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Conjecture: For any integers n >= m > 0, there are infinitely many positive integers s > p_n such that the number sum_{k=m}^n p_k*s^{n-k} (i.e., [p_m,...,p_n] in base s) is prime; moreover the smallest such an integer s (denoted by s(m,n)) does not exceed (n+1)*(m+n+1).

Note that s(1,n) = a(n) and s(4,21) = 546 < (21+1)*(21+4+1) = 572.

A related conjecture of the author states that for each n=2,3,... the polynomial sum_{k=1}^n p_k*x^(n-k) is irreducible modulo some prime. See also the author's comments on A000040.

The conjecture can be further extended as follows: If a_1 < ... < a_n are distinct integers with a_n prime, then there are infinitely many integers b > a_n such that [a_1,a_2,...,a_n] in base b is prime.

For example, [2,3,...,210,211] in base 55272 and[17,19,27,34,38,41] in base 300 are both prime.

See A224197 for a more general conjecture.

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

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

Conjecture: a(n) <= n^2+12 for all n>0.

Such a phenomenon happens quite often. In fact, for many interesting integer sequences a(k) (k=1,2,3,...), each of the polynomials x^n + sum_{k=0}^n a(k)*x^{n-k} (n>0) is irreducible modulo some prime not exceeding a*n^2+b*n+c, where a, b, c are suitable nonnegative constants.

It is well known that (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible over the rationals for any prime p and positive integer n.

We have the following "Reciprocity Law": For any positive integer n and primes p > 2 and q, the cyclotomic polynomial (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible modulo q if and only if q is a primitive root modulo p^n.

This can be proved as follows: As any monic irreducible polynomial over F_q=Z/qZ of degree k divides x^{q^k}-x in the ring F_q[x], the polynomial f(x)= (x^{p^n}-1)/(x^{p^{n-1}}-1) in F_q[x] has an irreducible factor of degree k < deg f if and only if f(x) is not coprime to x^{q^k}-x for some k < p^n-p^{n-1}. Note that gcd(x^{p^n}-1,x^{q^k-1}-1) = x^{gcd(p^n,q^k-1)}-1. If p^n | q^k-1, then x^{p^n}-1 | x^{q^k}-x and hence f(x) divides x^{q^k}-x; if p^n does not divide q^k-1, then gcd(x^{p^n}-1,x^{q^k-1}-1) divides x^{p^{n-1}}-1 and hence f(x) is coprime to x^{q^k}-x. Thus, f(x) is irreducible modulo q, if and only if p^n | q^k-1 for no 0 < k < p^n-p^{n-1}, i.e., q is a primitive root modulo p^n.

By the above "Reciprocity Law" in the case n=1, we have a(k) = A002233(k) for all k > 1.

Conjecture: a(n) <= sqrt(7*p_n) for all n>0.

Conjecture: a(n) <= n^2+22 for all n>0.

We have similar conjectures with 2k+1 in the definition replaced by (2k+1)^m (m=2,3,...).