A014945 Numbers k such that k divides 4^k - 1.

1, 3, 9, 21, 27, 63, 81, 147, 171, 189, 243, 441, 513, 567, 657, 729, 903, 1029, 1197, 1323, 1539, 1701, 1971, 2187, 2667, 2709, 3087, 3249, 3591, 3969, 4599, 4617, 5103, 5913, 6321, 6561, 7077, 7203, 8001, 8127, 8379, 9261, 9747, 10773, 11907, 12483

Offset

1,2

Comments

This sequence is closed under multiplication. - Charles R Greathouse IV, Nov 03 2016

Conjecture: if k divides 4^k - 1, then (4^k - 1)/k is squarefree. - Thomas Ordowski, Dec 24 2018

Following Greathouse's comment, see A323203 for the primitive terms. - Bernard Schott, Jan 03 2019

All terms except 1 are divisible by 3. Proof: suppose n>1 is in the sequence, and let p be its smallest prime factor. Of course p is odd. Since 4^n-1 is divisible by p, n is divisible by the multiplicative order of 4 mod p, which is less than p. But since n has no prime factors < p, that multiplicative order can only be 1, which means p=3. - Robert Israel, Jan 24 2019

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..872 from Muniru A Asiru, terms 873..2000 from Alois P. Heinz)

Formula

a(n) = A014741(n+1)/2.

Maple

select(n->modp(4^n-1, n)=0, [$1..13000]); # Muniru A Asiru, Dec 28 2018

Mathematica

Select[Range[12500], Divisible[4^#-1, #]&]  (* Harvey P. Dale, Mar 23 2011 *)

Prog

(PARI) is(n)=Mod(4, n)^n==1 \\ Charles R Greathouse IV, Nov 03 2016
(GAP) a:=Filtered([1..13000], n->(4^n-1) mod n=0);; Print(a); # Muniru A Asiru, Dec 28 2018
(Magma) [n: n in [1..12500] | (4^n-1) mod n eq 0 ]; // Vincenzo Librandi, Dec 29 2018
(Python)
for n in range(1, 1000):
    if (4**n-1) % n ==0:
        print(n, end=', ') # Stefano Spezia, Jan 05 2019

Crossrefs

Cf. A014741, A323203.

Sequence in context: A044055 A029542 A014962 * A045590 A369768 A242740

Adjacent sequences: A014942 A014943 A014944 * A014946 A014947 A014948

Keyword

nonn

Author

Olivier Gérard

Extensions

More terms and better description from Benoit Cloitre, Mar 05 2002

Status

approved