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 1-6 of 6 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

Views

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

A066099 Triangle read by rows, in which row n lists the compositions of n in reverse lexicographic order.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 3, 1, 2, 2, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 5, 4, 1, 3, 2, 3, 1, 1, 2, 3, 2, 2, 1, 2, 1, 2, 2, 1, 1, 1, 1, 4, 1, 3, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 6, 5, 1, 4, 2, 4, 1, 1, 3, 3, 3, 2, 1, 3, 1, 2, 3, 1, 1, 1, 2, 4, 2, 3
Offset: 1

Views

Author

Alford Arnold, Dec 30 2001

Keywords

Comments

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed lexicographic; see the example by Omar E. Pol. - Joerg Arndt, Sep 03 2013
This is the standard ordering for compositions in this database; it is similar to the Mathematica ordering for partitions (A080577). Other composition orderings include A124734 (similar to the Abramowitz & Stegun ordering for partitions, A036036), A108244 (similar to the Maple partition ordering, A080576), etc (see crossrefs).
Factorize each term in A057335; sequence records the values of the resulting exponents. It also runs through all possible permutations of multiset digits.
This can be regarded as a table in two ways: with each composition as a row, or with the compositions of each integer as a row. The first way has A000120 as row lengths and A070939 as row sums; the second has A001792 as row lengths and A001788 as row sums. - Franklin T. Adams-Watters, Nov 06 2006
This sequence includes every finite sequence of positive integers. - Franklin T. Adams-Watters, Nov 06 2006
Compositions (or ordered partitions) are also generated in sequence A101211. - Alford Arnold, Dec 12 2006
The equivalent sequence for partitions is A228531. - Omar E. Pol, Sep 03 2013
The sole partition of zero has no components, not a single component of length one. Hence the first nonempty row is row 1. - Franklin T. Adams-Watters, Apr 02 2014 [Edited by Andrey Zabolotskiy, May 19 2018]
See sequence A261300 for another version where the terms of each composition are concatenated to form one single integer: (0, 1, 2, 11, 3, 21, 12, 111,...). This also shows how the terms can be obtained from the binary numbers A007088, cf. Arnold's first Example. - M. F. Hasler, Aug 29 2015
The k-th composition in the list is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This is described as the standard ordering used in the OEIS, although the sister sequence A228351 is also sometimes considered to be canonical. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, May 19 2020
First differences of A030303 = positions of bits 1 in the concatenation A030190 (= A030302) of numbers written in binary (A007088). - Indices of record values (= first occurrence of n) are given by A005183: a(A005183(n)) = n, cf. FORMULA for more. - M. F. Hasler, Oct 12 2020
The geometric mean approaches the Somos constant (A112302). - Jwalin Bhatt, Feb 10 2025

Examples

			A057335 begins 1 2 4 6 8 12 18 30 16 24 36 ... so we can write
  1 2 1 3 2 1 1 4 3 2 2 1 1 1 1 ...
  . . 1 . 1 2 1 . 1 2 1 3 2 1 1 ...
  . . . . . . 1 . . . 1 . 1 2 1 ...
  . . . . . . . . . . . . . . 1 ...
and the columns here gives the rows of the triangle, which begins
  1
  2; 1 1
  3; 2 1; 1 2; 1 1 1
  4; 3 1; 2 2; 2 1 1; 1 3; 1 2 1; 1 1 2; 1 1 1 1
  ...
From _Omar E. Pol_, Sep 03 2013: (Start)
Illustration of initial terms:
  -----------------------------------
  n  j       Diagram   Composition j
  -----------------------------------
  .               _
  1  1           |_|   1;
  .             _ _
  2  1         |  _|   2,
  2  2         |_|_|   1, 1;
  .           _ _ _
  3  1       |    _|   3,
  3  2       |  _|_|   2, 1,
  3  3       | |  _|   1, 2,
  3  4       |_|_|_|   1, 1, 1;
  .         _ _ _ _
  4  1     |      _|   4,
  4  2     |    _|_|   3, 1,
  4  3     |   |  _|   2, 2,
  4  4     |  _|_|_|   2, 1, 1,
  4  5     | |    _|   1, 3,
  4  6     | |  _|_|   1, 2, 1,
  4  7     | | |  _|   1, 1, 2,
  4  8     |_|_|_|_|   1, 1, 1, 1;
(End)
		

Crossrefs

Lists of compositions of integers: this sequence (reverse lexicographic order; minus one gives A108730), A228351 (reverse colexicographic order - every composition is reversed; minus one gives A163510), A228369 (lexicographic), A228525 (colexicographic), A124734 (length, then lexicographic; minus one gives A124735), A296774 (length, then reverse lexicographic), A337243 (length, then colexicographic), A337259 (length, then reverse colexicographic), A296773 (decreasing length, then lexicographic), A296772 (decreasing length, then reverse lexicographic), A337260 (decreasing length, then colexicographic), A108244 (decreasing length, then reverse colexicographic), also A101211 and A227736 (run lengths of bits).
Cf. row length and row sums for different splittings into rows: A000120, A070939, A001792, A001788.
Cf. lists of partitions of integers, or multisets of integers: A026791 and crosserfs therein, A112798 and crossrefs therein.
See link for additional crossrefs pertaining to standard compositions.
A related ranking of finite sets is A048793/A272020.

Programs

  • Haskell
    a066099 = (!!) a066099_list
    a066099_list = concat a066099_tabf
    a066099_tabf = map a066099_row [1..]
    a066099_row n = reverse $ a228351_row n
    -- (each composition as a row)
    -- Peter Kagey, Aug 25 2016
    
  • Mathematica
    Table[FactorInteger[Apply[Times, Map[Prime, Accumulate @ IntegerDigits[n, 2]]]][[All, -1]], {n, 41}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)
    stc[n_] := Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]] // Reverse;
    Table[stc[n], {n, 0, 20}] // Flatten (* Gus Wiseman, May 19 2020 *)
    Table[Reverse @ LexicographicSort @ Flatten[Permutations /@ Partitions[n], 1], {n, 10}] // Flatten (* Eric W. Weisstein, Jun 26 2023 *)
  • PARI
    arow(n) = {local(v=vector(n),j=0,k=0);
       while(n>0,k++; if(n%2==1,v[j++]=k;k=0);n\=2);
       vector(j,i,v[j-i+1])} \\ returns empty for n=0. - Franklin T. Adams-Watters, Apr 02 2014
    
  • Python
    from itertools import islice
    from itertools import accumulate, count, groupby, islice
    def A066099_gen():
        for i in count(1):
            yield [len(list(g)) for _,g in groupby(accumulate(int(b) for b in bin(i)[2:]))]
    A066099 = list(islice(A066099_gen(), 120))  # Jwalin Bhatt, Feb 28 2025
  • Sage
    def a_row(n): return list(reversed(Compositions(n)))
    flatten([a_row(n) for n in range(1,6)]) # Peter Luschny, May 19 2018
    

Formula

From M. F. Hasler, Oct 12 2020: (Start)
a(n) = A030303(n+1) - A030303(n).
a(A005183(n)) = n; a(A005183(n)+1) = n-1 (n>1); a(A005183(n)+2) = 1. (End)

Extensions

Edited with additional terms by Franklin T. Adams-Watters, Nov 06 2006
0th row removed by Andrey Zabolotskiy, May 19 2018

A228351 Triangle read by rows in which row n lists the compositions (ordered partitions) of n (see Comments lines for definition).

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 4, 1, 3, 2, 2, 1, 1, 2, 3, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 5, 1, 4, 2, 3, 1, 1, 3, 3, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 4, 1, 1, 3, 1, 2, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 5, 2, 4, 1, 1, 4
Offset: 1

Views

Author

Omar E. Pol, Aug 30 2013

Keywords

Comments

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed co-lexicographic. - Joerg Arndt, Sep 02 2013
Dropping the "(list-)reversed" in the comment above gives A228525.
The equivalent sequence for partitions is A026792.
This sequence lists (without repetitions) all finite compositions, in such a way that, if [P_1, ..., P_r] denotes the composition occupying the n-th position in the list, then (((2*n/2^(P_1)-1)/2^(P_2)-1)/...)/2^(P_r)-1 = 0. - Lorenzo Sauras Altuzarra, Jan 22 2020
The k-th composition in the list is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, and taking first differences. Reversing again gives A066099, which is described as the standard ordering. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, Apr 01 2020
It follows from the previous comment that A000120(k) is the length of the k-th composition that is listed by this sequence (recall that A000120(k) is the number of 1's in the binary expansion of k). - Lorenzo Sauras Altuzarra, Sep 29 2020

Examples

			Illustration of initial terms:
-----------------------------------
n  j     Diagram     Composition j
-----------------------------------
.         _
1  1     |_|         1;
.         _ _
2  1     |_  |       2,
2  2     |_|_|       1, 1;
.         _ _ _
3  1     |_    |     3,
3  2     |_|_  |     1, 2,
3  3     |_  | |     2, 1,
3  4     |_|_|_|     1, 1, 1;
.         _ _ _ _
4  1     |_      |   4,
4  2     |_|_    |   1, 3,
4  3     |_  |   |   2, 2,
4  4     |_|_|_  |   1, 1, 2,
4  5     |_    | |   3, 1,
4  6     |_|_  | |   1, 2, 1,
4  7     |_  | | |   2, 1, 1,
4  8     |_|_|_|_|   1, 1, 1, 1;
.
Triangle begins:
[1];
[2],[1,1];
[3],[1,2],[2,1],[1,1,1];
[4],[1,3],[2,2],[1,1,2],[3,1],[1,2,1],[2,1,1],[1,1,1,1];
[5],[1,4],[2,3],[1,1,3],[3,2],[1,2,2],[2,1,2],[1,1,1,2],[4,1],[1,3,1],[2,2,1],[1,1,2,1],[3,1,1],[1,2,1,1],[2,1,1,1],[1,1,1,1,1];
...
For example [1,2] occupies the 5th position in the corresponding list of compositions and indeed (2*5/2^1-1)/2^2-1 = 0. - _Lorenzo Sauras Altuzarra_, Jan 22 2020
12 --binary expansion--> [1,1,0,0] --reverse--> [0,0,1,1] --positions of 1's--> [3,4] --prepend 0--> [0,3,4] --first differences--> [3,1]. - _Lorenzo Sauras Altuzarra_, Sep 29 2020
		

Crossrefs

Row n has length A001792(n-1). Row sums give A001787, n >= 1.
Cf. A000120 (binary weight), A001511, A006519, A011782, A026792, A065120.
A related ranking of finite sets is A048793/A272020.
All of the following consider the k-th row to be the k-th composition, ignoring the coarser grouping by sum.
- Indices of weakly increasing rows are A114994.
- Indices of weakly decreasing rows are A225620.
- Indices of strictly decreasing rows are A333255.
- Indices of strictly increasing rows are A333256.
- Indices of reversed interval rows A164894.
- Indices of interval rows are A246534.
- Indices of strict rows are A233564.
- Indices of constant rows are A272919.
- Indices of anti-run rows are A333489.
- Row k has A124767(k) runs and A333381(k) anti-runs.
- Row k has GCD A326674(k) and LCM A333226(k).
- Row k has Heinz number A333219(k).
Equals A163510+1, termwise.
Cf. A124734 (increasing length, then lexicographic).
Cf. A296774 (increasing length, then reverse lexicographic).
Cf. A337243 (increasing length, then colexicographic).
Cf. A337259 (increasing length, then reverse colexicographic).
Cf. A296773 (decreasing length, then lexicographic).
Cf. A296772 (decreasing length, then reverse lexicographic).
Cf. A337260 (decreasing length, then colexicographic).
Cf. A108244 (decreasing length, then reverse colexicographic).
Cf. A228369 (lexicographic).
Cf. A066099 (reverse lexicographic).
Cf. A228525 (colexicographic).

Programs

  • Haskell
    a228351 n = a228351_list !! (n - 1)
    a228351_list = concatMap a228351_row [1..]
    a228351_row 0 = []
    a228351_row n = a001511 n : a228351_row (n `div` 2^(a001511 n))
    -- Peter Kagey, Jun 27 2016
    
  • Maple
    # Program computing the sequence:
    A228351 := proc(n) local c, k, L, N: L, N := [], [seq(2*r, r = 1 .. n)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), op(c)]: k := k-1: c := 0: fi: od: od: L[n]: end: # Lorenzo Sauras Altuzarra, Jan 22 2020
    # Program computing the list of compositions:
    List := proc(n) local c, k, L, M, N: L, M, N := [], [], [seq(2*r, r = 1 .. 2^n-1)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), c]: k := k-1: c := 0: fi: od: M := [op(M), L]: L := []: od: M: end: # Lorenzo Sauras Altuzarra, Jan 22 2020
  • Mathematica
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    Table[Differences[Prepend[bpe[n],0]],{n,0,30}] (* Gus Wiseman, Apr 01 2020 *)
  • Python
    from itertools import count, islice
    def A228351_gen(): # generator of terms
        for n in count(1):
            k = n
            while k:
                yield (s:=(~k&k-1).bit_length()+1)
                k >>= s
    A228351_list = list(islice(A228351_gen(),30)) # Chai Wah Wu, Jul 17 2023

A100922 k appears A000120(k) times (appearances equal number of 1-bits).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 29, 29, 29, 29, 30, 30, 30, 30
Offset: 0

Views

Author

Rick L. Shepherd, Nov 21 2004

Keywords

Comments

Clearly every positive integer appears at least once in this sequence.

Examples

			The binary representation of 16 is 10000, which has one 1-bit (and four 0-bits), hence 16 appears once in this sequence (and four times in A100921).
		

Crossrefs

Cf. A100921 (n's appearances equal its number of 0-bits), A030530 (n's appearances equal its total number of bits), A227737 (n's appearances equal its total number of runs), A000069, A000120, A000788, A163510, A243067.

Programs

  • Maple
    T:= n-> n$add(i, i=Bits[Split](n)):
    seq(T(n), n=1..30);  # Alois P. Heinz, Nov 11 2024
  • Mathematica
    Table[Table[n,DigitCount[n,2,1]],{n,30}]//Flatten (* Harvey P. Dale, Aug 31 2017 *)
  • Python
    def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1)
    def A100922(n):
        if n == 0: return 1
        m, k = 1, 1
        while A000788(m)<=n: m<<=1
        while m-k>1:
            r = m+k>>1
            if A000788(r)>n:
                m = r
            else:
                k = r
        return m # Chai Wah Wu, Nov 11 2024

Formula

a(n) = the least k such that A000788(k) > n. - Antti Karttunen, Jun 20 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = Sum_{n>=1} (-1)^(n+1)/A000069(n) = 0.67968268... . - Amiram Eldar, Feb 18 2024

A244215 a(0)=0, and for n>=1, if n is the k-th Catalan number (A000108(k)), a(n) = k, otherwise the difference of the indices of the two largest Catalan numbers whose sum is less than or equal to n: a(n) = A244160(n) - A244160(n-A081290(n)).

Original entry on oeis.org

0, 1, 2, 1, 0, 3, 2, 1, 1, 1, 0, 0, 0, 0, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6
Offset: 0

Views

Author

Antti Karttunen, Jun 23 2014

Keywords

Comments

After a(0), if the leftmost digit in the Greedy Catalan Base representation of n [= A014418(n)] is larger than 1, then a(n) = 0, otherwise one more than the distance to the next nonzero digit to the right, or to the end of the numeral, if no more nonzero digits are present (i.e., if n is one of the Catalan numbers).
When searching for the two largest Catalan numbers whose sum is less than or equal to n, we first maximize the larger of those two numbers, which is A081290(n) = A000108(A244160(n)), after which we will find the next largest Catalan number that still "fits into" n. - Antti Karttunen, Mar 21 2015

Crossrefs

A244216 gives the positions of zeros, A244217 the positions of nonzero terms.

Programs

Formula

a(n) = A244160(n) - A244160(n-A081290(n)).
For all n >= 1, a(A000108(n)) = n.

Extensions

Name improved by Antti Karttunen, Mar 21 2015

A243067 Integers from 0 to A000120(n)-1 followed by integers from 0 to A000120(n+1)-1 and so on, starting with n=1.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 2, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0
Offset: 1

Views

Author

Antti Karttunen, Jun 19 2014

Keywords

Examples

			For n=1, also 1 in binary notation, so the count of its 1-bits is 1 (A000120(1)=1), we list numbers from 0 to 0, thus just 0.
For n=2, 10 in binary, thus A000120(2)=1, we list numbers from 0 to 0, thus just 0.
For n=3, 11 in binary, thus A000120(3)=2, we list numbers from 0 to 1, and so we have the first four terms of the sequence: 0; 0; 0, 1;
		

Crossrefs

Programs

Formula

a(n) = n - (1 + A000788(A100922(n-1)-1)).
Showing 1-6 of 6 results.