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A264907 a(n) is the smallest "cyclic" integer, k, that has exactly n prime factors.

%I A264907 #27 May 24 2021 00:53:01
%S A264907 1,2,15,255,5865,146965,3380195,125067215,7378965685,494390700895,
%T A264907 36090521165335,2571956263189313,187752807212819849,
%U A264907 18212022299643525353,1839414252263996060653,196817324992247578489871,21453088424154986055395939,2981979290957543061700035521
%N A264907 a(n) is the smallest "cyclic" integer, k, that has exactly n prime factors.
%C A264907 The cyclic numbers are given in A003277.
%C A264907 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.
%C A264907 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
%C A264907 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
%e A264907 The prime factorizations for terms a(1)..a(12) are:
%e A264907 2
%e A264907 3, 5
%e A264907 3, 5, 17
%e A264907 3, 5, 17, 23
%e A264907 5, 7, 13, 17, 19
%e A264907 5, 7, 13, 17, 19, 23
%e A264907 5, 7, 13, 17, 19, 23, 37
%e A264907 5, 7, 13, 17, 19, 23, 37, 59
%e A264907 5, 7, 13, 17, 19, 23, 37, 59, 67
%e A264907 5, 7, 13, 17, 19, 23, 37, 59, 67, 73
%e A264907 7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61
%e A264907 7, 11, 13, 17, 19, 31, 37, 41, 47, 59, 61, 73
%e A264907 Corrected by _Jeppe Stig Nielsen_, May 22 2021.
%e A264907 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
%o A264907 (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
%o A264907 (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
%Y A264907 Cf. A003277, A100564.
%K A264907 nonn
%O A264907 0,2
%A A264907 _Geoffrey Critzer_, Nov 28 2015
%E A264907 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
%E A264907 a(17) from _Jeppe Stig Nielsen_, May 22 2021