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.

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.

Links

Antti Karttunen, Table of n, a(n) for n = 1..102753

Index entries for sequences related to sigma(n)

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

Adjacent sequences: A347242 A347243 A347244 * A347246 A347247 A347248

Keyword

nonn

Author

Antti Karttunen, Aug 28 2021

Status

approved