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A326382
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Numbers m such that beta(m) = tau(m)/2 + 2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
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8
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32767, 65535, 67053, 2097151, 4381419, 7174453, 9808617, 13938267, 14348906, 19617234, 21523360, 29425851, 39234468, 43046720, 48686547, 49043085, 58851702, 61035156, 68660319, 71270178, 78468936, 88277553, 98086170, 107894787, 115174101, 117703404, 134217727, 142540356, 175965517
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OFFSET
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1,1
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COMMENTS
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As tau(m) = 2 * (beta(m) - 2) , the terms of this sequence are not squares.
There are 2 subsequences which realize a partition of this sequence (see array in link and examples):
1) Non-oblong composites which have exactly three Brazilian representations with three digits or more, they are in A326389.
2) Oblong numbers that have exactly four Brazilian representations with three digits or more. These integers have been found through b-file of Rémy Sigrist in A290869. These oblong integers are a subsequence of A309062.
There are no primes that satisfy this relation.
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LINKS
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EXAMPLE
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One example for each type:
1) The divisors of 32767 are {1, 7, 31, 151, 217, 1057, 4681, 32767} and tau(32767) = 8; also, 32767 = M_15 = R(15)_2 = 77777_8 = (31,31,31)_32 = (151,151)_216 = (31,31)_1056 = 77_4680 so beta(32767) = 6 with beta'(32767) = 3 and beta"(32767)= 3. The relation is beta(32767) = tau(32767)/2 + 2 = 6.
2) 61035156 = 7812 * 7813 is oblong with tau(61035156) = 144. The four Brazilian representations with three digits or more are 61035156 = R(12)_5 = 666666_25 = (31,31,31,31)_125 = (156,156,156)_625, so beta"(61035156) = 4 and beta(61035156) = tau(61035156)/2 + 2 = 74.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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