%I #37 Nov 04 2017 23:26:46
%S 0,1,1,2,1,2,1,6,2,2,1,36,1,2,2,24,1,36,1,36,2,2,1,1440,2,2,6,36,1,
%T 348,1,120,2,2,2,10560,1,2,2,1440,1,348,1,36,36,2,1,100800,2,36,2,36,
%U 1,1440,2,1440,2,2,1,2218560,1,2,36,720,2,348,1,36,2,348,1,9737280,1,2,36,36,2,348,1,100800,24,2,1,2218560,2,2,2,1440,1,2218560,2,36,2,2,2,10886400,1,36,36,10560
%N Number of permutations of the divisors of n that are greater than 1, in which consecutive elements are not coprime.
%C a(n) depends only on prime signature of n (cf. A025487). So a(60) = a(90) since 60 = 2^2 * 3 * 5 and 90 = 2 * 3^2 * 5 both have prime signature (2,1,1). - _Antti Karttunen_, Oct 22 2017
%C As a consequence of the comment above, a(n) = a(A046523(n)). - _David A. Corneth_, Oct 22 2017
%H David A. Corneth, <a href="/A163820/b163820.txt">Table of n, a(n) for n = 1..359</a> (first 119 terms from Antti Karttunen)
%H Antti Karttunen, <a href="/A163820/a163820.txt">Scheme-program for computing this sequence</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F a(p) = 1 for all primes p. a(p*q) = 2 for all pairs of (not necessarily distinct) primes p and q.
%F From _Antti Karttunen_, Oct 22 2017: (Start)
%F a(p^n) = A000142(n), for all primes p.
%F a(n) = A293900(n)*A293902(n).
%F (End)
%e The divisors of 12 that are > 1 are 2,3,4,6,12. In the permutations that are counted, 3 cannot be next to 2 or 4. However, a permutation that is among those counted is 6,2,4,12,3. The GCDs of adjacent pairs in this permutation are gcd(6,2)=2, gcd(2,4)=2, gcd(4,12)=4, gcd(12,3)=3. Note that all of these GCDs are > 1.
%t Array[Count[Permutations@ Rest@ Divisors[#], _?(NoneTrue[Partition[#, 2, 1], CoprimeQ @@ # &] &)] - Boole[# == 1] &, 59] (* _Michael De Vlieger_, Nov 04 2017 *)
%Y Cf. also A046523, A114717, A119842, A122977, A122978, A293900, A293902.
%K nonn
%O 1,4
%A _Leroy Quet_, Aug 04 2009
%E Definition corrected by _Leroy Quet_, Aug 15 2009
%E Edited and extended by _Max Alekseyev_, Jun 13 2011