cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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
Offset: 1

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Author

Bernard Schott, Sep 14 2019

Keywords

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 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.

Examples

			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.
		

Crossrefs

Cf. A326707 (tau(m)-3)/2, this sequence (tau(m)-1)/2.
Subsequence of A000290.

Programs

  • 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 *)

Formula

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