  • Mathematica
    PQ[n_]:=PrimeQ[Prime[n]-n+1]
p[k_,n_]:=PQ[Prime[k]]&&PQ[Prime[k*n]]
a[n_]:=Sum[If[p[k,n],1,0],{k,1,n}]
Table[a[n],{n,1,80}]

A238814 Primes p with prime(p) - p + 1 and prime(q) - q + 1 both prime, where q is the first prime after p.

Original entry on oeis.org

2, 3, 5, 13, 41, 83, 199, 211, 271, 277, 293, 307, 349, 661, 709, 743, 751, 823, 907, 1117, 1447, 1451, 1741, 1747, 2203, 2371, 2803, 2819, 2861, 2971, 3011, 3251, 3299, 3329, 3331, 3691, 3877, 4021, 4027, 4049, 4051, 4093, 4129, 4157, 4447, 4513, 4549, 4561, 4751, 4801, 5179, 5479, 5519, 5657, 5813, 6007, 6011, 6571, 7057, 7129
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 05 2014

Keywords

Comments

Conjecture: The sequence is infinite, in other words, A234695 contains infinitely many consecutive prime pairs prime(k) and prime(k+1).
This is motivated by the comments in A238766 and A238776, and the sequence is a subsequence of A234695.

Examples

			a(1) = 2 since prime(2) - 2 + 1 = 3 - 1 = 2 and prime(3) - 3 + 1 = 5 - 2 = 3 are both prime.
a(2) = 3 since prime(3) - 3 + 1 = 5 - 2 = 3 and prime(5) - 5 + 1 = 11 - 4 = 7 are both prime.

Links

Crossrefs

Cf. A000040, A234694, A234695, A238766, A238776.

Programs

  • Mathematica
    p[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]
n=0
Do[If[p[k]&&p[k+1],n=n+1;Print[n," ",Prime[k]]],{k,1,914}]
Select[Prime[Range[1000]],AllTrue[{Prime[#]-#+1,Prime[NextPrime[#]]-NextPrime[ #]+1},PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Aug 24 2019 *)
  • PARI
    step(p,k)=k++;while(k--,p=nextprime(p+1)); p
p=0;forprime(r=2,1e6,if(isprime(p++) && isprime(r-p+1), q=nextprime(p+1); if(isprime(step(r,q-p)-q+1), print1(p", ")))) \\ Charles R Greathouse IV, Mar 06 2014

A237715 Number of ordered ways to write n = p + q (q > 0) with p, prime(p) - p + 1 and prime(prime(q)) - prime(q) + 1 all prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 06 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3.

Examples

			a(3) = 1 since 3 = 2 + 1 with 2, prime(2) - 2 + 1 = 3 - 1 = 2 and prime(prime(1)) - prime(1) + 1 = prime(2) - 2 + 1  = 2 all prime.
a(7) = 2 since 7 = 3 + 4 with 3, prime(3) - 3 + 1 = 5 - 2 = 3 and prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 are all prime, and 7 = 5 + 2 with 5, prime(5) - 5 + 1 = 11 - 4 = 7 and prime(prime(2)) - prime(2) + 1 = prime(3) - 3 + 1 = 5 - 2 = 3 all prime.

Links

Crossrefs

Cf. A000040, A234694, A234695, A238766, A238776, A238814.

Programs

  • Mathematica
    pq[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]
a[n_]:=Sum[If[pq[k]&&pq[n-Prime[k]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]
Showing 1-4 of 4 results.

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