OFFSET
1,3
COMMENTS
Note the indexing: the domain starts from 1, while the range includes also zero.
For the question whether this sequence and A285332 are permutations of natural numbers, see comments in A285332 and the conjecture stated in A019565.
As a practical problem, it seems next-to-impossible to compute even the value of a(38). Even though we know that 38 certainly is not in a finite cycle of A019565, because A048675(38) = 129, A048675(129) = 8194 and A048675(8194) = 4503599627370561 which factorizes as 3^2 * 37 * 71 * 190483425427 (thus is not squarefree and A285320(38) = 3), the value of a(38) is most likely so huge that it will not fit into the data section or even into a b-file. The same problem applies to all numbers that share prime factors with 38, namely 76, 152, 304, 608, 722, ...
Terms a(39) .. a(61) are [632, 51, 8190, 60, 16382, 505, 17, 72057594037927932, 32766, 159, 29, 103, 1016, 153, 65534, 319, 50, 43, 16376, 131014, 131070, 57, 262142].
The name is slightly misleading. The given definition of a(n) is not always very helpful to compute the terms (cf. example of n = 38), it is actually not clear whether the sequence is well defined. - M. F. Hasler, Mar 01 2018
LINKS
FORMULA
EXAMPLE
a(1) = 0 and a(2) = 1 by definition.
a(3) = a(prime(2)) = a(A019565(2^1)) = 2*a(2) = 2.
a(4) = a(2^2) = a(A065642(2)) = 1 + 2*a(2) = 3.
a(5) = a(prime(3)) = a(A019565(2^2)) = 2*a(4) = 6.
a(9) = a(3^2) = a(A065642(3)) = 1 + 2*a(3) = 5.
a(10) = a(2*5) = a(prime(1)*prime(3)) = a(A019565(2^0+2^2)) = 2*a(1+4) = 12.
PROG
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Antti Karttunen, Apr 17 2017, comments edited Apr 19 2017
STATUS
approved