A293220 Sum of all partial fractions in the algorithm used for calculation of A053446(n).

1, 1, 1, 7, 12, 1, 8, 10, 1, 30, 12, 91, 108, 8, 6, 44, 157, 1, 45, 271, 300, 73, 164, 91, 162, 234, 1, 125, 588, 122, 175, 225, 684, 368, 65, 919, 373, 45, 512, 443, 206, 630, 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

Offset: 1

Author

Vladimir Shevelev and Antti Karttunen, Oct 08 2017

Keywords

• nonn

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 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

Crossrefs

Cf. A001651, A053446, A292270, A293445, A293446.

Cf. A038754 (seems to give the positions of ones).

Formula

Let n = 12. According to the comment, a(12) = 2 + 4 + 7 + 8 + 14 + 16 + 11 + 5 + 13 + 10 + 1 = 91.