A204983 a(n) = 2^(k-1)-2^(j-1), where (2^(k-1),2^(j-1)) is the least pair of distinct positive powers of 2 for which n divides 2^(k-1)-2^(j-1).

1, 2, 3, 4, 15, 6, 7, 8, 63, 30, 1023, 12, 4095, 14, 15, 16, 255, 126, 262143, 60, 63, 2046, 2047, 24, 1048575, 8190, 262143, 28, 268435455, 30, 31, 32, 1023, 510, 4095, 252, 68719476735, 524286, 4095, 120, 1048575, 126, 16383, 4092, 4095

Offset: 1

Author

Clark Kimberling, Jan 21 2012

Keywords

• nonn

Comments

For a guide to related sequences, see A204892.

(Conjecture) Equivalently, the solution set of 2^p * (2^q - 1) = x * y, OR 2^q - 1 = 2^p * x * y, for at most one of the naturals x and y being given; unknown p and q in the integers; then a(n) = 2^p * (2^q - 1) where p and q are directly related to n (see formula). - Andrew T. Porter, Dec 20 2022

Crossrefs

Cf. A204979, A204892, A204984.

Cf. A000079, A173787.

Cf. A007733, A007814.

Programs

Mathematica

(See the program at A204979.)

PARI

a(n) = for (k=1, oo, for (j=1, k-1, my(d=2^(k-1)-2^(j-1)); if (!(d % n), return(d)););); \\ Michel Marcus, Sep 16 2023

Formula

Conjecture: a(n) = 2^A007814(n) * (2^A007733(n) - 1). - Andrew T. Porter, Dec 20 2022