%I
%S 1,1,1,7,12,1,8,10,1,30,12,91,108,8,6,44,157,1,45,271,300,73,164,91,
%T 162,234,1,125,588,122,175,225,684,368,65,919,373,45,512,443,206,630,
%U 300,196,506,213,118,550,303,510,459,679,2028,1,208,941,286,1218,201,2611,62,691,751,724,1575,1374,540,3367,1004,36
%N Sum of all partial fractions in the algorithm used for calculation of A053446(n).
%C 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 n1. 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_2m_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 <= n1. For example, for n=12, A001651(12) = 17, we have exactly n1 = 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.
%H Antti Karttunen, <a href="/A293220/b293220.txt">Table of n, a(n) for n = 1..13122</a>
%H Antti Karttunen, <a href="/A293220/a293220.txt">Schemeprogram for computing this sequence and A053446</a>
%F Let n = 12. According to the comment, a(12) = 2 + 4 + 7 + 8 + 14 + 16 + 11 + 5 + 13 + 10 + 1 = 91.
%Y Cf. A001651, A053446, A292270, A293445, A293446.
%Y Cf. A038754 (seems to give the positions of ones).
%K nonn
%O 1,4
%A _Vladimir Shevelev_ & _Antti Karttunen_, Oct 08 2017
