A035095 Smallest prime congruent to 1 (mod prime(n)).

3, 7, 11, 29, 23, 53, 103, 191, 47, 59, 311, 149, 83, 173, 283, 107, 709, 367, 269, 569, 293, 317, 167, 179, 389, 607, 619, 643, 1091, 227, 509, 263, 823, 557, 1193, 907, 1571, 653, 2339, 347, 359, 1087, 383, 773, 3547, 797, 2111, 2677, 5449, 2749, 467

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

This is a version of the "least prime in special arithmetic progressions" problem.

Smallest numbers m such that largest prime factor of Phi(m) = prime(n), the n-th prime, also seems to be prime and identical to n-th term of A035095. See A068211, A068212, A065966: Min[x : A068211(x)=prime(n)] = A035095(n); e.g., Phi(a(7)) = Phi(103) = 2*3*17, of which 17 = p(7) is the largest prime factor, arising first here.

It appears that A035095, A066674, A125878 are probably all the same, but see the comments in A066674. - N. J. A. Sloane, Jan 05 2013

Minimum of the smallest prime factors of F(n,i) = (i^prime(n)-1)/(i-1), when i runs through all integers in [2, prime(n)]. Every prime factor of F(n,i) is congruent to 1 modulo prime(n). - Vladimir Shevelev, Nov 26 2014

Conjecture: a(n) is the smallest prime p such that gpf(p-1) = prime(n). See A023503. - Thomas Ordowski, Aug 06 2017

For n>1, a(n) is the smallest prime congruent to 1 mod (2*prime(n)). - Chai Wah Wu, Apr 28 2025

Examples

			a(8) = 191 because in the prime(8)k+1 = 19k+1 sequence, 191 is the smallest prime.
		

References

• E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen, Bd 1 (reprinted Chelsea 1953).
• E. C. Titchmarsh, A divisor problem, Renc. Circ. Math. Palermo, 54 (1930) pp. 414-429.
• P. Turan, Über Primzahlen der arithmetischen Progression, Acta Sci. Math. (Szeged), 8 (1936/37) pp. 226-235.

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• P. Erdős, On some application of Brun's method, Acta Sci. Math (Szeged), v. 13, 1949, pp. 57-63.
• A. Granville and C. Pomerance, On the least prime in certain arithmetic progressions
• A. Granville and C. Pomerance, On the least prime in certain arithmetic progressions J. Lond Math Soc s2-41 (2) (1990), pp. 193-200.
• D. R. Heath-Brown, almost-primes in arithmetic progressions in short intervals, Math Proc Cambr. Phil Soc v 83 (1978), pp. 357-375.
• D. R. Heath-Brown, Siegel zeros and the least prime in arithmetic progression, Quart. J. of Math 41 (49) (1990), pp. 405-418.
• H.-J. Kanold, Uber Primzahlen in arithmetischen Folgen, Math. Ann. v 156 (1964) pp. 393-395.
• U. V. Linnik, On the least prime in an arithmetic progression. I. The basic theorem, Rec. Math (N.S.) v 15 (57) (1944), pp 139-178. MR0012111
• C. Pomerance, A note on the least prime in an arithmetic progression J. Number Theory 12 (2) (1980), pp. 218-223.
• K. Prachar, Uber die kleinste Primzahl in einer arithmetischen Reihe, J Reine Angew Math. 206 (1961) pp. 3-4.
• A. Schinzel, Remark on the paper of K. Prachar Uber die kleinste.., J. Reine Angew Math. v 210 (1962) pp. 122-122.
• S. S. Wagstaff, Jr., The irregular primes to 125000, Math. Comp., 32 (1978) pp. 583-591.
• S. S. Wagstaff, Jr, Greatest of the Least Primes in Arithmetic Progressions Having a Given Modulus, Math. Comp., 33 (147) (1979) pp. 1073-1080.
• Index entries for sequences related to primes in arithmetic progressions

Crossrefs

Cf. A034694, A032448, A006530, A006093, A035096, A000040, A019434, A058383.

Cf. A068211, A068212, A065966, A000010, A070844-A070858, A061092.

Cf. A000040.

Programs

Mathematica

a[n_] := Block[{p = Prime[n]}, r = 1 + p; While[ !PrimeQ[r], r += p]; r]; Array[a, 51] (* Jean-François Alcover, Sep 20 2011, after PARI *)
a[n_]:=If[n<2,3,Block[{p=Prime[n]},r=1+2*p;While[!PrimeQ[r],r+=2*p]];r];Array[a,51] (* Zak Seidov, Dec 14 2013 *)

PARI

a(n)=local(p,r);p=prime(n);r=1;while(!isprime(r),r+=p);r

PARI

{my(N=66); forprime(p=2, , forprime(q=p+1,10^10, if((q-1)%p==0, print1(q,", "); N-=1; break)); if(N==0,break)); } \\ Joerg Arndt, May 27 2016

Python

from itertools import count
from sympy import prime, isprime
def A035095(n): return 3 if n==1 else next(filter(isprime,count((p:=prime(n)<<1)+1,p))) # Chai Wah Wu, Apr 28 2025

Formula

According to a long-standing conjecture (see the 1979 Wagstaff reference), a(n) <= prime(n)^2 + 1. This would be sufficient to imply that a(n) is the smallest prime such that greatest prime divisor of a(n)-1 is prime(n), the n-th prime: A006530(a(n)-1) = A000040(n). This in turn would be sufficient to imply that no value occurs twice in this sequence. - Franklin T. Adams-Watters, Jun 18 2010

a(n) = 1 + A035096(n)*A000040(n). - Zak Seidov, Dec 27 2013

Extensions

Edited by Franklin T. Adams-Watters, Jun 18 2010

Minor edits by N. J. A. Sloane, Jun 27 2010

Edited by N. J. A. Sloane, Jan 05 2013