OFFSET
1,1
COMMENTS
Conjecture: Any Carmichael number C divisible by p and 2p-1 (where p and 2p-1 are prime numbers) can be written as C = p*(2p-1)*(n*(2p-2)+p).
Checked for the first 30 Carmichael numbers divisible by p and 2p-1.
Note: we can see how easy is to obtain Carmichael numbers with this formula:
for n = 1 we get p*(2p-1)*(3p-2) and Carmichael numbers 1729, 172081, 294409 etc.
for n = 2 we get p*(2p-1)*(5p-4) and Carmichael numbers 2821, 63973, 488881 etc.
for n = 3 we get p*(2p-1)*(7p-6) and Carmichael numbers 399001, 53711113 etc.
LINKS
E. W. Weisstein, Carmichael Number
PROG
(PARI) Korselt(n)=my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]>1||(n-1)%(f[i, 1]-1), return(0))); 1
list(lim)=my(v=List(), q, t); forprime(p=3, round(solve(x=1, lim, 6*x^3-7*x^2+2*x-lim)), for(n=1, (lim\(2*p^2-p)-p)\(2*p-2), if(isprime(q=2*p-1)&&Korselt(t=p*q*(n*q-n+p)), listput(v, t)))); vecsort(Vec(v), , 8)
\\ Charles R Greathouse IV, Oct 02 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius Coman, Apr 18 2012
STATUS
approved