

A326710


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.


1



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

1,2


COMMENTS

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 NagellLjunggren equation y^q = (b^m1)/(b1) 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.


LINKS



FORMULA

a(n+1) = (A158235(n))^2 for n >= 1.


EXAMPLE

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.


MATHEMATICA

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 *)


CROSSREFS

Cf. A326707 (tau(m)3)/2, this sequence (tau(m)1)/2.


KEYWORD

nonn,base


AUTHOR



STATUS

approved



