A014945 - OEIS

%I #51 Sep 08 2022 08:44:39

%S 1,3,9,21,27,63,81,147,171,189,243,441,513,567,657,729,903,1029,1197,

%T 1323,1539,1701,1971,2187,2667,2709,3087,3249,3591,3969,4599,4617,

%U 5103,5913,6321,6561,7077,7203,8001,8127,8379,9261,9747,10773,11907,12483

%N Numbers k such that k divides 4^k - 1.

%C This sequence is closed under multiplication. - _Charles R Greathouse IV_, Nov 03 2016

%C Conjecture: if k divides 4^k - 1, then (4^k - 1)/k is squarefree. - _Thomas Ordowski_, Dec 24 2018

%C Following Greathouse's comment, see A323203 for the primitive terms. - _Bernard Schott_, Jan 03 2019

%C 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

%H Amiram Eldar, <a href="/A014945/b014945.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..872 from Muniru A Asiru, terms 873..2000 from Alois P. Heinz)

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

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

%t Select[Range[12500],Divisible[4^#-1,#]&]  (* _Harvey P. Dale_, Mar 23 2011 *)

%o (PARI) is(n)=Mod(4,n)^n==1 \\ _Charles R Greathouse IV_, Nov 03 2016

%o (GAP) a:=Filtered([1..13000],n->(4^n-1) mod n=0);; Print(a); # _Muniru A Asiru_, Dec 28 2018

%o (Magma) [n: n in [1..12500] | (4^n-1) mod n eq 0 ]; // _Vincenzo Librandi_, Dec 29 2018

%o (Python)

%o for n in range(1,1000):

%o     if (4**n-1) % n ==0:

%o         print(n, end=', ') # _Stefano Spezia_, Jan 05 2019

%Y Cf. A014741, A323203.

%K nonn

%O 1,2

%A _Olivier Gérard_

%E More terms and better description from _Benoit Cloitre_, Mar 05 2002