OFFSET
1,1
COMMENTS
Note that in this sequence, 30n-29, 60n-59, 90n-89 and 180n-179 do not have to be prime.
Conjecture: The number C = (30n-29)*(60n-59)*(90n-89)*(180n-179) is a Carmichael number if (but not only if) 30n-29, 60n-59, 90n-89 and 180n-179 are all four prime numbers.
The conjecture is checked for 0<n<150; the condition is satisfied just for n = 10, 52, 77, 143.
We got Carmichael numbers with more than three prime divisors for n = 2, 4, 72, 92, 95, 111.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
PROG
(PARI) K(n, c)=my(f=factor(c)); for(i=1, #f[, 1], if(f[i, 2]>1||(n-1)%(f[i, 1]-1), return(0))); 1
list(lim)=my(v=List(), C, n=1); while(n++ && (C=(30*n-29)*(60*n-59)*(90*n-89)*(180*n-179))<=lim, if(K(C, 30*n-29) && K(C, 60*n-59) && K(C, 90*n-89) && K(C, 180*n-179), listput(v, C))); Vec(v) \\ Charles R Greathouse IV, Jun 30 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius Coman, Apr 11 2012
STATUS
approved