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Largest m < n such that 2^m == 2^n (mod n).
0

%I #19 Nov 08 2016 17:59:03

%S 0,1,1,3,1,4,4,7,3,6,1,10,1,11,11,15,9,12,1,16,15,12,12,22,5,14,9,25,

%T 1,26,26,31,23,26,23,30,1,20,27,36,21,36,29,34,33,35,24,46,28,30,43,

%U 40,1,36,35,53,39,30,1,56,1,57,57,63,53,56,1,60,47,58,36,66,64,38,55,58,47,66,40,76,27,62,1

%N Largest m < n such that 2^m == 2^n (mod n).

%C If n is odd, then a(n) = n - A002326((n-1)/2).

%t Table[m = n - 1; While[Mod[2^m, n] != Mod[2^n, n], m--]; m, {n, 83}] (* _Michael De Vlieger_, Oct 02 2016 *)

%o (PARI) a(n) = {if(n==0,return(0));my(pt = valuation(n, 2), odd = n/2^pt, ul = odd-A002326(odd\2)); forstep(i = n-1, ul, -1, if(Mod(2,n)^i==Mod(2,n)^n,return(i)))} \\ _David A. Corneth_, Oct 01 2016

%o A002326(n)=if(n<0, 0, znorder(Mod(2, 2*n+1)))

%Y Cf. A002326, A015910, A270096.

%K nonn

%O 1,4

%A _Thomas Ordowski_, Oct 01 2016

%E More terms from _Altug Alkan_, Oct 01 2016