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A326710
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Squares m such that beta(m) = (tau(m) - 1)/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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1
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1, 121, 400, 1521, 1600, 2401, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361, 47155689
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OFFSET
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1,2
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COMMENTS
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As tau(m) = 2 * beta(m) + 1 is odd, the terms of this sequence are squares.
There are 3 classes of terms in this sequence (see examples):
1) The singleton {1} with 1^2 = 1.
2) The singleton {121}. Indeed, 121 is the only known square of prime that is Brazilian because 121 is a solution y^q of the Nagell-Ljunggren equation y^q = (b^m-1)/(b-1) with y = 11, q =2, b = 3, m = 5 (see A208242).
3) Squares of composites which have one Brazilian representation with three digits or more. These integers form A326711. We don't know if there exist squares of composites which have two or more Brazilian representations with three digits or more, consequently, there is no sequence with beta(m) = (tau(m) + k)/2, with k odd >= 1.
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LINKS
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FORMULA
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a(n+1) = (A158235(n))^2 for n >= 1.
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EXAMPLE
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One example for each type:
1) 1 is not Brazilian, tau(1) = 1 and beta(1) = (tau(1) - 1)/2 = 0.
2) 121 = 11^2 = 11111_3, tau(121) = 3 and beta(121) = (tau(121) - 1)/2 = 1.
3) 1521 = 39^2 = 333_22 = (13,13)_116 = 99_168 = 33_506. The divisors of 1521 are {1, 3, 9, 13, 39, 117, 169, 507, 1521} so tau(1521) = 9 and beta(1521) = (tau(1521) - 1)/2 = 4.
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MATHEMATICA
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brazQ[n_, b_] := Length@Union@IntegerDigits[n, b] == 1; beta[n_] := Sum[Boole @ brazQ[n, b], {b, 2, n - 2}]; aQ[n_] := beta[n] == (DivisorSigma[0, n] - 1)/2; Select[Range[6867]^2, aQ] (* Amiram Eldar, Sep 14 2019 *)
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CROSSREFS
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Cf. A326707 (tau(m)-3)/2, this sequence (tau(m)-1)/2.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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