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.

A181815 a(n) = largest integer such that, when any of its divisors divides A025487(n), the quotient is a member of A025487.

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 12, 5, 32, 9, 24, 10, 64, 18, 48, 20, 128, 36, 15, 96, 7, 27, 40, 256, 72, 30, 192, 14, 54, 80, 512, 144, 60, 384, 28, 108, 25, 160, 1024, 45, 288, 21, 81, 120, 768, 56, 216, 50, 320, 2048, 90, 576, 11, 42, 162, 240, 1536, 112, 432, 100, 640, 4096, 180, 1152
Offset: 1

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Author

Matthew Vandermast, Nov 30 2010

Keywords

Comments

A permutation of the natural numbers.
The number of divisors of a(n) equals the number of ordered factorizations of A025487(n) as A025487(j)*A025487(k). Cf. A182762.
From Antti Karttunen, Dec 28 2019: (Start)
Rearranges terms of A108951 into ascending order, as A108951(a(n)) = A025487(n).
The scatter plot looks like a curtain of fractal spray, which is typical for many of the so-called "entanglement permutations". Indeed, according to the terminology I use in my 2016-2017 paper, this sequence is obtained by entangling the complementary pair (A329898, A330683) with the complementary pair (A005843, A003961), which gives the following implicit recurrence: a(A329898(n)) = 2*a(n) and a(A330683(n)) = A003961(a(n)). An explicit form is given in the formula section.
(End)

Examples

			For any divisor d of 9 (d = 1, 3, 9), 36/d (36, 12, 4) is a member of A025487. 9 is the largest number with this relationship to 36; therefore, since 36 = A025487(11), a(11) = 9.
		

Crossrefs

If this sequence is considered the "primorial deflation" of A025487(n) (see first formula), the primorial inflation of n is A108951(n), and the primorial inflation of A025487(n) is A181817(n).
A181820(n) is another mapping from the members of A025487 to the positive integers.

Programs

  • Mathematica
    (* First, load the program at A025487, then: *)
    Map[If[OddQ@ #, 1, Times @@ Prime@ # &@ Rest@ NestWhile[Append[#1, {#3, Drop[#, -LengthWhile[Reverse@ #, # == 0 &]] &[#2 - PadRight[ConstantArray[1, #3], Length@ #2]]}] & @@ {#1, #2, LengthWhile[#2, # > 0 &]} & @@ {#, #[[-1, -1]]} &, {{0, TakeWhile[If[# == 1, {0}, Function[g, ReplacePart[Table[0, {PrimePi[g[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, g]]@ FactorInteger@ #], # > 0 &]}}, And[FreeQ[#[[-1, -1]], 0], Length[#[[-1, -1]] ] != 0] &][[All, 1]] ] &, Union@ Flatten@ f@ 6] (* Michael De Vlieger, Dec 28 2019 *)
  • PARI
    A181815(n) = A329900(A025487(n)); \\ Antti Karttunen, Dec 24 2019

Formula

If A025487(n) is considered in its form as Product A002110(i)^e(i), then a(n) = Product p(i)^e(i). If A025487(n) is instead considered as Product p(i)^e(i), then a(n) = Product (p(i)/A008578(i))^e(i).
a(n) = A122111(A181820(n)). - Matthew Vandermast, May 21 2012
From Antti Karttunen, Dec 24-29 2019: (Start)
a(n) = Product_{i=1..A051282(n)} A000040(A304886(i)).
a(n) = A329900(A025487(n)) = A319626(A025487(n)).
a(n) = A163511(A329905(n)).
For n > 1, if A330682(n) = 1, then a(n) = A003961(a(A329904(n))), otherwise a(n) = 2*a(A329904(n)).
A252464(a(n)) = A329907(n).
A330690(a(n)) = A050378(n).
a(A306802(n)) = A329902(n).
(End)

A329904 Position of A329899(n) in A025487.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Dec 24 2019

Keywords

Comments

Restricted growth sequence transform of A329899.
Each positive natural number occurs exactly twice.
For all i, j > 1:
a(i) = a(j) => A329907(i) = A329907(j).

Crossrefs

Programs

Formula

a(n) = A085089(A329899(n)).

A329906 a(0) = 1; a(1) = 2; after which a(2n) = A329898(a(n)), a(2n+1) = A330683(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 11, 6, 9, 7, 23, 15, 38, 8, 20, 13, 22, 10, 44, 30, 110, 19, 69, 49, 128, 12, 41, 27, 72, 17, 43, 29, 54, 14, 79, 56, 272, 37, 181, 136, 482, 26, 118, 86, 307, 61, 208, 156, 424, 16, 73, 52, 190, 34, 123, 89, 242, 24, 77, 55, 147, 36, 93, 66, 114, 18, 131, 97, 596, 68, 416, 323, 1448, 48, 286, 218, 990, 164, 711
Offset: 0

Views

Author

Antti Karttunen, Dec 24 2019

Keywords

Comments

Note the indexing: domain begins from zero, but the range does not include it.

Examples

			This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A329898 the parent, and each child to the right is obtained by applying A330683 to the parent:
                                      1
                                      |
                   ...................2...................
                  3                                       4
        5......../ \........11                  6......../ \........9
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    7       23         15       38          8       20         13       22
  10 44   30  110    19  69    49 128     12 41   27  72     17  43   29  54
etc.
		

Crossrefs

Cf. A329905 (inverse permutation).

Programs

Formula

a(0) = 1; a(1) = 2; after which a(2n) = A329898(a(n)), a(2n+1) = A330683(a(n)).
a(n) = A329901(A163511(n)).

A329907 Number of iterations of A329904 needed to reach 1.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 3, 5, 3, 5, 4, 6, 4, 6, 5, 7, 5, 4, 7, 4, 4, 6, 8, 6, 5, 8, 5, 5, 7, 9, 7, 6, 9, 6, 6, 4, 8, 10, 5, 8, 5, 5, 7, 10, 7, 7, 5, 9, 11, 6, 9, 5, 6, 6, 8, 11, 8, 8, 6, 10, 12, 7, 10, 6, 7, 7, 5, 9, 12, 5, 6, 9, 9, 7, 6, 11, 6, 13, 8, 11, 7, 8, 8, 6, 10, 13, 6, 7, 10, 10, 6, 8, 7, 12, 7, 14, 9, 12, 8, 9, 9, 7, 11
Offset: 1

Views

Author

Antti Karttunen, Dec 24 2019

Keywords

Comments

Equally, starting from A025487(n), number of iterations of A329899 needed to reach 1.
Any k > 0 occurs 2^(k-1) times in total in this sequence.

Crossrefs

Programs

Formula

a(1) = 0; for n > 1, a(n) = 1 + a(A329904(n)).
a(1) = 0; for n > 1, a(n) = A070939(A329905(n)).
a(n) = A252464(A181815(n)).
For all n >= 1, a(n) >= A061394(n).
Showing 1-4 of 4 results.