A249070 a(n+1) gives the number of occurrences of the maximum digit of a(n) in factorial base (i.e., A246359(a(n))) so far amongst the factorial base representations of all the terms up to and including a(n), with a(0)=0.
0, 1, 1, 2, 3, 5, 1, 7, 9, 12, 2, 13, 3, 16, 5, 6, 18, 1, 19, 2, 21, 3, 25, 27, 30, 32, 35, 7, 40, 9, 44, 4, 10, 11, 12, 13, 14, 15, 16, 18, 5, 19, 6, 56, 20, 7, 61, 22, 8, 64, 26, 66, 9, 69, 10, 29, 30, 76, 11, 32, 81, 12, 33, 88, 13, 36, 37, 38, 39, 40, 42, 14, 43, 15, 44, 16, 46, 17, 49, 50, 51, 52
Offset: 0
Examples
a(0) = 0 (by definition)
a(1) = 1 ('1' in factorial base), as 0 has occurred once in all the preceding terms.
a(2) = 1 as 1 has occurred once in all the preceding terms.
a(3) = 2 ('10' in factorial base), as digit '1' has occurred two times in total in all the preceding terms.
a(4) = 3 ('11' in factorial base), as '1' occurs once in each a(1) and a(2) and a(3).
a(5) = 5 ('21' in factorial base), as '1' occurs once in each of a(1), a(2) and a(3) and twice at a(4).
a(6) = 1 ('1' in factorial base), as '2' so far occurs only once at a(5)
a(7) = 7 = '101'
a(8) = 9 = '111'
a(9) = 12 = '200'
a(10) = 2 = '2'
a(11) = 13 = '201'
a(12) = 3 = '11'
a(12) = 3 = '11'
a(13) = 16 = '220'
a(14) = 5 = '21'
a(15) = 6 = '100'
a(16) = 18 = '300'
a(17) = 1 = '1'
a(18) = 19 = '301'
a(19) = 2 = '10'
a(20) = 21 = '311'
a(21) = 3 = '11'
a(22) = 25 = '1001'
a(23) = 27 = '1011'
a(24) = 30 = '1100'
a(25) = 32 = '1110'
a(26) = 35 = '1121'
a(27) = 7 (= '101' in factorial base), as the maximum digit in the factorial base representation of a(26), namely '2', has occurred in total 7 times in terms a(0) - a(26): once in each of a(5), a(9), a(11), a(14) and a(26), and twice in a(13).
Links
- Antti Karttunen, Table of n, a(n) for n = 0..10080