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%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
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