|
|
A328385
|
|
If n is of the form p^p, a(n) = n, otherwise a(n) is the first number found by iterating the map x -> A003415(x) that is different from n and either a prime, or whose degree (A051903) differs from the degree of n.
|
|
3
|
|
|
0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 7, 13, 1, 44, 10, 8, 27, 32, 1, 31, 1, 80, 9, 19, 12, 96, 1, 7, 16, 68, 1, 41, 1, 48, 39, 25, 1, 608, 14, 39, 20, 56, 1, 81, 16, 92, 13, 31, 1, 96, 1, 9, 51, 640, 18, 61, 1, 72, 8, 59, 1, 156, 1, 16, 55, 80, 18, 71, 1, 3424, 108, 43, 1, 128, 13, 45, 32, 140, 1, 123, 20, 96, 19
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
LINKS
|
|
|
FORMULA
|
a(1) = 0 [as here the degrees of 0 and 1 are considered different].
a(p) = 1 for all primes.
a(n) = A003415^(k)(n), when k = abs(A328384(n)). [Taking the abs(A328384(n))-th arithmetic derivative of n gives a(n)]
|
|
EXAMPLE
|
For n = 3, 3 is a prime, thus a(3) = 1.
For n = 4, A003415(4) = 4, thus as it is among the fixed points of A003415 and a(4) = 4.
For n = 8 = 2^3, its "degree" is A051903(33) = 3, but A003415(8) = 12 = 2^2 * 3, with degree 2, thus a(8) = 12.
For n = 21 = 3*7, A051903(21) = 1, the first derivative A003415(21) = 10 = 2*5 is of the same degree as A051903(10) = 1, but then continuing, we have A003415(10) = 7, which is a prime, thus a(21) = 7.
For n = 33 = 3*11, A051903(33) = 1, A003415(33) = 14 = 2*7, is of the same degree, but on the second iteration, A003415(14) = 9 = 3^2, with A051903(9) = 2, different from the initial degree, thus a(33) = 9.
|
|
PROG
|
(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A051903(n) = if((1==n), 0, vecmax(factor(n)[, 2]));
|
|
CROSSREFS
|
Cf. A328384 (the number of iterations needed to reach such a number).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|