OFFSET
1,1
COMMENTS
Note that in this sequence, 30k+7, 60k+13, and 150k+31 do not have to be prime. These numbers were found by taking the intersection of Carmichael numbers found by Pinch and numbers of the form (30k+7)*(60k+13)*(150k+31).
Conjecture: N = (30k+7)*(60k+13)*(150k+31) is a Carmichael number if (but not only if) 30k+7, 60k+13 and 150k+31 are all three prime numbers.
We checked the conjecture up to k = 256; we got Carmichael numbers with three prime divisors for k = 0, 1, 10, 12, 18, 24, 32, 43, 85, 102, 123, 129, 150, 201, 207, 256.
We got Carmichael numbers with more than three prime divisors for n = 14, 29, 109, 112.
All these numbers can be written as well as N = (n+1)*(2n+1)*(5n+1), where n = 30k+6.
The conjecture follows from Korselt's criterion. - Charles R Greathouse IV, Oct 02 2012
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Carmichael Number.
PROG
(PARI) test(lim)={
my(v=List(), n, f);
for(k=0, lim,
n=(30*k+7)*(60*k+13)*(150*k+31)-1;
f=factor(30*k+7);
for(i=1, #f[, 1], if(f[i, 2]>1 || n%(f[i, 1]-1), next(2)));
f=factor(60*k+13);
for(i=1, #f[, 1], if(f[i, 2]>1 || n%(f[i, 1]-1), next(2)));
f=factor(150*k+31);
for(i=1, #f[, 1], if(f[i, 2]>1 || n%(f[i, 1]-1), next(2)));
listput(v, n+1)
);
Vec(v)
}; \\ Charles R Greathouse IV, Oct 02 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius Coman, Apr 11 2012
EXTENSIONS
Extended and corrected by T. D. Noe, Apr 19 2012
STATUS
approved