A111303 Numbers n such that 2^tau(n) = n + 1 (where tau(n) = number of divisors of n).

1, 3, 15, 63, 255, 65535, 4294967295

Offset

1,2

Comments

It is clear that n+1 must be a power of 2. Hence n=2^k-1 for some k. Found k=1, 2, 4, 6, 8, 16, 32. No other k<150. - T. D. Noe, Nov 04 2005

Links

Table of n, a(n) for n=1..7.

Formula

Note that this is different from the sequence A019434 - 2.

Mathematica

Select[Range[10^6], 2^DivisorSigma[0, # ] == # + 1 &]
2^Select[Range[150], DivisorSigma[0, 2^#-1]==#&] - 1 (Noe)

Prog

(Python)
from sympy import divisor_count as tau
def afind(klimit, kstart=1):
    for k in range(kstart, klimit+1):
        m = 2**k - 1
        if 2**tau(m) == m + 1: print(m, end=", ")
afind(klimit=100) # Michael S. Branicky, Dec 16 2021

Crossrefs

Cf. A046801 (number of divisors of 2^n-1), A019434.

Sequence in context: A218282 A103454 A024036 * A118339 A083858 A151241

Adjacent sequences: A111300 A111301 A111302 * A111304 A111305 A111306

Keyword

nonn,hard,more

Author

Joseph L. Pe, Nov 02 2005

Extensions

One more term from T. D. Noe, Nov 04 2005

Status

approved