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A347245
Number of iterations of the map x -> A000593(x), when starting from x = n, needed to reach a number whose largest prime factor is at least as large as that of n itself (= A006530(n)). If 1 is reached without encountering such a number, then a(n) = 0.
6
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1
OFFSET
1,55
COMMENTS
For any hypothetical odd perfect number x, a(x) = 1.
The first occurrence of k = 0 .. 5 is at n = 0, 9, 55, 935, 102753, 262205.
FORMULA
For n >= 1, a(n) = 0 iff A347244(n) = 0.
For n > 1, a(n) = 1 iff A347246(n) = 1.
EXAMPLE
For n = 55 = 5*11, on the first iteration we get A000593(55) = 72 = 2^3 * 3^2, but both 2 and 3 are less than 11, thus on the second iteration, we get A000593(72) = 13, which is the first time when the largest prime factor is larger than that of 55 (13 > 11), thus a(55) = 2.
PROG
(PARI)
A000265(n) = (n >> valuation(n, 2));
A000593(n) = sigma(A000265(n));
A006530(n) = if(1==n, n, my(f=factor(n)); f[#f~, 1]);
A347245(n) = { my(gpf=A006530(n), k=0); while(n>1, n = A000593(n); k++; if(A006530(n)>=gpf, return(k))); (0); };
CROSSREFS
Cf. A000593, A006530, A336361, A347240, A347242 (positions of terms > 0), A347243 (positions of zeros), A347244, A347246.
Sequence in context: A327170 A024362 A370256 * A104488 A244413 A318655
KEYWORD
nonn
AUTHOR
Antti Karttunen, Aug 28 2021
STATUS
approved