%I #5 Mar 30 2012 18:50:54
%S 0,1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,6,7,7,7,7,7,7,
%T 7,7,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,
%U 11,12,13,14,15,16,17,18,19,20,20,20,20,20,20,20,20,20,20,20,21,21,22,22
%N Largest number of previous terms having at least one common digit in decimal representation.
%C a(n) = A003056(n) for n <= 63.
%e n=64: occurrences of digit d in the first 63 terms:
%e d=0:#{a(0),a(55),a(56),a(57),a(58),a(59),a(60),a(61),a(62),a(63)}=10,
%e d=1:#{a(1),a(2),a(55),a(56),a(57),a(58),a(59),a(60),a(61),a(62),a(63)}=11,
%e d=2:#{a(3),a(4),a(5)}=3,
%e d=3:#{a(6),a(7),a(8),a(9)}=4,
%e d=4:#{a(10),a(11),a(12),a(13),a(14)}=5,
%e d=5:#{a(15),a(16),a(17),a(18),a(19),a(20)}=6,
%e d=6:#{a(21),a(22),a(23),a(24),a(25),a(26),a(27)}=7,
%e d=7:#{a(28),a(29),a(30),a(31),a(32),a(33),a(34),a(35)}=8,
%e d=8:#{a(36),a(37),a(38),a(39),a(40),a(41),a(42),a(43),a(44)}=9,
%e d=9:#{a(45),a(46),a(47),a(48),a(49),a(50),a(51),a(52),a(53),a(54)}=10,
%e therefore a(64) = max{10, 11, 3, 4, 5, 6, 7, 8, 9, 10} = 11.
%K nonn,base
%O 0,4
%A _Reinhard Zumkeller_, Apr 13 2006