A064084 A multiplicative version of 2^n - 1 (A000225).

1, 3, 7, 15, 31, 21, 127, 255, 511, 93, 2047, 105, 8191, 381, 217, 65535, 131071, 1533, 524287, 465, 889, 6141, 8388607, 1785, 33554431, 24573, 134217727, 1905, 536870911, 651, 2147483647, 4294967295, 14329, 393213, 3937, 7665, 137438953471, 1572861, 57337

Offset: 1

Author

Jens Voß, Sep 04 2001

Keywords

• mult
• easy
• nonn

Comments

Since n -> 2^n - 1 is an embedding of the ordered structure N = {1, 2, 3, ...} (the order being the "divides" relation) into itself, a(n) always divides A000225(n); the sequence of quotients of A000225 and a is A064085.

Examples

			a(6) = (2^2 - 1) * (2^3 - 1) = 21 since 6 = 2 * 3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000225, A064085, A064086.

Programs

Maple

a:= n-> mul(2^(i[1]^i[2])-1, i=ifactors(n)[2]):
seq(a(n), n=1..50);  # Alois P. Heinz, Jun 09 2014

Mathematica

a[n_] := Times @@ (2^(Power @@@ FactorInteger[n]) - 1); Array[a, 40] (* Amiram Eldar, Aug 31 2023 *)

PARI

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,1] = 2^(f[i,1]^f[i,2])-1; f[i,2]=1); factorback(f); \\ Michel Marcus, Jun 09 2014

Formula

a(n) = (2^((p_1)^(e_1)) - 1) * ... * (2^((p_k)^(e_k)) - 1) where (p_1)^(e_1) * ... * (p_k)^(e_k) is the prime factorization of n.

Extensions

More terms from Michel Marcus, Jun 09 2014