OFFSET
0,2
COMMENTS
The cyclic numbers are given in A003277.
a(n) = k if k = p_1*p_2*...*p_n where the p_i are distinct primes and no p_j-1 is divisible by any p_i and k is the smallest such integer.
a(n) < a(n+1) for all n. For some n, also a(n) | a(n+1). Note that the divisors of any cyclic number are cyclic. - Jeppe Stig Nielsen, May 22 2021
Are the prime factors of a(n) always a "normal sequence of primes" in the sense of A100564? Equivalently, can you always find a(n) by starting from a suitable smallest prime p_1, and then pick the subsequent prime factors greedily (under the condition that p_j - 1 is not divisible by any p_i) until you have n primes? If yes, then it is easy to calculate a(n), as all one needs to do is find the optimal starting prime. - Jeppe Stig Nielsen, May 23 2021
EXAMPLE
The prime factorizations for terms a(1)..a(12) are:
2
3, 5
3, 5, 17
3, 5, 17, 23
5, 7, 13, 17, 19
5, 7, 13, 17, 19, 23
5, 7, 13, 17, 19, 23, 37
5, 7, 13, 17, 19, 23, 37, 59
5, 7, 13, 17, 19, 23, 37, 59, 67
5, 7, 13, 17, 19, 23, 37, 59, 67, 73
7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61
7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61, 73
Corrected by Jeppe Stig Nielsen, May 22 2021.
146965 = 5*7*13*17*19 is cyclic. Since it is the smallest example with 5 primes, 146965 = a(5). It is not a multiple of a(4) = 3*5*17*23. - Jeppe Stig Nielsen, May 22 2021
PROG
(PARI) n=0; for(m=1, +oo, if(gcd(m, eulerphi(m))==1&&omega(m)==n, print1(m, ", "); n++)) \\ slow, from Jeppe Stig Nielsen, May 22 2021
(PARI) N=0; for(n=0, +oo, a=+oo; forsubset([N, n], x, m=prod(j=1, n, prime(x[j])); m<a&&gcd(m, eulerphi(m))==1&&(a=m)); print1(a, ", "); a+=(n==1); for(p=1, +oo, m=a*prime(p); if(gcd(m, eulerphi(m))==1, N=p; break()))) \\ Jeppe Stig Nielsen, May 22 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Geoffrey Critzer, Nov 28 2015
EXTENSIONS
Wrong terms a(5), a(6), a(7), a(8), a(10), a(12) corrected, and more terms added, and a(0)=1 preprended by Jeppe Stig Nielsen, May 22 2021
a(17) from Jeppe Stig Nielsen, May 22 2021
STATUS
approved