%I #8 Aug 21 2014 14:46:20

%S 1,1,2,2,3,3,4,4,0,0,1,1,2,2,3,3,4,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,

%U 3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,0,0,0,0,0,0,0

%N 5*n digit-reversed mod 5.

%C The pattern of increasing frequency of repetition of digits is clear.

%F Contribution from _Enrique Pérez Herrero_, Jun 14 2010: (Start)

%F a(n)=mod(floor(5*n/10^(floor(log_10(5*n)))),5), this formula comes from the modulus 5 of the first digit of 5*n.

%F a(10^n)=1

%F (End)

%e a(61) =3 as, 61*5 = 305,digit reversed = 503 ==3 (mod 5)

%t Contribution from _Enrique Pérez Herrero_, Jun 14 2010: (Start)

%t A084054[n_Integer]:=Mod[FromDigits[Reverse[IntegerDigits[5*n]]],5];

%t (* Alternative formula *)

%t A084054[n_Integer]:=Mod[Floor[5*n/10^Floor[Log[10,5*n]]],5] (End)

%Y Cf. A084052, A084053, A084055, A084339, A084340, A084341.

%K base,easy,nonn

%O 2,3

%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 26 2003

%E More terms from _Ray Chandler_, May 27 2003