A117707 Largest number of previous terms having at least one common digit in decimal representation.

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, 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, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 22, 22

Offset: 0

Author

Reinhard Zumkeller, Apr 13 2006

Keywords

• nonn
• base

Comments

a(n) = A003056(n) for n <= 63.

Examples

			n=64: occurrences of digit d in the first 63 terms:
d=0:#{a(0),a(55),a(56),a(57),a(58),a(59),a(60),a(61),a(62),a(63)}=10,
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,
d=2:#{a(3),a(4),a(5)}=3,
d=3:#{a(6),a(7),a(8),a(9)}=4,
d=4:#{a(10),a(11),a(12),a(13),a(14)}=5,
d=5:#{a(15),a(16),a(17),a(18),a(19),a(20)}=6,
d=6:#{a(21),a(22),a(23),a(24),a(25),a(26),a(27)}=7,
d=7:#{a(28),a(29),a(30),a(31),a(32),a(33),a(34),a(35)}=8,
d=8:#{a(36),a(37),a(38),a(39),a(40),a(41),a(42),a(43),a(44)}=9,
d=9:#{a(45),a(46),a(47),a(48),a(49),a(50),a(51),a(52),a(53),a(54)}=10,
therefore a(64) = max{10, 11, 3, 4, 5, 6, 7, 8, 9, 10} = 11.