A122196 Fractal sequence: count down by 2's from successive integers.

1, 2, 3, 1, 4, 2, 5, 3, 1, 6, 4, 2, 7, 5, 3, 1, 8, 6, 4, 2, 9, 7, 5, 3, 1, 10, 8, 6, 4, 2, 11, 9, 7, 5, 3, 1, 12, 10, 8, 6, 4, 2, 13, 11, 9, 7, 5, 3, 1, 14, 12, 10, 8, 6, 4, 2, 15, 13, 11, 9, 7, 5, 3, 1, 16, 14, 12, 10, 8, 6, 4, 2, 17, 15, 13, 11, 9, 7, 5, 3, 1, 18, 16, 14, 12, 10, 8, 6, 4, 2, 19, 17

Offset: 1

Author

Franklin T. Adams-Watters, Aug 25 2006

Keywords

• easy
• nonn
• tabf

Comments

First differences of A076644. Fractal - deleting the first occurrence of each integer leaves the original sequence. Also, original sequence plus 1. 1's occur at square indices. New values occur at indices m^2+1 and m^2+m+1.

Ordinal transform of A122197.

Row sums give A002620. - Gary W. Adamson, Nov 29 2008

From Gary W. Adamson, Dec 05 2009: (Start)

A122196 considered as an infinite lower triangular matrix * [1,2,3,...] =

A006918 starting (1, 2, 5, 8, 14, 20, 30, 40, ...).

Let A122196 = an infinite lower triangular matrix M; then lim_{n->infinity} M^n = A171238, a left-shifted vector considered as a matrix. (End)

A122196 is the fractal sequence associated with the dispersion A082156; that is, A122196(n) is the number of the row of A082156 that contains n. - Clark Kimberling, Aug 12 2011

From Johannes W. Meijer, Sep 09 2013: (Start)

The alternating row sums lead to A004524(n+2).

The antidiagonal sums equal A001840(n). (End)

Examples

			The first few rows of the sequence a(n) as a triangle T(n, k):
n/k  1   2   3
1    1
2    2
3    3,  1
4    4,  2
5    5,  3,  1
6    6,  4,  2
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A076644, A122197, A000290, A033638, A002620, A006918, A171238, A082156, A000267.

Programs

Haskell

a122196 n = a122196_list !! (n-1)
a122196_list = concatMap (\x -> enumFromThenTo x (x - 2) 1) [1..]
-- Reinhard Zumkeller, Jul 19 2012

Maple

From Johannes W. Meijer, Sep 09 2013: (Start)
a := proc(n) local t: t:=floor((sqrt(4*n-3)-1)/2): floor(sqrt(4*n-1))-2*((n-1) mod (t+1)) end: seq(a(n), n=1..92); # End first program.
T := (n, k) -> n-2*k+2: seq(seq(T(n, k), k=1..floor((n+1)/2)), n=1..18); # End second program. (End)

Mathematica

Flatten@Range[Range[10], 1, -2] (* Birkas Gyorgy, Apr 07 2011 *)

Formula

From Boris Putievskiy, Sep 09 2013: (Start)

a(n) = 2*(1-A122197(n)) + A000267(n-1).

a(n) = floor(sqrt(4*n-1)) - 2*((n-1) mod (t+1)), where t = floor((sqrt(4*n-3)-1)/2). (End)

From Johannes W. Meijer, Sep 09 2013: (Start)

T(n, k) = n - 2*k + 2, for n >= 1 and 1 <= k <= floor((n+1)/2).

T(n, k) = A002260(n, n-2*k+2). (End)