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%I #19 Jul 23 2012 10:08:24
%S 2,3,3,47,1531,4073,5081,537661,5538947,5981567,148871869,5545986967,
%T 28511128379,85185688439
%N The least prime p such that sum(k=0...m, p^(k)) is divisible by m+1 for all m < n; where p^(k) denotes the k-th next larger prime: p^(0)=p, p^(1)=nextprime(p), etc.
%H C. Rivera (Ed.), <a href="http://www.primepuzzles.net/puzzles/puzz_415.htm">Puzzle 415. Sets of consecutive primes such that...</a> on primepuzzles.net.
%e {47, 53, 59, 61} are 4 consecutive primes, 47+53 is a multiple of 2, 47+53+59 is a multiple of 3 and 47+53+59+61 is a multiple of 4. Since this is the least such set, a(4)=47.
%o (PARI) A214219(n)={ n<2 & return(2); my(p=vector(n,k,prime(k)),s=sum(k=1,n,p[k]),t);
%o for(i=0,9e9,(s += -p[i%n+1] + p[i%n+1]=nextprime(p[(i-1)%n+1]+1))%n & next; (t=s-p[i%n+1])%(n-1) & next; for(j=2,n-2, (t -= p[(i-j+1)%n+1])%(n-j) & next(2)); return(p[(i+1)%n+1]))}
%K nonn
%O 1,1
%A _M. F. Hasler_, Jul 22 2012
%E a(13)-a(14) from _Jens Kruse Andersen_, Jul 22 2012
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