OFFSET
1,4
COMMENTS
This sequence gives an additional insight (cf. A292270) into the algorithm for the calculation of A053446(m), where m=A001651(n). Let us estimate how many steps are required before (the first) 1 will appear. Note that all partial fractions (which are indeed, integers) are residues modulo A001651(n) not divisible by 3 from the interval [1, A001651(n)-1]. So, if there is no a repetition, then the number of steps does not exceed n-1. Suppose then that there is a repetition before the appearance of 1. Then for a not divisible by 3 residue k from [1, A001651(n)-1], 3^m_1 == 3^m_2 == k (mod A001651(n)) such that m_2 > m_1. But then 3^(m_2-m_1) == 1 (mod A001651(n)). So, since m_2 - m_1 < m_2, it means that 1 should appear earlier than the repetition of k, which is a contradiction. So the number of steps <= n-1. For example, for n=12, A001651(12) = 17, we have exactly n-1 = 11 steps with all other not divisible by 3 residues <= 17 - 1 = 16 modulo 17 appearing before the final 1: 2, 4, 7, 8, 14, 16, 11, 5, 13, 10 , 1.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..13122
Antti Karttunen, Scheme-program for computing this sequence and A053446
FORMULA
Let n = 12. According to the comment, a(12) = 2 + 4 + 7 + 8 + 14 + 16 + 11 + 5 + 13 + 10 + 1 = 91.
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev & Antti Karttunen, Oct 08 2017
STATUS
approved