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.

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A233567 Number of ways to write n = p + q (q > 0) with p and p^4 + phi(q)^4 both prime, where phi(.) is Euler's totient function (A000010).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 13 2013

Keywords

Comments

Conjecture: If n > 2 is not equal to 5, then we have a(n) > 0, also there is a prime p < n with p^2 + phi(n-p)^2 prime.
We have verified this for n up to 10^7. The first assertion in the conjecture implies that there are infinitely many primes of the form p^4 + q^4, where p is a prime and q is a positive integer.

Examples

			a(7) = 1 since 7 = 3 + 4 with 3 and 3^4 + phi(4)^4 = 81 + 16 = 97 both prime.
a(12) = 1 since 12 = 7 + 5 with 7 and 7^4 + phi(5)^4 = 7^4 + 4^4 = 2657 both prime.
a(31) = 1 since 31 = 23 + 8 with 23 and 23^4 + phi(8)^4 = 23^4 + 4^4 = 280097 both prime.
a(36) = 1 since 36 = 3 + 33 with 3 and 3^4 + phi(33)^4 = 3^4 + 20^4 = 160081 both prime.
a(90) = 1 since 90 = 79 + 11 with 79 and 79^4 + phi(11)^4 = 79^4 + 10^4 = 38960081 both prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Prime[k]^4+EulerPhi[n-Prime[k]]^4],1,0],{k,1,PrimePi[n-1]}]
    Table[a[n],{n,1,100}]

A234808 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*n - p are both prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 2, 0, 3, 1, 2, 5, 2, 1, 5, 1, 2, 7, 2, 1, 4, 1, 2, 1, 4, 1, 4, 2, 4, 11, 4, 2, 3, 1, 5, 2, 3, 2, 6, 1, 5, 15, 4, 2, 9, 1, 6, 2, 5, 4, 6, 4, 4, 3, 8, 3, 6, 4, 7, 21, 2, 4, 7, 1, 7, 4, 6, 4, 6, 4, 8, 22, 7, 3, 13, 1, 10, 5, 3, 5, 7, 4, 9, 5, 10, 5, 8, 7, 7, 6, 8, 5, 6, 3, 8, 6, 7, 4, 8, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 30 2013

Keywords

Comments

Conjecture: a(n) > 0 except for n = 1, 8.
Clearly, this implies Goldbach's conjecture.

Examples

			a(3) = 1 since 2 + phi(1) = 3 and 2*3 - 3 = 3 are both prime.
a(20) = 1 since 11 + phi(9) = 17 and 2*20 - 17 = 23 are both prime.
a(22) = 1 since 1 + phi(21) = 13 and 2*22 - 13 = 31 are both prime.
a(24) = 1 since 9 + phi(15) = 17 and 2*24 - 17 = 31 are both prime.
a(76) = 1 since 67 + phi(9) = 73 and 2*76 - 73 = 79 are both prime.
		

Crossrefs

Programs

  • Mathematica
    f[n_,k_]:=k+EulerPhi[n-k]
    p[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[2n-f[n,k]]
    a[n_]:=a[n]=Sum[If[p[n,k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]

A234308 a(n) = |{0 < k <= n/2: phi(k^2)*phi(n-k) - 1 is a Sophie Germain prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 22 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) If n > 3, then phi(k^2)*phi(n-k) - 1 and phi(k^2)*phi(n-k) + 1 are both prime for some 0 < k < n, and also phi(j)^2*phi(n-j) - 1 and phi(j)^2*phi(n-j) + 1 are both prime for some 0 < j < n.
(iii) If n > 9 is not equal to 14, then |phi(k) - phi(n-k)|/2 is prime for some 0 < k < n, and also |phi(j) - phi(n-j)| - 1 and |phi(j) - phi(n-j)| + 1 are both prime for some 0 < j < n.
(iv) If n > 5, then sigma(k)*phi(n-k) + 1 is a square for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.
Note that part (i) of the conjecture implies that there are infinitely many Sophie Germain primes. We have verified part (i) for n up to 3*10^6.

Examples

			a(5) = 1 since phi(2^2)*phi(3) - 1 = 3 is a Sophie Germain prime.
a(10) = 1 since phi(1^2)*phi(9) - 1 = 5 is a Sophie Germain prime.
a(12) = 1 since phi(6^2)*phi(6) - 1 = 23 is a Sophie Germain prime.
a(30) = 1 since phi(2^2)*phi(28) - 1 = 23 is a Sophie Germain prime.
a(60) = 1 since phi(4^2)*phi(56) - 1 = 191 is a Sophie Germain prime.
a(75) = 1 since phi(14^2)*phi(61) - 1 = 5039 is a Sophie Germain prime.
a(95) = 1 since phi(30^2)*phi(65) - 1 = 11519 is a Sophie Germain prime.
a(106) = 1 since phi(22^2)*phi(84) - 1 = 5279 is a Sophie Germain prime.
a(110) = 1 since phi(9^2)*phi(101) - 1 = 5399 is a Sophie Germain prime.
a(156) = 1 since phi(27^2)*phi(129) - 1 = 40823 is a Sophie Germain prime.
		

Crossrefs

Programs

  • Mathematica
    SG[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
    a[n_]:=Sum[If[SG[EulerPhi[k^2]*EulerPhi[n-k]-1],1,0],{k,1,n/2}]
    Table[a[n],{n,1,100}]

A237531 a(n) = |{0 < k < n/2: phi(k*(n-k)) - 1 and phi(k*(n-k)) + 1 are both prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 09 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
Clearly, this implies the twin prime conjecture.

Examples

			a(12) = 1 since 12 = 3 + 9 with phi(3*9) - 1 = 17 and phi(3*9) + 1 = 19 both prime.
a(19) = 1 since 19 = 1 + 18 with phi(1*18) - 1 = 5 and phi(1*18) + 1 = 7 both prime.
a(86) = 1 since 86 = 8 + 78 with phi(8*78) - 1 = 191 and phi(8*78) + 1 = 193 both prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1]
    a[n_]:=Sum[If[p[k(n-k)],1,0],{k,1,(n-1)/2}]
    Table[a[n],{n,1,80}]
Previous Showing 11-14 of 14 results.