OFFSET
1,1
COMMENTS
Square of terms of A053182 are in this sequence. - Michel Marcus, May 08 2014
From Amiram Eldar, Nov 08 2024: (Start)
The possible values of the number of even divisors of even terms of this sequence is restricted by the number of odd divisors.
Let k be a term and d_odd(k) = A001227(k) and d_even(k) = A183063 be its number of odd divisors and number of even divisors, respectively. When k is even, in a valid permutation of its divisors there must be two even divisors between two odd divisors, at most 2 before the first odd divisor, and at most 2 after the last odd divisor.
Therefore, d_even(k) - 2*(d_odd(k) - 1) <= 4. Let d(k) = A000005(k) = d_odd(k) + d_even(k), and let e = A007814(k) and m = A000265(k). Then, k = 2^e * m, d(k) = (e+1) * d(m) = (e+1) * d_odd(k), so d_even(k) = e * d_odd(k), and |e-2| * d_odd(k) <= 2.
If m = 1, then d_odd(k) = 1 and e <= 4, so 16 = 2^4 is the largest power of 2 in this sequence.
If m = p is a prime, then d_odd(k) = 2 and e <= 3, and therefore only terms of the form 2*p, 4*p or 8*p are possible. 2*p is a term if and only if p is a term of A106067.
If m is composite, then d_odd(k) > 2 and e <= 2, and therefore k is not divisible by 8. (End)
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..74
FORMULA
A096527(a(n)) > 0.
EXAMPLE
Divisors of 24 are {1,2,3,4,6,8,12,24}: [2,8,3,12,4,1,24,6] -> (2+8+3,8+3+12,3+12+4,12+4+1,4+1+24,1+24+6) = (13,23,19,17,29,31): therefore 24 is a term.
PROG
(PARI) isok(p) = {my(n = #p); if(n < 3, return(0)); for(k = 1, n-2, if(!isprime(p[k]+p[k+1]+p[k+2]), return(0))); 1; }
is2(n) = {my(d = divisors(n)); forperm(d, p, if(isok(p), return(1))); 0; }
is1(k) = {my(e = valuation(k, 2), o = k >> e); (e == 0) || (o == 1 && e <= 4) || (abs(e-2) * numdiv(o) <= 2); }
is(k) = is1(k) && is2(k); \\ Amiram Eldar, Nov 08 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Jun 23 2004
EXTENSIONS
a(30)-a(51) from Michel Marcus, May 03 2014
a(52) onwards from Amiram Eldar, Nov 08 2024
STATUS
approved