A249146 -id:A249146 - cp's OEIS frontend

A249145 Length of the maximal run in the binary representation of A241946: a(n) = A043276(A249146(n)).

1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 4, 1, 4, 1, 2, 3, 1, 1, 2, 3, 1, 5, 1, 3, 2, 2, 1, 2, 2, 2, 3, 2, 4, 2, 4, 2, 3, 1, 3, 2, 2, 2, 1, 3, 3, 2, 2, 3, 1, 2, 3, 2, 2, 2, 3, 2, 5, 1, 3, 2, 3, 3, 4, 1, 2, 2, 2, 3, 4, 2, 2, 1, 5, 2, 2, 2, 3, 3, 2, 4, 3, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 4, 3, 2, 5, 2, 6, 1, 2, 5, 1, 2, 4, 2, 4, 1, 3, 2, 3, 3, 4, 2, 3, 5, 2, 2, 2, 2, 4, 2, 2, 2, 1, 3, 5, 3, 4

Offset: 1

Antti Karttunen, Oct 24 2014

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A043276, A249146, A249143.

(define (A249145 n) (A043276 (A249146 n)))

a(n) = A043276(A249146(n)).

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.

Original entry on oeis.org

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

Antti Karttunen, Oct 20 2014

			   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).

Antti Karttunen, Table of n, a(n) for n = 0..10080

Cf. A007623, A246359, A249146, A249150.

Differs from A249069 for the first time at n=27, where a(27) = 7, while A249069(27) = 38.

A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

Original entry on oeis.org

0, 1, 2, 4, 1, 6, 7, 1, 8, 2, 11, 3, 4, 5, 17, 19, 7, 4, 8, 5, 25, 26, 29, 31, 1, 32, 2, 35, 12, 14, 37, 41, 45, 49, 50, 52, 22, 57, 58, 61, 63, 1, 64, 2, 67, 25, 69, 73, 76, 32, 3, 33, 80, 4, 34, 87, 14, 92, 35, 36, 38, 99, 42, 105, 108, 47, 5, 114, 116, 49, 119, 23, 24, 25, 123, 54, 126, 127, 1, 128, 2

Offset: 0

Antti Karttunen, Oct 22 2014

Inspired by A248034.

			a(0) = 0 (by definition), and 0 is also '0' in binary.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the rightmost run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that the rightmost run of a(2) = 2 ('10' in binary) is one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the rightmost run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the rightmost run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the rightmost run of a(5) = 6 ('110' in binary) is one bit long, and so far there has occurred 7 such one bit runs in terms a(0) .. a(5), thus a(6) = 7.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A249143, A249146, A249148, A248034.

cp's OEIS Frontend

A249145 Length of the maximal run in the binary representation of A241946: a(n) = A043276(A249146(n)).

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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.

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A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

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