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 A206333 Smallest prime q such that, starting with q, there are prime(n)-1 consecutive primes = {1..prime(n)-1} modulo prime(n). 2
 3, 7, 251, 61223, 23700022897 (list; graph; refs; listen; history; text; internal format)
 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 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: A057619 A179859 A061422 * A088097 A064774 A089689 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

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Last modified May 6 21:35 EDT 2021. Contains 343597 sequences. (Running on oeis4.)