%I #18 Nov 20 2024 04:18:35

%S 1,1,2,3,3,5,6,1,4,9,3,55,90,9,14,5,30,1,18,3,10,21,6,161,84,2,130,45,

%T 9,29,30,3,2,33,11,35,90,15,5,351,27,82,28,7,22,15,90,3,120,3,50,51,6,

%U 53,18,9,154,33,12,11,110,25,50,7,7,195,18,9,34,69

%N Smallest m>0 such that (2*n)^2 - 1 divides (2^m)^(2*n) - 1.

%H Giovanni Resta and Chai Wah Wu, <a href="/A281363/b281363.txt">Table of n, a(n) for n = 1..10000</a> (terms for n = 1..1000 from Giovanni Resta)

%e a(3) = 2 because (2*3)^2 - 1 = 35 divides (2^2)^(2*3) - 1 = 4095.

%t Table[SelectFirst[Range@ 1200, Divisible[(2^#)^(2 n) - 1, (2 n)^2 - 1] &], {n, 84}] (* _Michael De Vlieger_, May 01 2016, Version 10 *)

%t a[n_] := Block[{m=1}, While[ PowerMod[2^m, 2*n, 4*n^2-1] != 1, m++]; m]; Array[a, 100] (* _Giovanni Resta_, May 05 2016 *)

%o (Python)

%o def A281363(n):

%o m, q = 1, 4*n**2-1

%o p = pow(2, 2*n, q)

%o r = p

%o while r != 1:

%o m += 1

%o r = (r*p) % q

%o return m # _Chai Wah Wu_, Jan 28 2017

%Y Cf. positive numbers n such that n^2 - 1 divides (2^k)^n - 1: A247219 (k=1), A271842 (k=2), A272062 (k=3).

%K nonn,changed

%O 1,3

%A _Juri-Stepan Gerasimov_, Apr 30 2016