A051464 Number of divisors of 4*(2^n-1) + 1.

2, 2, 2, 2, 4, 4, 2, 2, 4, 2, 4, 2, 4, 6, 4, 4, 16, 2, 4, 2, 4, 2, 4, 8, 12, 8, 2, 4, 8, 4, 4, 4, 4, 4, 4, 8, 8, 4, 8, 16, 8, 4, 8, 8, 6, 16, 8, 8, 8, 16, 8, 4, 32, 32, 8, 4, 8, 4, 4, 8, 16, 8, 8, 16, 48, 16, 16, 8, 4, 16, 4, 16, 16, 8, 8, 8, 16, 16, 8, 16, 32

Offset: 1

Author

Edwin D. Evans, eevans2(AT)pacbell.net

Keywords

• nonn

Comments

Create a table with tau(2^n-1) as the first row (A046801) and tau(m) as the first column (A000005). The second column is tau(A004760) and so on. Rows 2, 3 and 4 are easily described in terms of row 1. This sequence is row 5.

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..591

Crossrefs

Cf. A000005, A036563.

Programs

Mathematica

Array[DivisorSigma[0, 4*(2^# - 1) + 1] &, 81] (* Michael De Vlieger, Sep 15 2021 *)

PARI

a(n) = numdiv(4*(2^n-1) + 1); \\ Michel Marcus, Sep 16 2021

Formula

a(n) = tau(4*(2^n -1)+1), where d(n) = A000005(n).

Extensions

a(81) corrected by Sean A. Irvine, Sep 15 2021