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Then 2p+1 is called a safe prime: see A005385.

Primes p such that the equation phi(x) = 2p has solutions, where phi is the totient function. See A087634 for another such collection of primes. - T. D. Noe, Oct 24 2003

Subsequence of A117360. - Reinhard Zumkeller, Mar 10 2006

Let q = 2n+1. For these n (and q), the difference of two cyclotomic polynomials can be written as a cyclotomic polynomial in x^2: Phi(q,x) - Phi(2q,x) = 2x Phi(n,x^2). - T. D. Noe, Jan 04 2008

A Sophie Germain prime p is 2, 3 or of the form 6k-1, k >= 1, i.e., p = 5 (mod 6). A prime p of the form 6k+1, k >= 1, i.e., p = 1 (mod 6), cannot be a Sophie Germain prime since 2p+1 is divisible by 3. - Daniel Forgues, Jul 31 2009

Also solutions to the equation: floor(4/A000005(2*n^2+n)) = 1. - Enrique Pérez Herrero, May 03 2012

In the spirit of the conjecture related to A217788, we conjecture that for any integers n >= m > 0 there are infinitely many integers b > a(n) such that the number Sum_{k=m..n} a(k)*b^(n-k) is prime. - Zhi-Wei Sun, Mar 26 2013

If k is the product of a Sophie Germain prime p and its corresponding safe prime 2p+1, then a(n) = (k-phi(k))/3, where phi is Euler's totient function. - Wesley Ivan Hurt, Oct 03 2013

Giovanni Resta found the first Sophie Germain prime which is also a Brazilian number (A125134), 28792661 = 1 + 73 + 73^2 + 73^3 + 73^4 = (11111)73. - _Bernard Schott, Mar 07 2019

For all Sophie Germain primes p >= 5, 2*p + 1 = min(A, B) where A is the smallest prime factor of 2^p - 1 and B the smallest prime factor of (2^p + 1) / 3. - Alain Rocchelli, Feb 01 2023

Consider a pair of numbers (p, 2*p+1), with p >= 3. Then p is a Sophie Germain prime iff (p-1)!^2 + 6*p == 1 (mod p*(2*p+1)). - Davide Rotondo, May 02 2024

Conjecture: Let n be any positive integer. Then a(n) exists, moreover there are infinitely many integers b > 2n+1 such that [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime. Also, the polynomial S_n(x) = sum_{k=0}^n (2k+1)*x^{n-k} is irreducible modulo some prime p < (n+1)(n+2), and the Galois group of S_n(x) over the field of rational numbers is isomorphic to the symmetric group S_n.

This conjecture can be extended by replacing 2k+1 by (2k+1)^m. For example, [1^2,3^2,5^2,...,61^2,63^2] and [63^2,61^2,...,3^2,1^2] in base b=241784 are both prime.

Conjecture: For any n>0, we have a(n) <= n*(n+1), and the Galois group of SF_n(x) = sum_{k=0}^n A005117(k+1)*x^{n-k} over the rationals is isomorphic to the symmetric group S_n.

For another related conjecture, see the author's comment on A005117.

Conjecture: a(n) < n*(n+3)/2 for all n>1.

Note that a(20) = 229 < 20*(20+3)/2 = 230.

The conjecture was motivated by E. S. Selmer's result that for any n>1 the polynomial x^n-x-1 is irreducible over the field of rational numbers.

We also conjecture that for every n=2,3,... there is a positive integer z not exceeding the (2n-2)-th prime such that z^n-z-1 is prime, and the Galois group of x^n-x-1 over the field of rationals is isomorphic to the symmetric group S_n.

Conjecture: (i) For any positive integer k and distinct positive integers a_1< a_2 < ... < a_n with a_n prime, there are infinitely many integers b > a_n^k such that [a_1^k,a_2^k,...,a_n^k] in base b is prime.

(ii) For positive integers k, m and n>m, let s_k(m,n) denote the smallest integer b > p_n^k such that [p_m^k,p_{m+1}^k,...,p_n^k] in base b is prime. Then we have the inequality s_k(m,n) <= (n+1)^k*(m+n+1)^k.

This is the k-th power version of the author's conjecture related to A217788. Note that s(m,n) defined there is identical with s_1(m,n). It seems that s_2(m,n) < p_{n+1}*p_{m+n+1}.

For example, [2^2,6^2,9^2,20^2,29^2] in base 900 and [37^2,38^2,60^2,90^2,101^2] in base 10268 are both prime. Also, s_3(1,15) = 103960 and s_5(3,5) = 161098.

Note that for any integer b>13^2 the number [2^2,5,6,156,13^2] in base b is composite since

4x^4+5x^3+6x^2+156x+169 = (4x+13)*(x^3-2x^2+8x+13).

Although 1, 2, 3, 113, 115 are pairwise relatively prime, [1,2,3,113,115] in any base b>115 is composite since x^4+2x^3+3x^2+113x+115 = (x+5)*(x^3-3x^2+18x+23).

Conjecture: a(n) does not exceed the (4n-3)-th prime for each n>0. Moreover, for any integers m>1 and n>0 the polynomial sum_{k=0}^n (k+1)^m*x^{n-k} is irreducible modulo some prime, and its Galois group over the rationals is isomorphic to the symmetric group S_n. Also, for m,n=2,3,... there are infinitely many integers b > n^m such that [n^m,...,2^m,1^m] in base b is prime.

We have a similar conjecture with the above (k+1)^m replaced by (2k+1)^m.

Conjecture: (i) a(n) does not exceed n^2+n+5 for each n>0, and the Galois group of sum_{k=0}^n C_k*x^{n-k} over the rationals is isomorphic to the symmetric group S_n.

(ii) For any positive integer n, the polynomial sum_{k=0}^n binomial(2k,k)*x^{n-k} is irreducible modulo some prime if and only if n is not of the form 2k(k+1), where k is a positive integer.

(iii) For any positive integer n, the polynomial sum_{k=0}^n T_k*x^{n-k} is irreducible modulo some prime not exceeding n^2+n+5, where T_k referes to the central trinomial coefficient A002426(k) which is the coefficient of x^k in the expansion of (x^2+x+1)^k.

Conjecture: a(n) < 4n^2-1 for all n>0.

Conjecture: a(n) < n^2 for all n > 1.

Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0.