|
| |
|
|
A153212
|
|
A permutation of the integers >= 2 which is constructed by swapping each prime's index difference (from the previous prime) and the prime's power in the factorization of a number. In order the "index difference" to make sense the we consider the factorization to be sorted with respect to the primes but not the powers they are raised to, that is, first comes the smallest prime and each subsequent prime is larger than the previous disregarding their powers.
|
|
0
| |
|
|
2, 4, 3, 8, 6, 16, 5, 9, 18, 32, 15, 64, 54, 12, 7, 128, 10
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 2,1
|
|
|
COMMENTS
| For every n it is true that a(a(n)) = n
|
|
|
FORMULA
| denote the i-th prime with p(i): p(1)=2, p(2)=3, p(3)=5, p(4)=7, etc. let n = p(a1)^b1 * p(a2)^b2 * ... * p(ak)^bk is the factorization of n where p(i)^j is the i-th prime raised to power j. As mentioned above, we assume that the primes are sorted, i.e. a1 < a2 < a3... < ak. Then a(n) = p(c1)^d1 * p(c2)^d2 * ... * p(ck)^dk where c1 = b1 and c(i) = b(i) + c(i-1) for i > 1 d1 = a1 and d(i) = a(i) - a(i-1) for i > 1
|
|
|
EXAMPLE
| for n = 10 we have 10 = 2^1 * 5^1 = p(1)^1 * p(3)^1 then a(10) = p(1)^1 * p(2)^2 = 2^1 * 3^2 = 18 for n = 18 we have 18 = 2^1 * 3^2 = p(1)^1 * p(2)^2 then a(18) = p(1)^1 * p(3)^1 = 2^1 * 5^1 = 10
|
|
|
CROSSREFS
| Sequence in context: A048672 A127301 A122111 * A124833 A181815 A168521
Adjacent sequences: A153209 A153210 A153211 * A153213 A153214 A153215
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Luchezar Belev (l_belev(AT)yahoo.com), Dec 20 2008
|
| |
|
|
