OFFSET
1,3
COMMENTS
It is known that there is a finite number of Carmichael numbers with k prime factors if k-2 of the factors are fixed. Here we consider the case k=3 and impose the additional condition that prime(n) be the smallest of the 3 factors.
The primes related to the zeros in this sequence are in A051663. - Jack Brennen, Jul 01 2008
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from M. F. Hasler)
FORMULA
a(n) = # { pqr | p=prime(n) < q=prime(n') < r=prime(n") ; p-1 | pqr-1 ; q-1 | pqr-1 ; r-1 | pqr-1 }
EXAMPLE
a(1)=0 since prime(1)=2 and there is no even Carmichael number.
a(2)=1 since prime(2)=3 and 561 is the only Carmichael number of the form 3pq with p,q prime.
a(3)=3 since prime(3)=5 and the only Carmichael numbers of the form 5pq are {1105, 2465, 10585}.
PROG
(PARI) A141703(n, verbose=0) = { /* based on code by J.Brennen (jb AT brennen.net) */ local( V=[], B, p=prime(n), q, r); for( A=1, p-1, B=ceil((p^2+1)/A); while( 1, r=(p*B-p+A*B-B)/(A*B-p*p); q=(A*r-A+1)/p; q<=p && break; denominator(q)==1 && denominator(r)==1 && r>q && isprime(q) && isprime(r) && (p*q*r)%(p-1)==1 && V=concat(V, [p*q*r]); B++ )); verbose && print1(V); #V }
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jul 01 2008
STATUS
approved