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 1-3 of 3 results.

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 25, 12, 15, 10, 7, 32, 81, 54, 125, 36, 75, 50, 49, 24, 45, 30, 35, 20, 21, 14, 11, 64, 243, 162, 625, 108, 375, 250, 343, 72, 225, 150, 245, 100, 147, 98, 121, 48, 135, 90, 175, 60, 105, 70, 77, 40, 63, 42, 55, 28, 33, 22, 13, 128
Offset: 0

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Author

Leroy Quet, Jul 29 2009

Keywords

Comments

This is a permutation of the positive integers.
From Antti Karttunen, Jun 20 2014: (Start)
Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.
Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.
Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:
1
|
...................2...................
4 3
8......../ \........9 6......../ \........5
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
16 27 18 25 12 15 10 7
32 81 54 125 36 75 50 49 24 45 30 35 20 21 14 11
etc.
Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.
(End)
Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020
From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)
This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).
Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?
Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)
From Antti Karttunen, Jul 25 2023: (Start)
(In the above question, it is assumed that the starting offset would be 1 instead of 0).
Questions:
Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?
It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.
(End)
If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023
Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.
See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

Examples

			For n=3, whose binary representation is "11", we have A000120(3)=2, with A163510(3,1) = A163510(3,2) = 0, thus a(3) = p(2) * p(1)^0 * p(2)^0 = 3*1*1 = 3.
For n=9, "1001" in binary, we have A000120(9)=2, with A163510(9,1) = 0 and A163510(9,2) = 2, thus a(9) = p(2) * p(1)^0 * p(2)^2 = 3*1*9 = 27.
For n=10, "1010" in binary, we have A000120(10)=2, with A163510(10,1) = 1 and A163510(10,2) = 1, thus a(10) = p(2) * p(1)^1 * p(2)^1 = 3*2*3 = 18.
For n=15, "1111" in binary, we have A000120(15)=4, with A163510(15,1) = A163510(15,2) = A163510(15,3) = A163510(15,4) = 0, thus a(15) = p(4) * p(1)^0 * p(2)^0 * p(3)^0 * p(4)^0 = 7*1*1*1*1 = 7.
[1], [2], [1,1], [3], [1,2], [2,1] ... -> [1], [2], [3], [1,2], ... -> [0], [1], [2], [0,1], ... -> 2^0, 2^1, 2^2, 2^0*3^1, ... = 1, 2, 4, 3, ... - _Lorenzo Sauras Altuzarra_, Nov 28 2020
		

Crossrefs

Inverse: A243071.
Cf. A007283 (known positions where a(n)=n), A029747, A029748, A364255 [= gcd(n,a(n))], A364258 [= a(n)-n], A364287 (where a(n) < n), A364292 (where a(n) <= n), A364494 (where n|a(n)), A364496 (where a(n)|n), A364963, A364297.
Cf. A365808 (positions of squares), A365801 (of cubes), A365802 (of fifth powers), A365805 [= A052409(a(n))], A366287, A366391.
Cf. A005940, A332214, A332817, A366275 (variants).

Programs

  • Mathematica
    f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; {1}~Join~
    Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}] (* Michael De Vlieger, Jul 25 2016 *)
  • Python
    from sympy import prime
    def A163511(n):
        if n:
            k, c, m = n, 0, 1
            while k:
                c += 1
                m *= prime(c)**(s:=(~k&k-1).bit_length())
                k >>= s+1
            return m*prime(c)
        return 1 # Chai Wah Wu, Jul 17 2023

Formula

For n >= 1, a(2n) is even, a(2n+1) is odd. a(2^k) = 2^(k+1), for all k >= 0.
From Antti Karttunen, Jun 20 2014: (Start)
a(0) = 1, a(1) = 2, a(2n) = 2*a(n), a(2n+1) = A003961(a(n)).
As a more general observation about the parity, we have:
For n >= 1, A007814(a(n)) = A135523(n) = A007814(n) + A209229(n). [This permutation preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is incremented by one.]
For n >= 1, A055396(a(n)) = A091090(n) = A007814(n+1) + 1 - A036987(n).
For n >= 1, a(A000225(n)) = A000040(n).
(End)
From Antti Karttunen, Oct 11 2014: (Start)
As a composition of related permutations:
a(n) = A005940(1+A054429(n)).
a(n) = A064216(A245612(n))
a(n) = A246681(A246378(n)).
Also, for all n >= 0, it holds that:
A161511(n) = A243503(a(n)).
A243499(n) = A243504(a(n)).
(End)
More linking identities from Antti Karttunen, Dec 30 2017: (Start)
A046523(a(n)) = A278531(n). [See also A286531.]
A278224(a(n)) = A285713(n). [Another filter-sequence.]
A048675(a(n)) = A135529(n) seems to hold for n >= 1.
A250245(a(n)) = A252755(n).
A252742(a(n)) = A252744(n).
A245611(a(n)) = A253891(n).
A249824(a(n)) = A275716(n).
A292263(a(n)) = A292264(n). [A292944(n) + A292264(n) = n.]
--
A292383(a(n)) = A292274(n).
A292385(a(n)) = A292271(n). [A292271(n) + A292274(n) = n.]
--
A292941(a(n)) = A292942(n).
A292943(a(n)) = A292944(n).
A292945(a(n)) = A292946(n). [A292942(n) + A292944(n) + A292946(n) = n.]
--
A292253(a(n)) = A292254(n).
A292255(a(n)) = A292256(n). [A292944(n) + A292254(n) + A292256(n) = n.]
--
A279339(a(n)) = A279342(n).
a(A071574(n)) = A269847(n).
a(A279341(n)) = A279338(n).
a(A252756(n)) = A250246(n).
(1+A008836(a(n)))/2 = A059448(n).
(End)
From Antti Karttunen, Jul 26 2023: (Start)
For all n >= 0, a(A007283(n)) = A007283(n).
A001222(a(n)) = A290251(n).
(End)

Extensions

More terms computed and examples added by Antti Karttunen, Jun 20 2014

A293430 Persistently squarefree numbers for base-2 shifting: Numbers n such that all terms in finite set [n, floor(n/2), floor(n/4), floor(n/8), ..., 1] are squarefree.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 21, 22, 23, 26, 29, 30, 31, 42, 43, 46, 47, 53, 58, 59, 61, 62, 85, 86, 87, 93, 94, 95, 106, 107, 118, 119, 122, 123, 170, 173, 174, 186, 187, 190, 191, 213, 214, 215, 237, 238, 239, 246, 247, 341, 346, 347, 349, 373, 374, 381, 382, 383, 426, 427, 429, 430, 431, 474, 478, 479
Offset: 1

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Author

Keywords

Comments

Question: Is this sequence infinite? (My guess: yes). This is equivalent to questions asked in A293230. See also comments at A293441 and A293517.
For any odd n that is present, 2n is also present.

Examples

			For 479 we see that 479 is prime (thus squarefree, in A005117), [479/2] = 239 is also a prime, [239/2] = 119 = 7*17 (a squarefree composite), [119/2] = 59 (a prime), [59/2] = 29 (a prime), [29/2] = 14 = 2*7 (a squarefree composite), [14/2] = 7 (a prime), [7/2] = 3 (a prime), [3/2] = 1 (the end of halving process 1 is also squarefree), thus all the values obtained by repeated halving were squarefree and 479 is a member of this sequence. Here [ ] stands for taking floor.
		

Crossrefs

Marked terms in the binary tree illustration of A293230.
Subsequence of A293427 (thus also of A003754 and of A005117).
Positions of nonzero terms in A293233.
Cf. A293441, A293517, A293523 (for floor(n/3^k) analog), A293437 (for a subsequence).

Programs

  • Mathematica
    With[{s = Fold[Append[#1, MoebiusMu[#2] #1[[Floor[#2/2]]]] &, {1}, Range[2, 480]]}, Flatten@ Position[s, ?(# != 0 &)]] (* _Michael De Vlieger, Oct 10 2017 *)
  • PARI
    is_persistently_squarefree(n,base) = { while(n>1, if(!issquarefree(n),return(0)); n \= base); (1); };
    isA293430(n) = is_persistently_squarefree(n,2);
    n=0; k=1; while(k <= 10000, n=n+1; if(isA293430(n),write("b293430.txt", k, " ", n);k=k+1)); \\ Antti Karttunen, Oct 11 2017

A280873 Numbers whose binary expansion does not begin 10 and does not contain 2 adjacent 0's; Ahnentafel numbers of X-chromosome inheritance of a male.

Original entry on oeis.org

0, 1, 3, 6, 7, 13, 14, 15, 26, 27, 29, 30, 31, 53, 54, 55, 58, 59, 61, 62, 63, 106, 107, 109, 110, 111, 117, 118, 119, 122, 123, 125, 126, 127, 213, 214, 215, 218, 219, 221, 222, 223, 234, 235, 237, 238, 239, 245, 246, 247, 250, 251, 253, 254, 255
Offset: 0

Views

Author

Floris Strijbos, Jan 09 2017

Keywords

Comments

The number of ancestors at generation m from whom a living individual may have received an X chromosome allele is F_m, the m-th term of the Fibonacci Sequence.
From Antti Karttunen, Oct 11 2017: (Start)
The starting offset is zero (with a(0) = 0) for the same reason that we have A003714(0) = 0. Indeed, b(n) = A054429(A003714(n)) for n >= 0 yields the terms of this sequence, but in different order.
A163511(a(n)) for n >= 0 gives a permutation of squarefree numbers (A005117). See also A277006.
(End)

Crossrefs

Intersection of A003754 and A004760.
Positions where A163511 obtains squarefree (A005117) values.
Cf. also A293437 (a subsequence).

Programs

  • Maple
    gen[0]:= {0,1,3}:
    gen[1]:= {6,7}:
    for n from 2 to 10 do
      gen[n]:= map(t -> 2*t+1, gen[n-1]) union
          map(t -> 2*t, select(type, gen[n-1],odd))
    od:
    sort(convert(`union`(seq(gen[i],i=0..10)),list)); # Robert Israel, Oct 11 2017
  • Mathematica
    male = {1, 3}; generations = 8;
    Do[x = male[[i - 1]]; If[EvenQ[x],
                              male = Append[ male,   2*x + 1] ,
                              male = Flatten[Append[male, {2*x, 2*x + 1}]]]
           , {i, 3, Fibonacci[generations + 1]}]; male
  • PARI
    isA003754(n) = { n=bitor(n, n>>1)+1; n>>=valuation(n, 2); (n==1); }; \\ After Charles R Greathouse IV's Feb 06 2017 code.
    isA004760(n) = (n<2 || (binary(n)[2])); \\ This function also from Charles R Greathouse IV, Sep 23 2012
    isA280873(n) = (isA003754(n) && isA004760(n));
    n=0; k=0; while(k <= 10946, if(isA280873(n),write("b280873.txt", k, " ", n);k=k+1); n=n+1;); \\ Antti Karttunen, Oct 11 2017
    
  • Python
    def A280873():
        yield 1
        for x in A280873():
            if ((x & 1) and (x > 1)):
                yield 2*x
            yield 2*x+1
    def take(n, g):
      '''Returns a list composed of the next n elements returned by generator g.'''
      z = []
      if 0 == n: return(z)
      for x in g:
        z.append(x)
        if n > 1: n = n-1
        else: return(z)
    take(120, A280873())
    # Antti Karttunen, Oct 11 2017, after the given Mathematica-code (by Floris Strijbos) and a similar generator-example for A003714 by David Eppstein (cf. "Self-recursive generators" link).

Formula

{a(n) : n >= 1} = {k >= 1 : A365538(A054429(k)) > 0}. - Peter Munn, Jan 22 2024

Extensions

a(0) = 0 prepended and more descriptive alternative name added by Antti Karttunen, Oct 11 2017
Showing 1-3 of 3 results.