A206333 Smallest prime q such that, starting with q, there are prime(n)-1 consecutive primes = {1..prime(n)-1} modulo prime(n).

3, 7, 251, 61223, 23700022897

Offset

1,1

Comments

Heuristically a(n) is around (p-1)^p log 2, where p is the n-th prime. - Charles R Greathouse IV, Jan 11 2013

Links

Table of n, a(n) for n=1..5.

Example

Let p(n) = prime(n), then
a(1)=3 because 3 is smallest prime = ( 1 modulo p(1)) = 1 mod 2;
a(2)=7 because 2 smallest consecutive primes {7,11}= {1,2} modulo p(2) = {1,2} mod 3;
a(3) = 251 because {251,257,263,269} = {1,2,3,4} modulo p(3)= {1,2,3,4} mod 5;
a(4) = 61223 because {61223,61231,61253,61261,61283,61291} = {1,2,3,4,5,6} modulo p(4) = {1,2,3,4,5,6} mod 7;
n=5: p(n) = 11, prime(1037632211..1037632220) = {23700022897, 23700022909, 23700022921, 23700022933, 23700022967, 23700022979, 23700022991, 23700023003, 23700023059, 23700023093} = {1,2,3,4,5,6,7,8,9,10} mod 11, d={12,12,12,34,12,12,12,56,34}. - Zak Seidov, Jan 05 2013

Mathematica

Table[n = 1; While[Mod[Prime[Range[n, n+p-2]], p] != Range[p-1], n++]; Prime[n], {p, Prime[Range[4]]}] (* T. D. Noe, Feb 07 2012 *)

Prog

(PARI) a(n)=my(i=1, q=prime(n)); forprime(p=2, , if(p%q==i, if(i++==q, for(i=3, q, p=precprime(p-1)); return(p)), i=if(p%q==1, 2, 1))) \\ Charles R Greathouse IV, Jan 11 2013

Crossrefs

Sequence in context: A349838 A179859 A061422 * A088097 A064774 A348376

Adjacent sequences: A206330 A206331 A206332 * A206334 A206335 A206336

Keyword

hard,more,nonn

Author

Zak Seidov, Feb 06 2012

Extensions

a(5) from Zak Seidov, Jan 05 2013

Status

approved