A194032 -id:A194032 - cp's OEIS frontend

A194033 Inverse permutation of A194032; contains every positive integer exactly once.

1, 3, 6, 2, 5, 9, 10, 15, 4, 8, 13, 14, 20, 21, 28, 7, 12, 18, 19, 26, 27, 35, 36, 45, 11, 17, 24, 25, 33, 34, 43, 44, 54, 55, 66, 16, 23, 31, 32, 41, 42, 52, 53, 64, 65, 77, 78, 91, 22, 30, 39, 40, 50, 51, 62, 63, 75, 76, 89, 90, 104, 105, 120, 29

Offset: 1

Clark Kimberling, Aug 12 2011

nonn

Zhuorui He, Table of n, a(n) for n = 1..10000

Cf. A194032.

Mathematica
```
(See A194032.)
```

A194033(n)={my(x=sqrtint(n), y=n-x^2, z=min(y,floor(y/2+1))) ;binomial(x+z,2)+y+1} \\ Zhuorui He, Jul 19 2025

a(n) = binomial(x + z,2) + y + 1 where x = floor(sqrt(n)), y = n - x^2 and z = min(y,floor(y/2 + 1)). - Zhuorui He, Jul 19 2025

a(63) corrected and a(64) added by Zhuorui He, Jul 07 2025

A194029 Natural fractal sequence of the Fibonacci sequence (1, 2, 3, 5, 8, ...).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34

Offset: 1

Clark Kimberling, Aug 12 2011

Suppose that c(1), c(2), c(3), ... is an increasing sequence of positive integers with c(1) = 1, and that the sequence c(k+1) - c(k) is strictly increasing. The natural fractal sequence f of c is defined by:
  If c(k) <= n < c(k+1), then f(n) = 1 + n - c(k).
This defines the present sequence a(n) = f(n) for c = A000045.
The natural interspersion of c is here introduced as the array given by T(n,k) =(position of k-th n in f).  Note that c = (row 1 of T).
As a different example from the one considered here (c = A000045), let c = A000217 = (1, 3, 6, 10, 15, ...), the triangular numbers, so that f = (1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, ...) = A002260, and a northwest corner of T = A194029 is:
                    1   3   6  10  15  ...
                    2   4   7  11  16  ...
                    5   8  12  17  23  ...
                    9  13  18  24  31  ...
                   ...
Since every number in the set N of positive integers occurs exactly once in this (and every) interspersion, a listing of the terms of T by antidiagonals comprises a permutation, p, of N; letting q denote the inverse of p, we thus have for each c a fractal sequence, an interspersion T, and two permutations of N:
       c         f       T / p       q
    A000045   A194029   A194030   A194031
    A000290   A071797   A194032   A194033
    A000217   A002260   A066182   A066181
    A028387   A074294   A194034   A194035
    A028872   A071797   A194036   A194037
    A034856   A002260   A194038   A194040
It appears that this is also a triangle read by rows in which row n lists the first A000045(n) positive integers, n >= 1 (see example). - Omar E. Pol, May 28 2012
This is true, because the sequence c = A000045 has the property that c(k+1)  - c(k) = c(k-1), so the number of integers {1, 2, 3, ...} to be filled in from index n = c(k) to n = c(k+1)-1 is equal to c(k-1); see also the first EXAMPLE. - M. F. Hasler, Apr 23 2022

			The sequence (1, 2, 3, 5, 8, 13, ...) is used to place '1's in positions numbered 1, 2, 3, 5, 8, 13, ...  Then gaps are filled in with consecutive counting numbers:
  1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 1, ...
From _Omar E. Pol_, May 28 2012: (Start)
Written as an irregular triangle the sequence begins:
  1;
  1;
  1, 2;
  1, 2, 3;
  1, 2, 3, 4, 5;
  1, 2, 3, 4, 5, 6, 7, 8;
  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13;
  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21; ...
The row lengths are A000045(n).
(End)

Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168.

Alois P. Heinz, Rows n = 1..20, flattened
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.

Cf. A000045 (Fibonacci numbers).
Cf. A066628, A194030, A194031 (natural interspersion of A000045 and inverse permutation).
Cf. A130853.

T:= n-> $1..(<<0|1>, <1|1>>^n)[1, 2]:
seq(T(n), n=1..10);  # Alois P. Heinz, Dec 11 2024

z = 40;
c[k_] := Fibonacci[k + 1];
c = Table[c[k], {k, 1, z}]  (* A000045 *)
f[n_] := If[MemberQ[c, n], 1, 1 + f[n - 1]]
f = Table[f[n], {n, 1, 800}]  (* A194029 *)
r[n_] := Flatten[Position[f, n]]
t[n_, k_] := r[n][[k]]
TableForm[Table[t[n, k], {n, 1, 8}, {k, 1, 7}]]
p = Flatten[Table[t[k, n - k + 1], {n, 1, 13}, {k, 1, n}]]  (* A194030 *)
q[n_] := Position[p, n]; Flatten[Table[q[n], {n, 1, 80}]]  (* A194031 *)
Flatten[Range[Fibonacci[Range[66]]]] (* Birkas Gyorgy, Jun 30 2012 *)

a(n) = A066628(n)+1. - Alan Michael Gómez Calderón, Oct 30 2023

Edited by M. F. Hasler, Apr 23 2022

cp's OEIS Frontend

A194033 Inverse permutation of A194032; contains every positive integer exactly once.

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