cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 06 2011

Keywords

Comments

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)
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

Examples

			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)
		

References

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

Crossrefs

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).

Programs

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

Formula

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)

Extensions

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

A057027 Triangle T read by rows: row n consists of the numbers C(n,2)+1 to C(n+1,2); numbers in odd-numbered places form an increasing sequence and the others a decreasing sequence.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 10, 8, 9, 11, 15, 12, 14, 13, 16, 21, 17, 20, 18, 19, 22, 28, 23, 27, 24, 26, 25, 29, 36, 30, 35, 31, 34, 32, 33, 37, 45, 38, 44, 39, 43, 40, 42, 41, 46, 55, 47, 54, 48, 53, 49, 52, 50, 51, 56, 66, 57, 65, 58, 64, 59, 63, 60, 62, 61, 67, 78, 68, 77, 69, 76
Offset: 1

Views

Author

Clark Kimberling, Jul 28 2000

Keywords

Comments

Arrange the quotients F(i)/F(j) of Fibonacci numbers, for 2<=i

Examples

			For n=6, the ordered quotients are 1/8, 1/5, 2/8, 1/3, 3/8, 2/5, 1/2, 3/5, 5/8, 2/3; the positions of 1/5, 2/5, 3/5 are 2, 6, 8 (first terms of diagonal T(i, i-1)).
Triangle starts:
  1;
  2, 3;
  4, 6, 5;
  7,10, 8, 9;
  ...
		

Crossrefs

Reflection of the array in A057028 about its central column, a permutation of the natural numbers.
Inverse permutation to A064578. Central column: A057029.
Column 1 is A000124, column 2 is A000217.
Row sums are A006003.

Programs

  • Mathematica
    nn= 12; t = Table[Range[Binomial[n, 2] + 1, Binomial[n + 1, 2]], {n, nn}]; Table[t[[n, If[OddQ@ k, Ceiling[k/2], -k/2] ]], {n, nn}, {k, n}] // Flatten (* Michael De Vlieger, Jul 02 2016 *)

Formula

From Werner Schulte, Sep 09 2024: (Start)
T(n, k) = (n^2 + (-1)^k * (n - k) + (3 + (-1)^k) / 2) / 2.
T(n, 1) = (n^2 - n + 2) / 2 = A000124(n).
T(n, 2) = (n^2 + n) / 2 = A000217(n) for n >= 2.
T(n, k) = T(n, k-2) - (-1)^k for 3 <= k <= n. (End)
G.f.: x*y*(1 + x*(y - 1) - x^4*(y - 1)*y^2 + x^5*y^3 + x^3*y*(y^2 - y - 1) - x^2*(y^2 + y - 1))/((1 - x)^3*(1 - x*y)^3*(1 + x*y)). - Stefano Spezia, Sep 10 2024

Extensions

Corrected and extended by Vladeta Jovovic, Oct 18 2001

A188568 Enumeration table T(n,k) by descending antidiagonals. The order of the list - if n is odd: T(n,1), T(2,n-1), T(n-2,3), ..., T(n-1,2), T(1,n); if n is even: T(1,n), T(n-1,2), T(3,n-2), ..., T(2,n-1), T(n,1).

Original entry on oeis.org

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

Author

Boris Putievskiy, Dec 27 2012

Keywords

Comments

Self-inverse permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Call a "layer" a pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1). This table read layer by layer clockwise is A194280. This table read by boustrophedonic ("ox-plowing") method - layer clockwise, layer counterclockwise and so on - is A064790. - Boris Putievskiy, Mar 14 2013

Examples

			The start of the sequence as table:
   1,  2,  6,  7, 15, 16, 28, ...
   3,  5,  9, 12, 20, 23, 35, ...
   4,  8, 13, 18, 26, 31, 43, ...
  10, 14, 19, 25, 33, 40, 52, ...
  11, 17, 24, 32, 41, 50, 62, ...
  21, 27, 34, 42, 51, 61, 73, ...
  22, 30, 39, 49, 60, 72, 85, ...
  ...
The start of the sequence as triangular array read by rows:
   1;
   2,  3;
   6,  5,  4;
   7,  9,  8, 10;
  15, 12, 13, 14, 11;
  16, 20, 18, 19, 17, 21;
  28, 23, 26, 25, 24, 27, 22;
  ...
Row number k contains permutation of the k numbers:
{ (k^2-k+2)/2, (k^2-k+2)/2 + 1, ..., (k^2+k-2)/2 + 1 }.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{t, i, j},
    t = Floor[(Sqrt[8n-7]-1)/2];
    i = n-t(t+1)/2;
    j = (t^2+3t+4)/2-n;
    ((i+j-1)(i+j-2) + ((-1)^Max[i,j]+1)i - ((-1)^Max[i,j]-1)j)/2];
    Array[a, 55] (* Jean-François Alcover, Jan 26 2019 *)
  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    m=((i+j-1)*(i+j-2)+((-1)**max(i,j)+1)*i-((-1)**max(i,j)-1)*j)/2

Formula

a(n) = ((i+j-1)*(i+j-2)+((-1)^max(i,j)+1)*i-((-1)^max(i,j)-1)*j)/2, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor[(-1+sqrt(8*n-7))/2].

A185180 Enumeration table T(n,k) by antidiagonals. The order of the list is symmetrical movement from center to edges diagonal.

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 9, 7, 8, 10, 14, 12, 11, 13, 15, 20, 18, 16, 17, 19, 21, 27, 25, 23, 22, 24, 26, 28, 35, 33, 31, 29, 30, 32, 34, 36, 44, 42, 40, 38, 37, 39, 41, 43, 45, 54, 52, 50, 48, 46, 47, 49, 51, 53, 55, 65, 63, 61, 59, 57, 56, 58, 60, 62, 64, 66, 77, 75
Offset: 1

Author

Boris Putievskiy, Dec 26 2012

Keywords

Comments

The natural numbers are grouped in chunks of 1, 2, 3, 4,... as (1), (2,3), (4,5,6), (7,8,9,10), etc and each group fills a diagonal in the table. The smallest number in a group is in A000124, the largest in A000217. Numbers in a group are placed on free spots as close as possible to the middle of the diagonal, given preference to the smaller row numbers in the table if there is a draw.
The resulting array is apparently a transposed version of A064789 (if this was also written as an array).
The order of the list table T(n,k):
if n is odd:
T (floor(n+1)/2,floor(n+1)/2), T(floor(n+1)/2-1,floor(n+1)/2+1), T(floor(n+1)/2+1,floor(n+1)/2-1),...T(1,n), T(n,1)
if n is even:
T(floor(n+1)/2-1,floor(n+1)/2+1), T(floor(n+1)/2+1,floor(n+1)/2-1),...T(1,n), T(n,1).
Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.

Examples

			The start of the sequence as table:
  1....2....5....9...14...20...27 ...
  3....4....7...12...18...25...33 ...
  6....8...11...16...23...31...40 ...
  10..13...17...22...29...38...48 ...
  15..19...24...30...37...46...57 ...
  21..26...32...39...47...56...67 ...
  28..34...41...49...58...68...79 ...
  ...
The start of the sequence as triangle array read by rows:
  1;
  2,   3;
  5,   4,  6;
  9,   7,  8, 10;
  14, 12, 11, 13, 15;
  20, 18, 16, 17, 19, 21;
  27, 25, 23, 22, 24, 26, 28;
. . .
Row number k (k > 1) of the triangle contains a permutation of the set of k numbers from (k^2-k+2)/2, (k^2-k+2)/2 + 1 ,...up to (k^2+k-2)/2 + 1, namely (k^2+k-2)/2, (k^2+k-2)/2 -2,...,(k^2-k+2)/2, (k^2-k+2)/2 + 2,..., (k^2+k-2)/2-1, (k^2+k-2)/2+1.
		

Programs

  • Mathematica
    a[n_] := Module[{i, j, t}, i = n - t(t+1)/2; j = (t^2 + 3t + 4)/2 - n; t = Floor[(-1 + Sqrt[8n - 7])/2]; If[j <= i, (i(i+1) + (j-1)(j + 2i - 4))/2, (i(i+1) + (j-1)(j + 2i - 4))/2 + 2(j-i) - 1]];
    Array[a, 68] (* Jean-François Alcover, Nov 21 2018, from Python *)
  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    if j<=i:
       m=(i*(i+1) + (j-1)*(j+2*i-4))/2
    else:
       m=(i*(i+1) + (j-1)*(j+2*i-4))/2 +2*(j-i)-1

Formula

a(n) = (i*(i+1) + (j-1)*(j+2*i-4))/2, if j<=i, a(n)=(i*(i+1) + (j-1)*(j+2*i-4))/2 +2*(j-i)-1, if j>i, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor[(-1+sqrt(8*n-7))/2].
Showing 1-4 of 4 results.