A105029 Write numbers in binary under each other, left justified, read diagonals in downward direction, convert to decimal.

0, 2, 6, 5, 4, 14, 13, 8, 11, 10, 9, 12, 30, 29, 24, 19, 18, 17, 20, 23, 22, 21, 16, 27, 26, 25, 28, 62, 61, 56, 51, 34, 33, 36, 39, 38, 37, 32, 43, 42, 41, 44, 47, 46, 45, 40, 35, 50, 49, 52, 55, 54, 53, 48, 59, 58, 57, 60, 126, 125, 120, 115, 98, 65, 68, 71, 70

Offset: 0

Author

Benoit Cloitre, Apr 03 2005

Keywords

• nonn
• base

Comments

All terms are distinct, but the numbers 2^m - 1 are missing.

a(n) = Sum_{k>=1} B(n+k-1,k)*2^(A103586(n)-k) where B(n,k) n>=1, k>=1 is the infinite array:

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

.......

where n-th row consists of binary expansion of n followed by 0's.

a(n) = A105025(n) iff A070939(n) = A103586(n), cf. A214489. - Reinhard Zumkeller, Jul 21 2012

Examples

			0
1
1 0
1 1
1 0 0
1 0 1
1 1 0
1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
and reading the diagonals downwards we get 0, 10, 110, 101, 100, 1110, 1101, etc.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A102370, A105030, A105025, A105026, A105027, A105028, A105033.

Cf. A000079, A070939, A103586.

Programs

Haskell

import Data.Bits ((.|.), (.&.))
a105029 n = foldl (.|.) 0 $ zipWith (.&.) a000079_list $
   map (\x -> (len + 1 - a070939 x) * x)
       (reverse $ enumFromTo n (n - 1 + len))  where len = a103586 n
-- Reinhard Zumkeller, Jul 21 2012