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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A273494 a(n) = A245325(n) + A245326(n).

Original entry on oeis.org

2, 3, 3, 5, 4, 5, 4, 8, 7, 7, 5, 8, 7, 7, 5, 13, 11, 12, 9, 11, 10, 9, 6, 13, 11, 12, 9, 11, 10, 9, 6, 21, 18, 19, 14, 19, 17, 16, 11, 18, 15, 17, 13, 14, 13, 11, 7, 21, 18, 19, 14, 19, 17, 16, 11, 18, 15, 17, 13, 14, 13, 11, 7, 34, 29, 31, 23, 30, 27, 25, 17, 31, 26, 29, 22, 25, 23, 20, 13, 29, 25, 26, 19, 27, 24, 23, 16, 23
Offset: 1

Views

Author

Yosu Yurramendi, May 23 2016

Keywords

Comments

The terms (n>0) may be written as a left-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 5, 4,
8, 7, 7, 5, 8, 7, 7, 5,
13,11,12, 9,11,10, 9, 6,13,11,12, 9,11,10, 9, 6,
21,18,19,14,19,17,16,11,18,15,17,13,14,13,11, 7,21,18,19,14,19,17,...
All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).
The terms (n>0) may also be written as a right-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 5, 4,
8, 7, 7, 5, 8, 7, 7, 5,
13,11,12, 9,11,10, 9, 6,13,11,12, 9,11,10, 9, 6,
..., 18,15,17,13,14,13,11, 7,21,18,19,14,19,17,16,11,18,15,17,13,14,13,11, 7,
Each column is an arithmetic sequence. The differences of the arithmetic sequences give the sequence A071585: a(2^(m+1)-1-k) - a(2^m-1-k) = A071585(k), m >= 0, 0 <= k < 2^m.
n > 1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), which is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A245325(n)/A245326(n) is also an enumeration system of all positive rationals, and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306.
The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), A086592 (A020650+A020651), A268087 (A162909+A162910).

Crossrefs

Programs

  • PARI
    a(n) = my(x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, y], [x, y]=[y, x+y])); x \\ Mikhail Kurkov, Mar 10 2023

Formula

a(n) = A273493(A059893(n)), a(A059893(n)) = A273493(n), n > 0. - Yosu Yurramendi, May 30 2017
a(n) = A007306(A059893(A180200(n))) = A007306(A059894(A154435(n))). - Yosu Yurramendi, Sep 20 2021

A324337 a(n) = A002487(A006068(n)).

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 1, 3, 4, 3, 2, 5, 1, 4, 5, 3, 5, 4, 3, 7, 2, 7, 8, 5, 1, 5, 7, 4, 7, 5, 3, 8, 6, 5, 4, 9, 3, 10, 11, 7, 2, 9, 12, 7, 11, 8, 5, 13, 1, 6, 9, 5, 10, 7, 4, 11, 9, 7, 5, 12, 3, 11, 13, 8, 7, 6, 5, 11, 4, 13, 14, 9, 3, 13, 17, 10, 15, 11, 7, 18, 2, 11, 16, 9, 17, 12, 7, 19, 14, 11, 8, 19, 5, 18, 21, 13, 1, 7, 11, 6, 13, 9, 5, 14, 13, 10
Offset: 0

Views

Author

Antti Karttunen, Feb 23 2019

Keywords

Comments

Like in A324338, a few terms preceding each position n = 2^k seem to be a batch of nearby Fibonacci numbers in some order.
For all n > 0 A324338(n)/A324337(n) constitutes an enumeration system of all positive rationals. For all n > 0 A324338(n) + A324337(n) = A071585(n). - Yosu Yurramendi, Oct 22 2019

Crossrefs

Programs

Formula

From Yosu Yurramendi, Oct 22 2019: (Start)
a(2^m+ k) = A324338(2^m+2^(m-1)+k), m > 0, 0 <= k < 2^(m-1)
a(2^m+2^(m-1)+k) = A324338(2^m+ k), m > 0, 0 <= k < 2^(m-1). (End)
a(n) = A324338(A063946(n)), n > 0. Yosu Yurramendi, Nov 04 2019
a(n) = A002487(A248663(A283477(n))). - Antti Karttunen, Nov 06 2019
a(n) = A002487(1+A233279(n)). - Yosu Yurramendi, Nov 08 2019
From Yosu Yurramendi, Nov 28 2019: (Start)
a(2^(m+1)+k) - a(2^m+k) = A324338(k), m >= 0, 0 <= k < 2^m.
a(A059893(2^(m+1)+A001969(k+1))) - a(A059893(2^m+A001969(k+1))) = A071585(k), m >= 0, 0 <= k < 2^(m-1).
a(A059893(2^(m+1)+ A000069(k+1))) = A071585(k), m >= 1, 0 <= k < 2^(m-1). (End)
From Yosu Yurramendi, Nov 29 2019: (Start)
For n > 0:
A324338(n) + A324337(n) = A071585(n).
A324338(2*A001969(n) )-A324337(2*A001969(n) ) = A071585(n-1)
A324338(2*A001969(n)+1)-A324337(2*A001969(n)+1) = -A324337(n-1)
A324338(2*A000069(n) )-A324337(2*A000069(n) ) = -A071585(n-1)
A324338(2*A000069(n)+1)-A324337(2*A000069(n)+1) = A324338(n-1) (End)
a(n) = A002487(1+A233279(n)). - Yosu Yurramendi, Dec 27 2019

A324338 a(n) = A002487(1+A006068(n)).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 3, 2, 1, 4, 5, 3, 4, 3, 2, 5, 1, 5, 7, 4, 7, 5, 3, 8, 5, 4, 3, 7, 2, 7, 8, 5, 1, 6, 9, 5, 10, 7, 4, 11, 9, 7, 5, 12, 3, 11, 13, 8, 6, 5, 4, 9, 3, 10, 11, 7, 2, 9, 12, 7, 11, 8, 5, 13, 1, 7, 11, 6, 13, 9, 5, 14, 13, 10, 7, 17, 4, 15, 18, 11, 11, 9, 7, 16, 5, 17, 19, 12, 3, 14, 19, 11, 18, 13, 8, 21, 7, 6, 5, 11, 4, 13, 14, 9, 3, 13
Offset: 0

Views

Author

Antti Karttunen, Feb 23 2019

Keywords

Comments

Like in A324337, a few terms preceding each 2^k-th term (here always 1) seem to consist of a batch of nearby Fibonacci numbers (A000045) in some order. For example, a(65533) = 987, a(65534) = 610 and a(65535) = 1597.
For all n > 0 A324338(n)/A324337(n) constitutes an enumeration system of all positive rationals. For all n > 0 A324338(n) + A324337(n) = A071585(n). - Yosu Yurramendi, Oct 22 2019

Crossrefs

Programs

Formula

a(n) = A002487(1+A006068(n)).
a(2^n) = 1 for all n >= 0.
From Yosu Yurramendi, Oct 22 2019: (Start)
a(2^m+2^(m-1)+k) = A324337(2^m+ k), m > 0, 0 <= k < 2^(m-1)
a(2^m+ k) = A324337(2^m+2^(m-1)+k), m > 0, 0 <= k < 2^(m-1). (End)
a(n) = A324337(A063946(n)), n > 0. Yosu Yurramendi, Nov 04 2019
a(n) = A002487(A233279(n)), n > 0. Yosu Yurramendi, Nov 08 2019
From Yosu Yurramendi, Nov 28 2019: (Start)
a(2^(m+1)+k) - a(2^m+k) = A324337(k), m >= 0, 0 <= k < 2^m.
a(A059893(2^(m+1)+A000069(k+1))) - a(A059893(2^m+A000069(k+1))) = A071585(k), m >= 1, 0 <= k < 2^(m-1).
a(A059893(2^m+ A001969(k+1))) = A071585(k), m >= 0, 0 <= k < 2^(m-1). (End)
From Yosu Yurramendi, Nov 29 2019: (Start)
For n > 0:
A324338(n) + A324337(n) = A071585(n).
A324338(2*A001969(n) )-A324337(2*A001969(n) ) = A071585(n-1)
A324338(2*A001969(n)+1)-A324337(2*A001969(n)+1) = -A324337(n-1)
A324338(2*A000069(n) )-A324337(2*A000069(n) ) = -A071585(n-1)
A324338(2*A000069(n)+1)-A324337(2*A000069(n)+1) = A324338(n-1) (End)
a(n) = A002487(A233279(n)). Yosu Yurramendi, Dec 27 2019

A341915 For any nonnegative number n with runs in binary expansion (r_1, ..., r_w), a(n) = Sum_{k = 1..w} 2^(r_1 + ... + r_k - 1).

Original entry on oeis.org

0, 1, 3, 2, 5, 7, 6, 4, 9, 13, 15, 11, 10, 14, 12, 8, 17, 25, 29, 21, 23, 31, 27, 19, 18, 26, 30, 22, 20, 28, 24, 16, 33, 49, 57, 41, 45, 61, 53, 37, 39, 55, 63, 47, 43, 59, 51, 35, 34, 50, 58, 42, 46, 62, 54, 38, 36, 52, 60, 44, 40, 56, 48, 32, 65, 97, 113
Offset: 0

Views

Author

Rémy Sigrist, Feb 23 2021

Keywords

Comments

This sequence is a permutation of the nonnegative integers with inverse A341916.
This sequence has connections with A003188; here we compute partials sums of runs from left to right, there from right to left.

Examples

			For n = 23,
- the binary representation of 23 is "10111",
- the corresponding run lengths are (1, 1, 3),
- so a(23) = 2^(1-1) + 2^(1+1-1) + 2^(1+1+3-1) = 19.
		

Crossrefs

Cf. A003188, A005811, A059893, A101211, A341916 (inverse), A341943 (fixed points).

Programs

  • Mathematica
    a[n_] := If[n == 0, 0, 2^((Length /@ Split[IntegerDigits[n, 2]] // Accumulate)-1) // Total];
    Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 02 2022 *)
  • PARI
    a(n) = { my (v=0); while (n, my (w=valuation(n+n%2,2)); n\=2^w; v=2^w*(1+v)); v/2 }

Formula

a(n) = A059893(A003188(n)).
a(n) = Sum_{k = 1..A005811(n)} 2^((Sum_{m = 1..k} A101211(m))-1).
a(n) < 2^k for any n < 2^k.
A000120(a(n)) = A000120(A003188(n)) = A005811(n).

A345253 Maximal Fibonacci tree: Arrangement of the positive integers as labels of a complete binary tree.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 13, 11, 14, 16, 21, 12, 15, 17, 22, 18, 23, 26, 34, 19, 24, 27, 35, 29, 37, 42, 55, 20, 25, 28, 36, 30, 38, 43, 56, 31, 39, 44, 57, 47, 60, 68, 89, 32, 40, 45, 58, 48, 61, 69, 90, 50, 63, 71, 92, 76, 97, 110, 144, 33, 41, 46, 59, 49
Offset: 1

Views

Author

J. Parker Shectman, Jun 12 2021

Keywords

Comments

Every positive integer occurs exactly once, so that, as a sequence, a(n) is a permutation of the positive integers.
Descending from the root node 1, generate tree by outer composition of L(n) = n + F(Finv(n)) and R(n) = n + F(Finv(n) + 1), respectively, according to left or right branching, where F(n) = A000045(n) are the Fibonacci numbers and Finv(n) = A130233(n) is the 'lower' Fibonacci inverse. This produces each number by maximal Fibonacci expansion (cf. example below of Method 2, entry A343152, and links).
(Level of tree): The number of terms in this expansion of n is the level of the tree on which n appears, A112310(n-1) + 1 = A200648(n+1). The number of terms in the expansion of a(n) is floor(log_2(n)) + 1 = A113473(n) = A070939(n) = A029837(n+1).
"Maximal Fibonacci expansion" maximizes the sum of coefficients over all Fibonacci numbers (of positive index), allowing both F(1) = 1 and F(2) = 1. Thus, it is just an expansion and not a representation (like "greedy" and "lazy"), as it "breaks the rule" by using two bits that correspond to elements of equal value, rather than using distinct basis elements (link). This reveals connections to the cf. sequences: Binary strings that emerge in lexicographic order from "maximal Fibonacci gaps" (example), binary trees of the positive integers, and I-D arrays "harvested" from the trees. To define the expansion uniquely, always include F(1), so that the expansion of positive integer n equals F(1) for n = 1 and F(1) prepended to the lazy Fibonacci representation of n-1 for n > 1. Hence, a(1) = 1, and for n > 1, a(n) = A095903(n-1) + 1. The "redundant" expansion arranges the positive integers in the single binary tree {T(n,k)}, rather than the two trees at A255773 and A255774 that result from representation (see link).
(Left-to-right order in tree): Each F(t)-sized block (F(t+1), ..., F(t+2) - 1) of successive positive integers ("Fibonacci cohort" t) appears in right-to-left order in the tree as reordered in A343152, where elements of each cohort appear consecutively (see link).
Descending from the root node 1, generate tree by the inner composition of A026351 and A026352, that is, one plus the sequences of lower and upper Wythoff numbers, A000201 and A001950, respectively, according to left or right branching (see example below of Method 1 and links).
Generate tree from (one plus) the number of (initial) zeros on the positive integers for the outer composition of sequences, A060143 and A060144, respectively, according to left or right branching descending from the identity (c.f example below of Method 3 and links).
The lower Wythoff numbers, A000201, appear exclusively in the 1st, 3rd, 5th, ... right clades of the tree, while the upper Wythoff numbers A001950, appear exclusively in the 2nd, 4th, 6th, ... right clades of the tree. Here, the k-th right clade comprises the nodes at positions 2^(k+1) and 2^k + 1, together with all descendants of the latter (link).
(Duality with tree A232560, and related arrays): Consider the labeled binary trees a(n) = A232560(A059893(n)) and A232560(n) = a(A059893(n)). Labels along maximal straight paths that always branch left in a(n) give rows of array A345252, while labels along maximal straight paths that always branch left in A232560 give rows of array A083047.
Sorting the labels from each successive right clade of the binary tree a(n) gives the successive columns of A083047, while sorting labels from each successive right clade of A232560 gives each successive column of A345252. This makes the trees a(n) and A232560 "blade-duals," blade being a contraction of branch-clade (see entry for A345254 and link). A200648(n)+1 gives the level of the tree on which elements of array first-columns A345252(n,1) and A083047(n,1) appear.
(Palindromes and coincidence of elements): Trees a(n) and A232560 coincide when the sequence of left and right branching is a palindrome: a(A329395(n)) = A232560(A329395(n)). As Kimberling notes (cf. A059893), this happens at fixed points of A059893(n) or, equivalently, at n for which A081242(n) is a palindrome.
The inverse permutation of a(n) as a sequence can be read from a "tetrangle" or "irregular triangle" tableau with F(t) (Fibonacci number) entries on each row t, for t = 1, 2, 3, ..., in which an entry on row t is 2*x the entry x immediately above it on row t-1, if such exists, or otherwise 2*x + 1 the entry x in the corresponding position on row t-2 (thus generating new rows as in A243571 but without sorting the numbers into increasing order, linked reference):
1,
2,
3, 4,
5, 6, 8,
7, 9, 10, 12, 16,
11, 13, 17, 14, 18, 20, 24, 32,
...
With the right-justified tableau substituted by a left-justified tableau, the same procedure yields the inverse permutation for the "minimal Fibonacci tree," A048680(A059893(n)), the "cohort-dual" tree of a(n), where "cohort" t is the F(t)-sized block of successive entries in the tableau (see entry for A345252, linked reference).
(Coincidence of elements): a(A020988(n)) = A048680(A059893(A020988(n))) = A099919(n) and a(A020989(n)) = A048680(A059893(A020989(n))) = A049651(n). Collectively, a(A061547(n)) = A048680(A059893(A061547(n))) = union(A049651(n), A099919(n)).
With two types of duality, the tree forms a quartet of binary-tree arrangements of the positive integers, together with its blade dual A232560, its cohort dual A048680(A059893), and blade dual A048680 of the latter.
Order in the tree is "memory-less": Let a(n) and a(m) label nodes at positions n and m, respectively. Let d1 and d2 be two descending paths, i.e., sequences branching left or right from a starting node. (Nodal positions for the left and right children of the node at position p are given by 2*p and 2*p + 1, resp., and d1 and d2 are compositions of these.) Then a(d1(n)) < a(d2(n)) if and only if a(d1(m)) < a(d2(m)) (linked reference).

Examples

			As a complete binary tree:
                    1
           /                 \
          2                   3
      /       \          /        \
     4         5        6          8
    / \       / \      / \        / \
   7    9    10   13   11   14   16   21
  / \  / \  /  \ /  \ /  \ /  \ /  \ /  \
  ...
By maximal Fibonacci expansion:
                                        F(1)
                      /                                       \
                F(1) + F(2)                               F(1) + F(3)
           /                    \                    /                  \
  F(1) + F(2) + F(3)   F(1) + F(2) + F(4)   F(1) + F(3) + F(4)   F(1) + F(3) + F(5)
  ...
"Fibonacci gaps," or differences between successive indices in maximal Fibonacci expansion above, are A007931(n-1) for n > 1 (see link):
                   *
          /                  \
         1                    2
     /       \           /        \
    11        12        21        22
   /  \      /  \      /  \      /  \
  111  112  121  122  211  212  221  222
  / \  / \  / \  / \  / \  / \  / \  / \
  ...
In examples of the three methods below:
Branch left-right-right down the tree to arrive at nodal position n = 2*(2*(2*1) + 1) + 1 = 11;
Branch right-left-left down the tree to arrive at nodal position n = 2*(2*(2*1 + 1)) = 12.
Tree by inner composition of (one plus) the lower and upper Wythoff sequences, A000201 and A001950 (Method 1):
a(11) = A000201(A001950(A001950(1) + 1) + 1) + 1 = 13.
a(12) = A001950(A000201(A000201(1) + 1) + 1) + 1 = 11.
Tree by (outer) composition of branching functions L(n) = n + F(Finv(n)) and R(n) = n + F(Finv(n) + 1), where F(n) = A000045(n) and Finv(n) = A130233(n) (Method 2):
a(11) = R(R(L(1))) = 13.
a(12) = L(R(R(1))) = 11.
Tree by outer composition of A060143 and A060144 (Wythoff inverse sequences) (Method 3):
a(11) = 13, position of first nonzero in A060144(A060144(A060143(m))) = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ..., for m = 1, 2, 3, ....
a(12) = 11, position of first nonzero in A060143(A060143(A060144(m))) = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ..., for m = 1, 2, 3, ....
		

Crossrefs

Programs

  • Mathematica
    (* For binary tree implementations, see supporting file under LINKS *)
    a[n_] := (x = 0; y = 0; BDn = Reverse[IntegerDigits[n, 2]]; imax = Length[BDn] - 1; For[i = 0, i <= imax, i++, {x, y} = {y + 1, x + y}; If[BDn[[i + 1]] == 1, {x, y} = {y, x + y}]]; y);
    (* Adapted from PARI code of Kevin Ryde *)
  • PARI
    a(n) = my(x=0,y=0); for(i=0,logint(n,2), [x,y]=[y+1,x+y]; if(bittest(n,i), [x,y]=[y,x+y])); y; \\ Kevin Ryde, Jun 19 2021

Formula

a(1) = 1 and for n > 1, a(n) = A095903(n-1) + 1.
a(n) = A232560(A059893(n)).

A351702 In the balanced ternary representation of n, reverse the order of digits other than the most significant.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 11, 6, 9, 12, 7, 10, 13, 14, 23, 32, 17, 26, 35, 20, 29, 38, 15, 24, 33, 18, 27, 36, 21, 30, 39, 16, 25, 34, 19, 28, 37, 22, 31, 40, 41, 68, 95, 50, 77, 104, 59, 86, 113, 44, 71, 98, 53, 80, 107, 62, 89, 116, 47, 74, 101, 56, 83, 110, 65, 92
Offset: 0

Views

Author

Kevin Ryde, Feb 19 2022

Keywords

Comments

Self-inverse permutation with swaps confined to terms of a given digit length (A134021) so within blocks n = (3^k+1)/2 .. (3^(k+1)-1)/2.
Can extend to negative n by a(-n) = -a(n).
A072998 is balanced ternary coded in decimal digits so that reversal except first digit of A072998(n) is at A072998(a(n)). Similarly its ternary equivalent A157671, and also A132141 ternary starting with 1.
These sequences all have a fixed initial digit followed by all ternary strings which is the reversed part. A007932 is such strings as decimal digits 1,2,3 but it omits the empty string so the whole reversal of A007932(n) is at A007932(a(n+1)-1).
Fixed points a(n) = n are where n in balanced ternary is a palindrome apart from its initial 1. These are the full balanced ternary palindromes with their least significant 1 removed, so all n = (A134027(m)-1)/3 for m>=2.

Examples

			n    = 224 = balanced ternary 1,  0, -1, 1,  0, -1
                      reverse     ^^^^^^^^^^^^^^^^
a(n) = 168 = balanced ternary 1, -1,  0, 1, -1,  0
		

Crossrefs

Cf. A059095 (balanced ternary), A134028 (full reverse), A134027 (palindromes).
In other bases: A059893 (binary), A343150 (Zeckendorf), A343152 (lazy Fibonacci).

Programs

  • PARI
    a(n) = if(n==0,0, my(k=if(n,logint(n<<1,3)), s=(3^k+1)>>1); s + fromdigits(Vec(Vecrev(digits(n-s,3)),k),3));

A372431 Positive integers k such that the prime indices of k are disjoint from the binary indices of k.

Original entry on oeis.org

1, 2, 4, 7, 8, 9, 10, 11, 12, 13, 16, 17, 19, 21, 23, 24, 25, 26, 29, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 43, 44, 46, 47, 48, 49, 50, 53, 57, 58, 59, 61, 62, 64, 65, 67, 69, 71, 72, 73, 74, 76, 79, 80, 81, 82, 83, 84, 86, 89, 92, 93, 94, 96, 97, 98, 101
Offset: 1

Views

Author

Gus Wiseman, May 03 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

Examples

			The binary indices of 65 are {1,7}, and the prime indices are {3,6}, so 65 is in the sequence.
The terms together with their prime indices begin:
     1: {}
     2: {1}
     4: {1,1}
     7: {4}
     8: {1,1,1}
     9: {2,2}
    10: {1,3}
    11: {5}
    12: {1,1,2}
    13: {6}
    16: {1,1,1,1}
The terms together with their binary expansions and binary indices begin:
   1:       1 ~ {1}
   2:      10 ~ {2}
   4:     100 ~ {3}
   7:     111 ~ {1,2,3}
   8:    1000 ~ {4}
   9:    1001 ~ {1,4}
  10:    1010 ~ {2,4}
  11:    1011 ~ {1,2,4}
  12:    1100 ~ {3,4}
  13:    1101 ~ {1,3,4}
  16:   10000 ~ {5}
		

Crossrefs

For subset instead of disjoint we have A372430.
The complement is A372432.
Equal lengths: A071814, zeros of A372441.
Equal sums: A372427, zeros of A372428.
Equal maxima: A372436, zeros of A372442.
A019565 gives Heinz number of binary indices, adjoint A048675.
A029837 gives greatest binary index, least A001511.
A048793 lists binary indices, length A000120, reverse A272020, sum A029931.
A061395 gives greatest prime index, least A055396.
A070939 gives length of binary expansion.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.

Programs

  • Mathematica
    bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[100],Intersection[bix[#],prix[#]]=={}&]

A081242 Left-to-right binary enumeration.

Original entry on oeis.org

1, 2, 11, 21, 12, 22, 111, 211, 121, 221, 112, 212, 122, 222, 1111, 2111, 1211, 2211, 1121, 2121, 1221, 2221, 1112, 2112, 1212, 2212, 1122, 2122, 1222, 2222, 11111, 21111, 12111, 22111, 11211, 21211, 12211, 22211, 11121, 21121, 12121, 22121
Offset: 2

Views

Author

Clark Kimberling, Mar 12 2003

Keywords

Comments

In A007931, the arithmetic is done from right to left, yielding reversals of the terms of A081242. In A007931, new wordlengths occur at 1,3,7,15,...; in A081242, they occur at 2,4,8,16,.... In A007931, indexing starts at 1 and the sequence is numerical; in A081242, indexing starts at 2, leaving room for the empty word at position 1 and the sequence consists of all binary words.

Examples

			Eleven in base 2 is 1011; remove initial 1 to have 011; add 1 to all digits to have 122; reverse those to have a(11)=221. Or, start with the empty word and add 1's successively, using digits 1 and 2, and doing the arithmetic from left to right:
e -> 1 -> 2 -> 11 -> 21 -> 12 -> 22 -> 111 -> 211 -> 121 -> 221
		

Crossrefs

Formula

Write n in base 2, remove initial 1, add 1 to remaining digits and reverse their order. Or, start with empty word e, follow by 1 and then successively add 1, always working from left to right.

Extensions

Example edited by Franklin T. Adams-Watters, Jan 25 2010

A334031 The smallest number whose unsorted prime signature is the reversed n-th composition in standard order.

Original entry on oeis.org

1, 2, 4, 6, 8, 18, 12, 30, 16, 54, 36, 150, 24, 90, 60, 210, 32, 162, 108, 750, 72, 450, 300, 1470, 48, 270, 180, 1050, 120, 630, 420, 2310, 64, 486, 324, 3750, 216, 2250, 1500, 10290, 144, 1350, 900, 7350, 600, 4410, 2940, 25410, 96, 810, 540, 5250, 360, 3150
Offset: 0

Views

Author

Gus Wiseman, Apr 17 2020

Keywords

Comments

All terms are normal (A055932), meaning their prime indices cover an initial interval of positive integers.
Unsorted prime signature is the sequence of exponents in a number's prime factorization.
The k-th composition in standard order (row k of A066099) 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 gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The sequence of terms together with their prime indices begins:
       1: {}
       2: {1}
       4: {1,1}
       6: {1,2}
       8: {1,1,1}
      18: {1,2,2}
      12: {1,1,2}
      30: {1,2,3}
      16: {1,1,1,1}
      54: {1,2,2,2}
      36: {1,1,2,2}
     150: {1,2,3,3}
      24: {1,1,1,2}
      90: {1,2,2,3}
      60: {1,1,2,3}
     210: {1,2,3,4}
      32: {1,1,1,1,1}
     162: {1,2,2,2,2}
For example, the 13th composition in standard order is (1,2,1), and the least number with prime signature (1,2,1) is 90 = 2^1 * 3^2 * 5^1, so a(13) = 90.
		

Crossrefs

The range is A055932.
The non-reversed version is A057335.
Unsorted prime signature is A124010.
Numbers whose prime signature is aperiodic are A329139.
Normal numbers with standard compositions as prime signature are A334032.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Aperiodic compositions are A328594.
- Normal compositions are A333217.
- Heinz number is A333219.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Product[Prime[i]^stc[n][[-i]],{i,DigitCount[n,2,1]}],{n,0,100}]

Formula

a(n) = A057335(A059893(n)).

A334033 The a(n)-th composition in standard order (graded reverse-lexicographic) is the reversed unsorted prime signature of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 6, 1, 3, 3, 8, 1, 5, 1, 6, 3, 3, 1, 12, 2, 3, 4, 6, 1, 7, 1, 16, 3, 3, 3, 10, 1, 3, 3, 12, 1, 7, 1, 6, 6, 3, 1, 24, 2, 5, 3, 6, 1, 9, 3, 12, 3, 3, 1, 14, 1, 3, 6, 32, 3, 7, 1, 6, 3, 7, 1, 20, 1, 3, 5, 6, 3, 7, 1, 24, 8, 3, 1
Offset: 1

Views

Author

Gus Wiseman, Apr 18 2020

Keywords

Comments

Unsorted prime signature (A124010) is the sequence of exponents in a number's prime factorization.
The k-th composition in standard order (row k of A066099) 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 gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The unsorted prime signature of 12345678 is (1,2,1,1), whose reverse (1,1,2,1) is the 29th composition in standard order, so a(12345678) = 29.
		

Crossrefs

Positions of first appearances are A334031.
The non-reversed version is A334032.
Unsorted prime signature is A124010.
Least number with reversed prime signature is A331580.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Aperiodic compositions are A328594.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.

Programs

  • Mathematica
    stcinv[q_]:=Total[2^Accumulate[Reverse[q]]]/2;
    Table[stcinv[Reverse[Last/@If[n==1,{},FactorInteger[n]]]],{n,100}]

Formula

a(A334031(n)) = n.
A334031(a(n)) = A071364(n).
a(A057335(n))= A059893(n).
A057335(a(n)) = A331580(n).
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