

A014945


Numbers k such that k divides 4^k  1.


27



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
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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^n1 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^n1, 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^n1) mod n=0);; Print(a); # Muniru A Asiru, Dec 28 2018
(Magma) [n: n in [1..12500]  (4^n1) mod n eq 0 ]; // Vincenzo Librandi, Dec 29 2018
(Python)
for n in range(1, 1000):
if (4**n1) % n ==0:
print(n, end=', ') # Stefano Spezia, Jan 05 2019


CROSSREFS

Cf. A014741, A323203.
Sequence in context: A044055 A029542 A014962 * A045590 A242740 A029536
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



