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.

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A349552 a(n) is the number of halving partitions of n (see Comments for definition).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 3, 1, 3, 3, 5, 2, 5, 3, 4, 1, 4, 3, 6, 3, 7, 5, 7, 2, 6, 5, 8, 3, 7, 4, 5, 1, 4, 4, 7, 3, 9, 6, 9, 3, 9, 7, 12, 5, 11, 7, 9, 2, 8, 6, 11, 5, 12, 8, 11, 3, 9, 7, 11, 4, 9, 5, 6, 1, 4, 4, 8, 4, 10, 7, 11, 3, 11, 9, 15, 6, 15, 9, 12, 3, 10, 9, 16, 7, 18, 12, 17, 5, 15, 11, 18, 7, 15, 9, 11, 2, 8, 8, 14, 6
Offset: 0

Views

Author

Clark Kimberling, Dec 26 2021

Keywords

Comments

For m >= 1, let S(m) = { f(m/2), c(m/2) }, where f = floor and c = ceiling. A halving partition of n is a partition p(1) + p(2) + ... + p(k) of n such that p(1) is in S(n) and p(i) is in S(p(i-1)) for i = 2, 3, ..., k. The basic idea is that each term after the first is about half of the preceding term.

Examples

			a(9) = 3 counts these partitions:
  c(9/2) + f(5/2) + (2/2) + c(1/2) = 5 + 2 + 1 + 1;
  c(9/2) + c(5/2) + f(3/2) = 5 + 3 + 1;
  f(9/2) + (4/2) + (2/2) + c(1/2) = 4 + 2 + 1 + 1.
a(13) = 5 counts these partitions:
  c(13/2) + c(7/2) + (4/2) = 7 + 4 + 2;
  c(13/2) + f(7/2) + c(3/2) + (2/2) = 7 + 3 + 2 + 1;
  c(13/2) + f(7/2) + f(3/2) + (2/2) + c(1/2) = 7 + 3 + 1 + 1 + 1;
  f(13/2) + (6/2) + c(3/2) + (2/2) + c(1/2) = 6 + 3 + 2 + 1 + 1;
  f(13/2) + (6/2) + f(3/2) + (2/2) + c(1/2) + c(1/2) = 6 + 3 + 1 + 1 + 1 + 1.
		

Crossrefs

Programs

  • PARI
    { a349552(n,p=n) = if(n==0,1,if(n<0||p==0,0,if(p%2,a(n-p\2-1,p\2+1))+a(n-p\2,p\2))); } \\ Max Alekseyev, Sep 30 2024

Formula

From Alois P. Heinz, Sep 30 2024: (Start)
a(A000079(n)) = 1.
a(A000225(n)) = A028310(n). (End)

Extensions

Corrected and extended by Max Alekseyev, Sep 30 2024

A349550 Meta-Wythoff array based on A097285: M = (M(n,k)), by downward antidiagonals; every row of M is eventually a row of the Wythoff array, W = A035513, and every row of W is a row of M; see Comments.

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 5, 4, 3, 1, 8, 7, 5, 4, 2, 13, 11, 8, 5, 4, 3, 21, 18, 13, 9, 6, 4, 1, 34, 29, 21, 14, 10, 7, 5, 2, 55, 47, 34, 23, 16, 11, 6, 5, 3, 89, 76, 55, 37, 26, 18, 11, 7, 5, 4, 144, 123, 89, 60, 42, 29, 17, 12, 8, 5, 1, 233, 199, 144, 97, 68, 47
Offset: 1

Views

Author

Clark Kimberling, Nov 21 2021

Keywords

Comments

Suppose that (s(1), s(2), ...) is a sequence satisfying s(k) = s(k-1) + s(k-2) for k >= 3. If s(1) and s(2) are positive integers, then there is an index n such that (s(n), s(n+1), ...) is a row of A035513. The n-th row of M is the sequence (s(1), s(2), ...), where (s(1), s(2)) are the n-th pair described in A097285.
Every row of W is a row of M; indeed, M consists of all tails of all rows of W.

Examples

			Corner:
  1, 2, 3,  5,  8, 13, 21, 34,  55,  89, 144, 233
  1, 3, 4,  7, 11, 18, 29, 47,  76, 123, 199, 322
  2, 3, 5,  8, 13, 21, 34, 55,  89, 144, 233, 377
  1, 4, 5,  9, 14, 23, 37, 60,  97, 157, 254, 411
  2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466
  3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521
  1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500
  2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555
  3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
  4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665
Example: The first 7 pairs in A097285 are (1,2), (1,3), (2,3), (1,4), (2,4), (3,4), (1,5), so that the first 7 rows of M are
(1,2,3,5,8,...) = (row 1 of W) = Fibonacci numbers, A000045;
(1,3 4,7,11,...), which includes row 2 of W, the Lucas numbers, A000032;
(2,3,5,8,13,...), a tail of row 1 of W;
(1,4,5,9,14,...), which includes row 4 of W;
(2,4,6,10,16,...), which includes row 3 of W;
(3,4,7,11,18,...), which includes row 2 of W;
(1,5,6,11,17,...), which includes row 7 of W.
		

Crossrefs

Programs

  • Mathematica
    z1 = 30; zc = 20; zr = 20;
    t1 = {1, 2}; Do[t1 = Join[t1, Riffle[Range[n - 1], n], {n}], {n, 3, z1}]; (* A097285 *)
    t = Partition[t1, 2];
    f[n_] := Fibonacci[n]; r = (1 + Sqrt[5])/2;
    s[h_, k_] := Table[h*f[n - 1] + k*f[n], {n, 2, zc}];
    w = Table[Join[{h = t[[n]][[1]], k = t[[n]][[2]]}, s[h, k]], {n, 1, zr}]
    TableForm[w]  (* A349550 array *)
    w1[n_, k_] := w[[n]][[k]];
    Table[w1[n - k + 1, k], {n, 13}, {k, n, 1, -1}] // Flatten (* A349550 sequence *)
Showing 1-2 of 2 results.