A234337 -id:A234337 - cp's OEIS frontend

A234388 Primes of the form 2^k*(2^{phi(m)} - 1) + 1, where k and m are positive integers, and phi(.) is Euler's totient function.

3, 5, 7, 13, 17, 31, 61, 97, 127, 193, 241, 257, 769, 1009, 1021, 2017, 4093, 7681, 8161, 8191, 12289, 15361, 16369, 16381, 32257, 61441, 64513, 65521, 65537, 131041, 131071, 523777, 524287, 786433, 1032193, 1048573, 4194301, 8257537, 8380417, 16515073, 16760833, 16776961, 16777153, 16777213, 67043329, 132120577, 134215681, 268369921, 536870401, 1073479681, 2013265921, 2113929217, 2146959361, 2147483137, 2147483647, 3221225473, 4293918721, 17175674881, 34359214081, 34359738337

Offset: 1

Zhi-Wei Sun, Dec 25 2013

nonn

Conjecture: (i) Any integer n > 1 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) + 1 is prime.
(ii) Each integer n > 2 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) - 1 is prime.
Part (i) of the conjecture implies that this sequence has infinitely many terms. See also A234399.
Note that the sequence contains all Fermat primes and Mersenne primes since 2^k + 1 = 2^k*(2^{phi(1)} - 1) + 1 and 2^p - 1 = 2*(2^{phi(p)} - 1) + 1, where k is a positive integer and p is a prime.

			a(1) = 3 since 2*(2^{phi(1)} - 1) + 1 = 3 is prime.
a(2) = 5 since 2^2*(2^{phi(1)} - 1) + 1 = 5 is prime.
a(3) = 7 since 2*(2^{phi(3)} - 1) + 1 = 7 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

Cf. A000010, A000040, A000079, A000668, A019434, A152449, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234399

S:=Intersection[Union[Table[EulerPhi[k],{k,1,5000}]],Table[k,{k,1,500}]]
n=0;Do[If[MemberQ[S,k]&&PrimeQ[2^m-2^(m-k)+1],n=n+1;Print[n," ",2^m-2^(m-k)+1]],{m,1,500},{k,1,m-1}]

A234399 a(n) = |{0 < k < n: 2^k*(2^phi(n-k) - 1) + 1 is prime}|, where phi(.) is Euler's totient function.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Dec 25 2013

nonn

Conjecture: a(n) > 0 for all n > 1.
See also the conjecture in A234388.
The conjecture is false. a(5962) = 0. - Jason Yuen, Nov 04 2024

			a(7) = 2 since 2^1*(2^phi(6)-1) + 1 = 2*3 + 1 = 7 and 2^2*(2^phi(5)-1) + 1 = 4*15 + 1 = 61 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000

Cf. A000010, A000040, A000079, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234388

f[n_,k_]:=f[n,k]=2^k*(2^(EulerPhi[n-k])-1)+1
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

a(n) = sum(k=1,n-1,ispseudoprime(2^k*(2^eulerphi(n-k)-1)+1)) \\ Jason Yuen, Nov 04 2024

A236483 a(n) = |{0 < k < n-2: p(k) + 2^(phi(n-k)/2) is prime}|, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 2, 3, 5, 4, 3, 6, 1, 6, 5, 7, 7, 5, 6, 4, 6, 5, 4, 7, 4, 6, 5, 5, 6, 9, 6, 13, 6, 10, 5, 6, 7, 4, 10, 9, 12, 6, 12, 3, 8, 8, 9, 11, 8, 11, 7, 11, 7, 8, 8, 9, 8, 10, 10, 9, 9, 14, 8, 15, 8, 11, 13, 12, 12, 15, 13, 12, 8, 11, 12, 10, 11, 12, 12, 13, 8, 12, 14, 8, 13, 10, 8, 8, 12, 8, 15, 11, 10, 11, 11, 13, 14, 10, 8

Offset: 1

Zhi-Wei Sun, Jan 27 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any integer n > 1, there is a positive integer k < n such that p(k)*2^(phi(n-k)) + 1 is prime.
(iii) For any integer n > 1 not among 6, 23, 42, there is a positive integer k < n such that 2*p(k) + p(n-k) is prime.
Clearly, part (i) of the conjecture implies that there are infinitely many primes of the form p(k) + 2^m, where k and m are positive integers.

			a(14) = 1 since p(7) + 2^(phi(7)/2) = 15 + 2^3 = 23 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000010, A000040, A000041, A000079, A234337.

a[n_]:=Sum[If[PrimeQ[PartitionsP[k]+2^(EulerPhi[n-k]/2)],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

A234388 Primes of the form 2^k*(2^{phi(m)} - 1) + 1, where k and m are positive integers, and phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A234399 a(n) = |{0 < k < n: 2^k*(2^phi(n-k) - 1) + 1 is prime}|, where phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A236483 a(n) = |{0 < k < n-2: p(k) + 2^(phi(n-k)/2) is prime}|, where p(.) is the partition function (A000041) and phi(.) is Euler's totient function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica