A330081 -id:A330081 - cp's OEIS frontend

A194959 Fractalization of (1 + floor(n/2)).

1, 1, 2, 1, 3, 2, 1, 3, 4, 2, 1, 3, 5, 4, 2, 1, 3, 5, 6, 4, 2, 1, 3, 5, 7, 6, 4, 2, 1, 3, 5, 7, 8, 6, 4, 2, 1, 3, 5, 7, 9, 8, 6, 4, 2, 1, 3, 5, 7, 9, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 12, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 13, 12, 10, 8, 6, 4, 2, 1, 3, 5

Offset: 1

Clark Kimberling, Sep 06 2011

Suppose that p(1), p(2), p(3), ... is an integer sequence satisfying 1 <= p(n) <= n for n >= 1. Define g(1)=(1) and for n > 1, form g(n) from g(n-1) by inserting n so that its position in the resulting n-tuple is p(n). The sequence f obtained by concatenating g(1), g(2), g(3), ... is clearly a fractal sequence, here introduced as the fractalization of p. The interspersion associated with f is here introduced as the interspersion fractally induced by p, denoted by I(p); thus, the k-th term in the n-th row of I(p) is the position of the k-th n in f. Regarded as a sequence, I(p) is a permutation of the positive integers; its inverse permutation is denoted by Q(p).
...
Example: Let p=(1,2,2,3,3,4,4,5,5,6,6,7,7,...)=A008619. Then g(1)=(1), g(2)=(1,2), g(3)=(1,3,2), so that
f=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,1,3,5,6,4,2,1,...)=A194959; and I(p)=A057027, Q(p)=A064578.
The interspersion I(P) has the following northwest corner, easily read from f:
  1  2  4  7 11 16 22
  3  6 10 15 21 28 36
  5  8 12 17 23 30 38
  9 14 20 27 35 44 54
  ...
Following is a chart of selected p, f, I(p), and Q(p):
   p         f        I(p)      Q(p)
A000027   A002260   A000027   A000027
A008619   A194959   A057027   A064578
A194960   A194961   A194962   A194963
A053824   A194965   A194966   A194967
A053737   A194973   A194974   A194975
A019446   A194968   A194969   A194970
A049474   A194976   A194977   A194978
A194979   A194980   A194981   A194982
A194964   A194983   A194984   A194985
A194986   A194987   A194988   A194989
Count odd numbers up to n, then even numbers down from n. - Franklin T. Adams-Watters, Jan 21 2012
This sequence defines the square array A(n,k), n > 0 and k > 0, read by antidiagonals and the triangle T(n,k) = A(n+1-k,k) for 1 <= k <= n read by rows (see Formula and Example). - Werner Schulte, May 27 2018

			The sequence p=A008619 begins with 1,2,2,3,3,4,4,5,5,..., so that g(1)=(1). To form g(2), write g(1) and append 2 so that in g(2) this 2 has position p(2)=2: g(2)=(1,2). Then form g(3) by inserting 3 at position p(3)=2: g(3)=(1,3,2), and so on. The fractal sequence A194959 is formed as the concatenation g(1)g(2)g(3)g(4)g(5)...=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,...).
From _Werner Schulte_, May 27 2018: (Start)
This sequence seen as a square array read by antidiagonals:
  n\k: 1  2  3  4  5   6   7   8   9  10  11  12 ...
  ===================================================
   1   1  2  2  2  2   2   2   2   2   2   2   2 ... (see A040000)
   2   1  3  4  4  4   4   4   4   4   4   4   4 ... (see A113311)
   3   1  3  5  6  6   6   6   6   6   6   6   6 ...
   4   1  3  5  7  8   8   8   8   8   8   8   8 ...
   5   1  3  5  7  9  10  10  10  10  10  10  10 ...
   6   1  3  5  7  9  11  12  12  12  12  12  12 ...
   7   1  3  5  7  9  11  13  14  14  14  14  14 ...
   8   1  3  5  7  9  11  13  15  16  16  16  16 ...
   9   1  3  5  7  9  11  13  15  17  18  18  18 ...
  10   1  3  5  7  9  11  13  15  17  19  20  20 ...
  etc.
This sequence seen as a triangle read by rows:
  n\k:  1  2  3  4  5   6   7   8   9  10  11  12  ...
  ======================================================
   1    1
   2    1  2
   3    1  3  2
   4    1  3  4  2
   5    1  3  5  4  2
   6    1  3  5  6  4   2
   7    1  3  5  7  6   4   2
   8    1  3  5  7  8   6   4   2
   9    1  3  5  7  9   8   6   4   2
  10    1  3  5  7  9  10   8   6   4   2
  11    1  3  5  7  9  11  10   8   6   4   2
  12    1  3  5  7  9  11  12  10   8   6   4   2
  etc.
(End)

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

Paul Lévy, Sur quelques classes de permutations, Compositio Mathematica, Volume 8, 1951, pages 1-48. P_n = g(n).
Eric Weisstein's World of Mathematics, Fractal sequence
Eric Weisstein's World of Mathematics, Interspersion
Wikipedia, Fractal sequence

Cf. A000142, A000217, A005408, A005843, A008619, A057027, A064578, A209229, A210535, A219977; A000012 (col 1), A157532 (col 2), A040000 (row 1), A113311 (row 2); A194029 (introduces the natural fractal sequence and natural interspersion of a sequence - different from those introduced at A194959).

Cf. A003558 (g permutation order), A102417 (index), A330081 (on bits), A057058 (inverse).

r = 2; p[n_] := 1 + Floor[n/r]
Table[p[n], {n, 1, 90}]  (* A008619 *)
g[1] = {1}; g[n_] := Insert[g[n - 1], n, p[n]]
f[1] = g[1]; f[n_] := Join[f[n - 1], g[n]]
f[20] (* A194959 *)
row[n_] := Position[f[30], n];
u = TableForm[Table[row[n], {n, 1, 5}]]
v[n_, k_] := Part[row[n], k];
w = Flatten[Table[v[k, n - k + 1], {n, 1, 13},
{k, 1, n}]]  (* A057027 *)
q[n_] := Position[w, n]; Flatten[
Table[q[n], {n, 1, 80}]]  (* A064578 *)
Flatten[FoldList[Insert[#1, #2, Floor[#2/2] + 1] &, {}, Range[10]]] (* Birkas Gyorgy, Jun 30 2012 *)

T(n,k) = min(k<<1-1,(n-k+1)<<1); \\ Kevin Ryde, Oct 09 2020

From Werner Schulte, May 27 2018 and Jul 10 2018: (Start)
Seen as a triangle: It seems that the triangle T(n,k) for 1 <= k <= n (see Example) is the mirror image of A210535.
Seen as a square array A(n,k) and as a triangle T(n,k):
A(n,k) = 2*k-1 for 1 <= k <= n, and A(n,k) = 2*n for 1 <= n < k.
A(n+1,k+1) = A(n,k+1) + A(n,k) - A(n-1,k) for k > 0 and n > 1.
A(n,k) = A(k,n) - 1 for n > k >= 1.
P(n,x) = Sum_{k>0} A(n,k)*x^(k-1) = (1-x^n)*(1-x^2)/(1-x)^3 for n >= 1.
Q(y,k) = Sum_{n>0} A(n,k)*y^(n-1) = 1/(1-y) for k = 1 and Q(y,k) = Q(y,1) + P(k-1,y) for k > 1.
G.f.: Sum_{n>0, k>0} A(n,k)*x^(k-1)*y^(n-1) = (1+x)/((1-x)*(1-y)*(1-x*y)).
Sum_{k=1..n} A(n+1-k,k) = Sum_{k=1..n} T(n,k) = A000217(n) for n > 0.
Sum_{k=1..n} (-1)^(k-1) * A(n+1-k,k) = Sum_{k=1..n} (-1)^(k-1) * T(n,k) = A219977(n-1) for n > 0.
Product_{k=1..n} A(n+1-k,k) = Product_{k=1..n} T(n,k) = A000142(n) for n > 0.
A(n+m,n) = A005408(n-1) for n > 0 and some fixed m >= 0.
A(n,n+m) = A005843(n) for n > 0 and some fixed m > 0.
Let A_m be the upper left part of the square array A(n,k) with m rows and m columns. Then det(A_m) = 1 for some fixed m > 0.
The P(n,x) satisfy the recurrence equation P(n+1,x) = P(n,x) + x^n*P(1,x) for n > 0 and initial value P(1,x) = (1+x)/(1-x).
Let B(n,k) be multiplicative with B(n,p^e) = A(n,e+1) for e >= 0 and some fixed n > 0. That yields the Dirichlet g.f.: Sum_{k>0} B(n,k)/k^s = (zeta(s))^3/(zeta(2*s)*zeta(n*s)).
Sum_{k=1..n} A(k,n+1-k)*A209229(k) = 2*n-1. (conjectured)
(End)
From Kevin Ryde, Oct 09 2020: (Start)
T(n,k) = 2*k-1 if 2*k-1 <= n, or 2*(n+1-k) if 2*k-1 > n. [Lévy, chapter 1 section 1 equations (a),(b)]
Fixed points T(n,k)=k for k=1 and k = (2/3)*(n+1) when an integer. [Lévy, chapter 1 section 2 equation (3)]
(End)

Name corrected by Franklin T. Adams-Watters, Jan 21 2012

A333776 Scan the binary representation of n from right to left; at each 1, reverse the bits to the right and excluding this 1. The resulting binary representation is that of a(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 5, 7, 8, 12, 10, 14, 9, 11, 13, 15, 16, 24, 20, 28, 18, 22, 26, 30, 17, 19, 21, 23, 25, 29, 27, 31, 32, 48, 40, 56, 36, 44, 52, 60, 34, 38, 42, 46, 50, 58, 54, 62, 33, 35, 37, 39, 41, 45, 43, 47, 49, 57, 53, 61, 51, 55, 59, 63, 64, 96, 80

Offset: 0

Rémy Sigrist, Apr 05 2020

This sequence is a permutation of the nonnegative integers (as it is injective and preserves the binary length); see A333777 for the inverse.

We can devise a variant of this sequence for any fixed base b > 1, by performing a reversal at each nonzero digit in base b.

			For n = 90:
- the binary representation of 90 is "1011010",
- this binary representation evolves as follows (parentheses indicate reversals):
    1 0 1 1 0 1(0)
    1 0 1 1(0 1 0)
    1 0 1(0 1 0 1)
    1(1 0 1 0 1 0)
- the resulting binary representation is "1101010"
- and a(90) = 106.
The binary plot of the first terms is as follows (#'s denote 1's):
                                  ################################
                  ################ # #  ##    ####        ########
          ######## # #  ##    ####  ## # #  ## # #    #### # #  ##
      #### # #  ##  ## # #  ## # #    #### # #  ##  ## # #  ## # #
    ## # #  ## # #    #### # #  ##        ######## # #  ##    ####
   # #  ##    ####        ########                ################
            1         2         3         4         5         6
  0123456789012345678901234567890123456789012345678901234567890123

See A333692 for a similar sequence.

Cf. A000120, A330081, A333777 (inverse), A333778 (fixed points).

a(n, base=2) = { my (d=digits(n, base), t=[]); forstep (k=#d, 1, -1, if (d[k], t=Vecrev(t)); t=concat(d[k], t)); fromdigits(t, base); }

a(2*n) <= 2*a(n) with equality iff n = 0 or n is a power of 2.

A000120(a(n)) = A000120(n).

A329303 If the run lengths in binary expansion of n are (r(1), ..., r(w)), then the run lengths in binary expansion of a(n) are (r(1), r(3), r(5), ..., r(6), r(4), r(2)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 10, 9, 12, 13, 14, 15, 16, 23, 20, 19, 22, 21, 18, 17, 24, 27, 26, 25, 28, 29, 30, 31, 32, 47, 40, 39, 44, 43, 36, 35, 46, 41, 42, 45, 38, 37, 34, 33, 48, 55, 52, 51, 54, 53, 50, 49, 56, 59, 58, 57, 60, 61, 62, 63, 64, 95, 80, 79

Offset: 0

Rémy Sigrist, Dec 01 2019

This sequence is a permutation of the nonnegative integers that preserves the binary length as well as the number of runs. See A330091 for the inverse.

			For n = 19999:
- the binary representation of 19999 is "100111000011111",
- the corresponding run lengths are (1, 2, 3, 4, 5),
- hence the run lengths of a(n) are (1, 3, 5, 4, 2),
- and its binary representation is "100011111000011",
- so a(n) = 18371.

See A330081 for a similar sequence.

Cf. A003558, A194959, A330091 (inverse).

torl(n) = { my (rr=[]); while (n, my (r=valuation(n+(n%2),2)); rr = concat(r, rr); n\=2^r); rr }
shuffle(v) = { my (w=vector(#v), o=0, e=#v+1); for (k=1, #v, w[if (k%2, o++, e--)]=v[k]); w }
fromrl(rr) = { my (v=0); for (k=1, #rr, v = (v+(k%2))*2^rr[k]-(k%2)); v }
a(n) = fromrl(shuffle(torl(n)))

If n has w binary runs, then a^A003558(w-1)(n) = n (where a^k denotes the k-th iterate of the sequence).

cp's OEIS Frontend

A330090 Inverse permutation to A330081.

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A194959 Fractalization of (1 + floor(n/2)).

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A333776 Scan the binary representation of n from right to left; at each 1, reverse the bits to the right and excluding this 1. The resulting binary representation is that of a(n).

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A329303 If the run lengths in binary expansion of n are (r(1), ..., r(w)), then the run lengths in binary expansion of a(n) are (r(1), r(3), r(5), ..., r(6), r(4), r(2)).

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