|
%I #22 Oct 30 2018 10:31:02
%S 3,7,251,61223,23700022897
%N Smallest prime q such that, starting with q, there are prime(n)-1 consecutive primes = {1..prime(n)-1} modulo prime(n).
%C Heuristically a(n) is around (p-1)^p log 2, where p is the n-th prime. - _Charles R Greathouse IV_, Jan 11 2013
%e Let p(n) = prime(n), then
%e a(1)=3 because 3 is smallest prime = ( 1 modulo p(1)) = 1 mod 2;
%e a(2)=7 because 2 smallest consecutive primes {7,11}= {1,2} modulo p(2) = {1,2} mod 3;
%e a(3) = 251 because {251,257,263,269} = {1,2,3,4} modulo p(3)= {1,2,3,4} mod 5;
%e 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;
%e 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
%t 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 *)
%o (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
%K hard,more,nonn
%O 1,1
%A _Zak Seidov_, Feb 06 2012
%E a(5) from _Zak Seidov_, Jan 05 2013
|
