A047980 a(n) is smallest difference d of an arithmetic progression dk+1 whose first prime occurs at the n-th position.

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

Offset: 1

Author

Labos Elemer

Keywords

• nonn

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)).

Index of the first occurrence of n in A034693. - Amarnath Murthy, May 08 2003

Examples

			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
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..150 (terms 1..72 from Robert Israel)
• Jon E. Schoenfield, Terms <= 5*10^8: Table of n, a(n) for n = 1..406, with -1 for each term > 5*10^8
• Index entries for sequences related to primes in arithmetic progressions

Crossrefs

Cf. A034693, A034694, A034780, A034782, A034783, A034784, A006093, A047981, A047982.

Programs

MATLAB

function [ A ] = A047980( P, N )
%   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
end % Robert Israel, Jan 25 2016

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:
seq(A[i],i=1..N); # Robert Israel, Jan 25 2016

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 *)

Formula

a(n) = min{k | A034693(k) = n}.