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73, 97, 103, 193, 229, 241, 277, 283, 313, 331, 367, 373, 397, 433, 457, 463, 547, 607, 619, 643, 661, 709, 727, 733, 739, 757, 823, 859, 883, 907, 967, 997, 1021, 1033, 1069, 1087, 1093, 1123, 1129, 1171, 1237, 1249, 1303, 1423, 1447, 1453, 1483, 1489, 1543, 1579, 1597
COMMENTS
a(n) = P(n,3) = 1 + 3*K(n,3) = 1 + 3* A034782(n). P(n,3) are special primes of the form 3k+1. The relevant values of k are given by A034782. Note that, e.g., 13, 19, 31, 5, 13 are not in this sequence.
PROG
(PARI) a034693(n) = my(s=1); while(!isprime(s*n+1), s++); s;
isok(n) = a034693(n) == 3;
lista(nn) = {for (n=1, nn, if (isok(n), print1(3*n+1, ", ")); ); } \\ Michel Marcus, May 13 2018
EXTENSIONS
Corrected (wrong term 769 removed) and extended by Michel Marcus, May 13 2018
Smallest prime == 1 (mod n).
+10
61
2, 3, 7, 5, 11, 7, 29, 17, 19, 11, 23, 13, 53, 29, 31, 17, 103, 19, 191, 41, 43, 23, 47, 73, 101, 53, 109, 29, 59, 31, 311, 97, 67, 103, 71, 37, 149, 191, 79, 41, 83, 43, 173, 89, 181, 47, 283, 97, 197, 101, 103, 53, 107, 109, 331, 113, 229, 59, 709, 61, 367, 311
COMMENTS
Thangadurai and Vatwani prove that a(n) <= 2^(phi(n)+1)-1. - T. D. Noe, Oct 12 2011 Eric Bach and Jonathan Sorenson show that, assuming GRH, a(n) <= (1 + o(1))*(phi(n)*log(n))^2 for n > 1. See the abstract of their paper in the Links section. - Jianing Song, Nov 10 2019 a(n) is the smallest prime p such that the multiplicative group modulo p has a subgroup of order n. - Joerg Arndt, Oct 18 2020
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, section 2.12, pp. 127-130.
P. Ribenboim, The Book of Prime Number Records. Chapter 4,IV.B.: The Smallest Prime In Arithmetic Progressions, 1989, pp. 217-223.
EXAMPLE
If n = 7, the smallest prime in the sequence 8, 15, 22, 29, ... is 29, so a(7) = 29.
MATHEMATICA
a[n_] := Block[{k = 1}, If[n == 1, 2, While[Mod[Prime@k, n] != 1, k++ ]; Prime@k]]; Array[a, 64] (* Robert G. Wilson v, Jul 08 2006 *) With[{prs=Prime[Range[200]]}, Flatten[Table[Select[prs, Mod[#-1, n]==0&, 1], {n, 70}]]] (* Harvey P. Dale, Sep 22 2021 *)
PROG
(PARI) a(n)=if(n<0, 0, s=1; while((prime(s)-1)%n>0, s++); prime(s))
(Haskell)
a034694 n = until ((== 1) . a010051) (+ n) (n + 1)
CROSSREFS
Cf. A034693, A034780, A034782, A034783, A034784, A034785, A034846, A034847, A034848, A034849, A038700, A085420.
a(n) is smallest difference d of an arithmetic progression dk+1 whose first prime occurs at the n-th position.
+10
3
1, 3, 24, 7, 38, 17, 184, 71, 368, 19, 668, 59, 634, 167, 512, 757, 1028, 197, 1468, 159, 3382, 799, 4106, 227, 10012, 317, 7628, 415, 11282, 361, 38032, 521, 53630, 3289, 37274, 2633, 63334, 1637, 34108, 1861, 102296, 1691, 119074, 1997, 109474, 2053
COMMENTS
Definition involves two minimal conditions: (1) the first prime (as in A034693) and (2) dk+1 sequences were searched with minimal d. Present terms are the first ones in sequences analogous to A034780, A034782- A034784, A006093 (called there K(n,m)).
EXAMPLE
For n=2, the sequence with d=1 is 2,3,4,5,... with the prime 2 for k=1. The sequence with d=2 is 3,5,7,9,... with the prime 3 for k=1. The sequence with d=3 is 4,7,10,13,... with the prime 7 for k=2. So a(n)=3. - Michael B. Porter, Mar 18 2019
MAPLE
N:= 40: # to get a(n) for n <= N
count:= 0:
p:= 0:
Ds:= {1}:
while count < N do
p:= nextprime(p);
ds:= select(d -> (p-1)/d <= N, numtheory:-divisors(p-1) minus Ds);
for d in ds do
n:= (p-1)/d;
if not assigned(A[n]) then
A[n]:= d;
count:= count+1;
fi
od:
Ds:= Ds union ds;
od:
MATHEMATICA
With[{s = Table[k = 1; While[! PrimeQ[k n + 1], k++]; k, {n, 10^6}]}, TakeWhile[#, # > 0 &] &@ Flatten@ Array[FirstPosition[s, #] /. k_ /; MissingQ@ k -> {0} &, Max@ s]] (* Michael De Vlieger, Aug 01 2017 *)
PROG
(MATLAB)
% Get values a(i) for i <= N with a(i) <= P/i
% using primes <= P.
% Returned entries A(n) = 0 correspond to unknown a(n) > P/n
Primes = primes(P);
A = zeros(1, N);
Ds = zeros(1, P);
for p = Primes
ns = [1:N];
ns = ns(mod((p-1) * ones(1, N), ns) == 0);
newds = (p-1) ./ns;
ns = ns(A(ns) == 0);
ds = (p-1) ./ ns;
q = (Ds(ds) == 0);
A(ns(q)) = ds(q);
Ds(newds) = 1;
end
3, 7, 17, 71, 19, 59, 167, 757, 197, 159, 799, 227, 317, 415, 361, 521, 3289, 2633, 1637, 1861, 1691, 1997, 2053, 4097, 6437, 5731, 9199, 11603, 5641, 3833, 26885, 6637, 26815, 32117, 18637, 29933, 31667, 5227, 19891, 47303, 54973, 5207, 59537
FORMULA
a(n) = min {k}: A034693(a(n)) is an even number such that in a(n)*k+1 progression the first prime occurs at even 2n=k position.
EXAMPLE
First example: a(1)=3 since in 3k+1 sequence, the first term is 3, a prime and the d=2 is the smallest such difference. The next such progression is 5k+1 because 5*2+1=11 is prime. 2nd example: here at n=6 a(6)=59. This means that 2n=12 occurs first in A034693 at its position 59, which means that its first prime is 12*59+1=709. arises as 12th term (such progressions are: 59k+1,85k+1,133k+1, etc.)
1, 24, 38, 184, 368, 668, 634, 512, 1028, 1468, 3382, 4106, 10012, 7628, 11282, 38032, 53630, 37274, 63334, 34108, 102296, 119074, 109474, 117206, 60664, 410942, 204614, 127942, 125618, 595358, 517882, 304702, 352022, 1549498, 651034, 506732, 5573116, 1379216, 1763144
FORMULA
a(n) = min {d}: A034693(a(n)) is an odd number k such that in a(n)*k+1 progression the first prime occurs at k=2n+1 position.
EXAMPLE
a(2)=38 because A034693(38) = 2*2+1 = 5 is the first 5; 5*38+1 = 191 is the first prime. The successive progressions in which the first prime appears at position 5 are as follows: 38k+1, 62k+1, 164k+1. 2nd example: a(20)=102296 because. The first 41 appears in A034693 at this index. Also 102296*(2*20+1)+1 = 102296*41+1 = 4194137 is the first prime in {102296k+1}. The next progression with this position of prime emergence is 109946k+1 (the corresponding prime is 4507787).
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