A281363 Smallest m>0 such that (2*n)^2 - 1 divides (2^m)^(2*n) - 1.
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, 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, 53, 18, 9, 154, 33, 12, 11, 110, 25, 50, 7, 7, 195, 18, 9, 34, 69
Offset: 1
Keywords
Examples
a(3) = 2 because (2*3)^2 - 1 = 35 divides (2^2)^(2*3) - 1 = 4095.
Links
- Giovanni Resta and Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms for n = 1..1000 from Giovanni Resta)
Crossrefs
Programs
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Mathematica
Table[SelectFirst[Range@ 1200, Divisible[(2^#)^(2 n) - 1, (2 n)^2 - 1] &], {n, 84}] (* Michael De Vlieger, May 01 2016, Version 10 *) 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 *) -
Python
def A281363(n): m, q = 1, 4*n**2-1 p = pow(2, 2*n, q) r = p while r != 1: m += 1 r = (r*p) % q return m # Chai Wah Wu, Jan 28 2017