OFFSET
1,1
COMMENTS
This sequence assumes prime(0) = 1.
Heuristically, we should expect a(n) -> infinity as n -> infinity, i.e. for any k there should be only finitely many terms <= k. It seems likely that 3 is the only n for which a(n) = 2. There are at least 14 occurrences of a(n) = 3, including 1, 4, 6, 8, 21, 29, 44, 84, 191, 378, 13006, 39420, 62947, 78156. - Robert Israel, Aug 14 2015
LINKS
Harvey P. Dale and Robert Israel, Table of n, a(n) for n = 1..10000(1..1000 from Harvey P. Dale)
FORMULA
MAPLE
Primes:= select(isprime, [2, (2*i+1 $ i=1..1000)]):
3, seq(max(numtheory:-factorset(Primes[i]*Primes[i+1]+1)), i=1..nops(Primes)-1); # Robert Israel, Aug 14 2015
MATHEMATICA
Join[{3}, FactorInteger[#][[-1, 1]]&/@(Times@@@Partition[Prime[ Range[ 60]], 2, 1]+1)] (* Harvey P. Dale, Apr 12 2013 *)
PROG
(PARI) gpf(n)=my(f=factor(n)[, 1]~); f[#f];
myprime(n)=if(n==0, 1, prime(n));
first(m)=vector(m, i, gpf(1+myprime(i)*myprime(i-1))); \\ Anders Hellström, Aug 13 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved