A122197 Fractal sequence: count up to successive integers twice.

1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5

Offset: 1

Author

Franklin T. Adams-Watters, Aug 25 2006

Keywords

• easy
• nonn
• tabf

Comments

Fractal - deleting the first occurrence of each integer leaves the original sequence. Also, deleting all the 1's leaves the original sequence plus 1. New values occur at square indices. 1's occur at indices m^2+1 and m^2+m+1. Ordinal transform of A122196.

Except for its initial 1, A122197 is the natural fractal sequence of A002620; that is, A122197(n+1) is the number of the row of A194061 that contains n. See A194029 for definition of natural fractal sequence. - Clark Kimberling, Aug 12 2011

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

Triangle read by rows formed from antidiagonals of triangle A002260.

The row sums equal A008805(n-1) and the antidiagonal sums equal A211534(n+5). (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    1
  3    1,  2
  4    1,  2
  5    1,  2,  3
  6    1,  2,  3
		

Links

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

Crossrefs

Cf. A122196, A000290, A033638, A002260, A001477, A000194, A339399.

Programs

Haskell

import Data.List (transpose, genericIndex)
a122197 n k = genericIndex (a122197_row n) (k - 1)
a122197_row n = genericIndex a122197_tabf (n - 1)
a122197_tabf = concat $ transpose [a002260_tabl, a002260_tabl]
a122197_list = concat a122197_tabf
-- Reinhard Zumkeller, Aug 07 2015, 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): (n-1) mod (t+1) + 1 end: seq(a(n), n=1..105); # End first program
T := proc(n, k): if n < 1 then return(0) elif k < 1 or k> floor((n+1)/2) then return(0) else k fi: end: seq(seq(T(n, k), k=1..floor((n+1)/2)), n=1..19); # End second program. (End)

Mathematica

With[{c=Table[Range[n],{n,10}]},Flatten[Riffle[c,c]]] (* Harvey P. Dale, Apr 19 2013 *)

PARI

a(n)=n - (sqrtint(4*n) + 1)\2*sqrtint(n-1) \\ Charles R Greathouse IV, Jun 08 2020

Python

from math import isqrt
def A122197(n): return 1 if n<=1 else 1+((n-1)%((m:=isqrt(n-1))+int(n-1>m*(m+1)))) # Chai Wah Wu, Jun 05 2025

Formula

From Boris Putievskiy, Sep 09 2013: (Start)

a(n) = (A001477(n-1) mod A000194(n-1)) + 1 for n >= 2 with a(1) = 1.

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

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

T(n, k) = k for n >= 1 and 1 <= k <= (n+1)/2; T(n, k) = 0 elsewhere.

T(n, k) = A002260(n-k, k). (End)

a(n) = n - floor(sqrt(n) + 1/2)*floor(sqrt(n-1)). - Ridouane Oudra, Jun 08 2020

a(n) = A339399(2n-1). - Wesley Ivan Hurt, Jan 09 2022