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A293220
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Sum of all partial fractions in the algorithm used for calculation of A053446(n).
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2
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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
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OFFSET
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1,4
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COMMENTS
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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.
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LINKS
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FORMULA
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Let n = 12. According to the comment, a(12) = 2 + 4 + 7 + 8 + 14 + 16 + 11 + 5 + 13 + 10 + 1 = 91.
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CROSSREFS
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Cf. A038754 (seems to give the positions of ones).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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