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A253595
Least Carmichael number that is divisible by the n-th cyclic number A003277(n), or 0 if no such number exists.
2
561, 1105, 1729, 561, 1105, 62745, 561, 1729, 6601, 2465, 2821, 561, 825265, 29341, 6601, 334153, 62745, 561, 2433601, 74165065, 29341, 1105, 8911, 116150434401, 10024561, 10585, 41041, 2508013, 55462177, 1105, 11921001
OFFSET
3,1
COMMENTS
Has any odd cyclic number at least one Carmichael multiple?
LINKS
Amiram Eldar, Table of n, a(n) for n = 3..2747 (calculated using data from Claude Goutier; terms 3..291 from Tim Johannes Ohrtmann, terms 292..853 from Max Alekseyev)
Open Problem Garden, Michon's conjecture.
EXAMPLE
a(8) = 62745 because this is the least Carmichael number which is divisible by 15 (the 8th cyclic 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
isA002997(n)=n%2 && !isprime(n) && Korselt(n) && n>1
a(n) = {on = odd cyclic number(n); cn = 1; until (isA002997(cn) && (cn % on == 0), cn++); cn; }
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
a(292)-a(853) from Max Alekseyev, Apr 26 2015
Escape clause added by Jianing Song, Dec 12 2021
STATUS
approved