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.

A007337 Signature sequence of sqrt(3).

Original entry on oeis.org

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

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Author

Keywords

Comments

Arrange the numbers i+j*x (i,j >= 1) in increasing order; the sequence of i's is the signature of x.

References

  • C. Kimberling, "Fractal Sequences and Interspersions", Ars Combinatoria, vol. 45, 1997, p. 157.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First occurrence of n is given in A022778.

Programs

  • Mathematica
    Take[ Transpose[ Sort[ Flatten[ Table[{i + j*Sqrt[3], i}, {i, 25}, {j, 17}], 1], #1[[1]] < #2[[1]] &]][[2]], 95] (* Robert G. Wilson v, Jul 24 2004 *)

Extensions

More terms from Robert G. Wilson v, Jul 24 2004

A255977 The number of numbers j+k*r <= n, where r = golden ratio and j and k are nonnegative integers.

Original entry on oeis.org

1, 2, 4, 6, 9, 13, 17, 22, 27, 33, 40, 47, 55, 64, 73, 83, 93, 104, 116, 128, 141, 154, 168, 183, 198, 214, 231, 248, 266, 284, 303, 323, 343, 364, 386, 408, 431, 454, 478, 503, 528, 554, 580, 607, 635, 663, 692, 722, 752, 783, 814, 846, 879, 912, 946, 980
Offset: 1

Views

Author

Clark Kimberling, Mar 14 2015

Keywords

Comments

The difference sequence is A019446.
From Thomas Anton, Oct 22 2018: (Start)
It appears that this sequence (apart from the first term) can be obtained through the following sieving process. Start with the positive integers. Then, at each stage, circle the first remaining number that has not already been circled, and delete all terms in the subsequence of terms that were not circled in previous stages with circled indices that have not yet been deleted. E.g., the first few iterations are
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
We circle 1, and take the subsequence of previously uncircled numbers, which is the entire sequence, and delete all terms with circled indices that have not been deleted, in this case, just the 1st, 1.
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
We circle 2, and take the previously uncircled subsequence, which is again the entire sequence, and delete all terms with circled indices that have not been deleted, just the 2nd, 3.
(2), 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
We circle 4, and take the previously uncircled subsequence, which is all terms of the sequence except 2, and delete all terms in that subsequence with circled indices (as terms of the subsequence) that have not been deleted, the 2nd and 4th, respectively the 3rd and 5th terms of the entire sequence, 5 and 7.
(2), (4), 6, 8, 9, 10, 11, 12, 13, 14, ...
etc.
(End)

Crossrefs

Programs

  • Maple
    t:=(1+sqrt(5))/2: a:=n->n+1+add(floor((n-k)/t),k=0..n): seq(a(n),n=0..55); # Muniru A Asiru, Oct 24 2018
  • Mathematica
    Table[n + 1 + Sum[Floor[(n - k)/GoldenRatio], {k, 0, n}], {n, 0, 200}]

Formula

a(n) = n + 1 + Sum{floor[(n - k)/tau], k = 0..n}, where tau = (1 + sqrt(5))/2.
a(n) = A054347(n-1) - (n^2 - 3*n)/2. - Alan Michael Gómez Calderón, Nov 21 2024

A022780 Place where n-th 1 occurs in A023118.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 13, 17, 21, 26, 31, 36, 42, 48, 55, 62, 70, 78, 87, 96, 105, 115, 125, 136, 147, 159, 171, 184, 197, 210, 224, 238, 253, 268, 284, 300, 317, 334, 351, 369, 387, 406, 425, 445, 465, 486, 507, 529, 551, 573, 596, 619, 643, 667, 692
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n + 1 + Sum[Floor[(n - k)/Sqrt[5]], {k, 0, n}], {n, 0, 200}] (* A022780 *)
    (* Clark Kimberling, Mar 14 2015 *)

Formula

a(n) = n + 1 + Sum{floor[(n - k)/sqrt(5)], k = 0..n}.

A283940 Interspersion of the signature sequence of sqrt(3).

Original entry on oeis.org

1, 3, 2, 7, 5, 4, 13, 10, 8, 6, 20, 17, 14, 11, 9, 29, 25, 22, 18, 15, 12, 40, 35, 31, 27, 23, 19, 16, 53, 47, 42, 37, 33, 28, 24, 21, 67, 61, 55, 49, 44, 39, 34, 30, 26, 83, 76, 70, 63, 57, 51, 46, 41, 36, 32, 101, 93, 86, 79, 72, 65, 59, 54, 48, 43, 38
Offset: 1

Views

Author

Clark Kimberling, Mar 19 2017

Keywords

Comments

Row n is the ordered sequence of numbers k such that A007337(k)=n. As a sequence, A283940 is a permutation of the positive integers. This is a transposable interspersion; i.e., every row intersperses all other rows, and every column intersperses all other columns.

Examples

			Northwest corner:
  1   3    7    13   20   29   40   53
  2   5    10   17   25   35   47   61
  4   8    14   22   31   42   55   70
  6   11   18   27   37   49   63   79
  9   15   23   33   44   57   72   89
  12  19   28   39   51   65   81   99
  16  24   34   46   59   74   91   110
  21  30   41   54   68   84   102  122
		

Crossrefs

Programs

  • Mathematica
    r = Sqrt[3]; z = 100;
    s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
    u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A022778, col 1 of A283940 *)
    v = Table[s[n], {n, 0, z}] (* A022777, row 1 of A283940*)
    w[i_, j_] := u[[i]] + v[[j]] + (i - 1)*(j - 1) - 1;
    Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283940, array *)
    Flatten[Table[w[k, n - k + 1], {n, 1, 20}, {k, 1, n}]] (* A283940, sequence *)
  • PARI
    r = sqrt(3);
    z = 100;
    s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));
    p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
    u = v = vector(z + 1);
    for(n=1, 101, (v[n] = s(n - 1)));
    for(n=1, 101, (u[n] = p(n - 1)));
    w(i, j) = u[i] + v[j] + (i - 1) * (j - 1) - 1;
    tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(k, n - k + 1), ", "); );print(); ); };
    tabl(10) \\ Indranil Ghosh, Mar 21 2017
    
  • Python
    r = 3 ** 0.5
    def s(n): return 1 if n<1 else s(n - 1) + 1 + int(n*r)
    def p(n): return n + 1 + sum([int((n - k)/r) for k in range(0, n+1)])
    v=[s(n) for n in range(0, 101)]
    u=[p(n) for n in range(0, 101)]
    def w(i,j): return u[i - 1] + v[j - 1] + (i - 1) * (j - 1) - 1
    for n in range(1, 11):
        print ([w(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 21 2017
    
  • Python
    import numpy as np
    r = np.sqrt(3)
    x = np.arange(11)
    u = np.cumsum(np.ceil(x / r)).astype(int)
    v = np.cumsum(np.ceil(x * r)).astype(int)
    print(*[1 + u[k] + v[n-k] + k*(n-k) for n in range(11) for k in range(n+1)], sep=', ')
    # David Radcliffe, May 10 2025
Showing 1-4 of 4 results.