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 1-10 of 24 results. Next

A272919 Numbers of the form 2^(n-1)*(2^(n*m)-1)/(2^n-1), n >= 1, m >= 1.

Original entry on oeis.org

1, 2, 3, 4, 7, 8, 10, 15, 16, 31, 32, 36, 42, 63, 64, 127, 128, 136, 170, 255, 256, 292, 511, 512, 528, 682, 1023, 1024, 2047, 2048, 2080, 2184, 2340, 2730, 4095, 4096, 8191, 8192, 8256, 10922, 16383, 16384, 16912, 18724, 32767, 32768, 32896, 34952, 43690, 65535, 65536, 131071
Offset: 1

Views

Author

Ivan Neretin, May 10 2016

Keywords

Comments

In other words, numbers whose binary representation consists of one or more repeating blocks with only one 1 in each block.
Also, fixed points of the permutations A139706 and A139708.
Each a(n) is a term of A064896 multiplied by some power of 2. As such, this sequence must also be a subsequence of A125121.
Also the numbers that uniquely index a Haar graph (i.e., 5 and 6 are not in the sequence since H(5) is isomorphic to H(6)). - Eric W. Weisstein, Aug 19 2017
From Gus Wiseman, Apr 04 2020: (Start)
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. This sequence lists all positive integers k such that the k-th composition in standard order is constant. For example, the sequence together with the corresponding constant compositions begins:
0: () 136: (4,4)
1: (1) 170: (2,2,2,2)
2: (2) 255: (1,1,1,1,1,1,1,1)
3: (1,1) 256: (9)
4: (3) 292: (3,3,3)
7: (1,1,1) 511: (1,1,1,1,1,1,1,1,1)
8: (4) 512: (10)
10: (2,2) 528: (5,5)
15: (1,1,1,1) 682: (2,2,2,2,2)
16: (5) 1023: (1,1,1,1,1,1,1,1,1,1)
31: (1,1,1,1,1) 1024: (11)
32: (6) 2047: (1,1,1,1,1,1,1,1,1,1,1)
36: (3,3) 2048: (12)
42: (2,2,2) 2080: (6,6)
63: (1,1,1,1,1,1) 2184: (4,4,4)
64: (7) 2340: (3,3,3,3)
127: (1,1,1,1,1,1,1) 2730: (2,2,2,2,2,2)
128: (8) 4095: (1,1,1,1,1,1,1,1,1,1,1,1)
(End)

Crossrefs

Cf. A137706 (smallest number indexing a new Haar graph).
Compositions in standard order are A066099.
Strict compositions are ranked by A233564.

Programs

  • Maple
    N:= 10^6: # to get all terms <= N
    R:= select(`<=`,{seq(seq(2^(n-1)*(2^(n*m)-1)/(2^n-1), m = 1 .. ilog2(2*N)/n), n = 1..ilog2(2*N))},N):
    sort(convert(R,list)); # Robert Israel, May 10 2016
  • Mathematica
    Flatten@Table[d = Reverse@Divisors[n]; 2^(d - 1)*(2^n - 1)/(2^d - 1), {n, 17}]

Formula

From Gus Wiseman, Apr 04 2020: (Start)
A333381(a(n)) = A027750(n).
For n > 0, A124767(a(n)) = 1.
If n is a power of two, A333628(a(n)) = 0, otherwise = 1.
A333627(a(n)) is a power of 2.
(End)

A233564 c-squarefree numbers: positive integers which in binary are concatenation of distinct parts of the form 10...0 with nonnegative number of zeros.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 16, 17, 18, 20, 24, 32, 33, 34, 37, 38, 40, 41, 44, 48, 50, 52, 64, 65, 66, 68, 69, 70, 72, 80, 81, 88, 96, 98, 104, 128, 129, 130, 132, 133, 134, 137, 140, 144, 145, 152, 160, 161, 176, 192, 194, 196, 200, 208, 256, 257, 258, 260, 261
Offset: 1

Views

Author

Vladimir Shevelev, Dec 13 2013

Keywords

Comments

Number of terms in interval [2^(n-1), 2^n) is the number of compositions of n with distinct parts (cf. A032020). For example, if n=6, then interval [2^5, 2^6) contains 11 terms {32,...,52}. This corresponds to 11 compositions with distinct parts of 6: 6, 5+1, 1+5, 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3.
From Gus Wiseman, Apr 06 2020: (Start)
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. This sequence lists all numbers k such that the k-th composition in standard order is strict. For example, the sequence together with the corresponding strict compositions begins:
0: () 38: (3,1,2) 98: (1,4,2)
1: (1) 40: (2,4) 104: (1,2,4)
2: (2) 41: (2,3,1) 128: (8)
4: (3) 44: (2,1,3) 129: (7,1)
5: (2,1) 48: (1,5) 130: (6,2)
6: (1,2) 50: (1,3,2) 132: (5,3)
8: (4) 52: (1,2,3) 133: (5,2,1)
9: (3,1) 64: (7) 134: (5,1,2)
12: (1,3) 65: (6,1) 137: (4,3,1)
16: (5) 66: (5,2) 140: (4,1,3)
17: (4,1) 68: (4,3) 144: (3,5)
18: (3,2) 69: (4,2,1) 145: (3,4,1)
20: (2,3) 70: (4,1,2) 152: (3,1,4)
24: (1,4) 72: (3,4) 160: (2,6)
32: (6) 80: (2,5) 161: (2,5,1)
33: (5,1) 81: (2,4,1) 176: (2,1,5)
34: (4,2) 88: (2,1,4) 192: (1,7)
37: (3,2,1) 96: (1,6) 194: (1,5,2)
(End)

Examples

			49 in binary has the following parts of the form 10...0 with nonnegative number of  zeros: (1),(1000),(1). Two of them are the same. So it is not in the sequence. On the other hand, 50 has distinct parts (1)(100)(10), thus it is a term.
		

Crossrefs

A subset of A333489 and superset of A333218.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Weighted sum is A029931.
- Partial sums from the right are A048793.
- Sum is A070939.
- Runs are counted by A124767.
- Reversed initial intervals A164894.
- Initial intervals are A246534.
- Constant compositions are A272919.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are counted by A333381.
- Anti-runs are A333489.

Programs

  • Mathematica
    bitPatt[n_]:=bitPatt[n]=Split[IntegerDigits[n,2],#1>#2||#2==0&];
    Select[Range[0,300],bitPatt[#]==DeleteDuplicates[bitPatt[#]]&] (* Peter J. C. Moses, Dec 13 2013 *)
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],UnsameQ@@stc[#]&] (* Gus Wiseman, Apr 04 2020 *)

Extensions

More terms from Peter J. C. Moses, Dec 13 2013
0 prepended by Gus Wiseman, Apr 04 2020

A333217 Numbers k such that the k-th composition in standard order covers an initial interval of positive integers.

Original entry on oeis.org

0, 1, 3, 5, 6, 7, 11, 13, 14, 15, 21, 22, 23, 26, 27, 29, 30, 31, 37, 38, 41, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 75, 77, 78, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101, 102, 105, 106, 107, 108, 109, 110, 111, 114, 116, 117, 118
Offset: 1

Views

Author

Gus Wiseman, Mar 15 2020

Keywords

Comments

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.

Examples

			The sequence of terms together with the corresponding compositions begins:
    0: ()              37: (3,2,1)           75: (3,2,1,1)
    1: (1)             38: (3,1,2)           77: (3,1,2,1)
    3: (1,1)           41: (2,3,1)           78: (3,1,1,2)
    5: (2,1)           43: (2,2,1,1)         83: (2,3,1,1)
    6: (1,2)           44: (2,1,3)           85: (2,2,2,1)
    7: (1,1,1)         45: (2,1,2,1)         86: (2,2,1,2)
   11: (2,1,1)         46: (2,1,1,2)         87: (2,2,1,1,1)
   13: (1,2,1)         47: (2,1,1,1,1)       89: (2,1,3,1)
   14: (1,1,2)         50: (1,3,2)           90: (2,1,2,2)
   15: (1,1,1,1)       52: (1,2,3)           91: (2,1,2,1,1)
   21: (2,2,1)         53: (1,2,2,1)         92: (2,1,1,3)
   22: (2,1,2)         54: (1,2,1,2)         93: (2,1,1,2,1)
   23: (2,1,1,1)       55: (1,2,1,1,1)       94: (2,1,1,1,2)
   26: (1,2,2)         58: (1,1,2,2)         95: (2,1,1,1,1,1)
   27: (1,2,1,1)       59: (1,1,2,1,1)      101: (1,3,2,1)
   29: (1,1,2,1)       61: (1,1,1,2,1)      102: (1,3,1,2)
   30: (1,1,1,2)       62: (1,1,1,1,2)      105: (1,2,3,1)
   31: (1,1,1,1,1)     63: (1,1,1,1,1,1)    106: (1,2,2,2)
		

Crossrefs

Sequences covering an initial interval are counted by A000670.
Composition in standard order are A066099.
The case of strictly increasing initial intervals is A164894.
The case of strictly decreasing initial intervals is A246534.
The case of permutations is A333218.
The weakly increasing version is A333379.
The weakly decreasing version is A333380.

Programs

  • Mathematica
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],normQ[stc[#]]&]

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

A225620 Indices of partitions in the table of compositions of A228351.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 10, 12, 14, 15, 16, 20, 24, 26, 28, 30, 31, 32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63, 64, 72, 80, 84, 96, 100, 104, 106, 112, 116, 120, 122, 124, 126, 127, 128, 136, 144, 160, 164, 168, 170, 192, 200, 208, 212, 224, 228, 232, 234, 240, 244, 248, 250, 252, 254, 255
Offset: 1

Views

Author

Omar E. Pol, Aug 03 2013

Keywords

Comments

Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).
In order to construct this sequence as a triangle we use the following rules:
- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.
- Then replace this new set of parts by the concatenation of its parts.
- Then replace this string by its mirror version which is a binary number.
T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).
The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.
First differs from A065609 at a(23).
Conjecture: this sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023

Examples

			T(6,8) = 58 because 58 in base 2 is 111010 whose mirror is 010111 which is the concatenation of 01, 01, 1, 1, whose number of digits are 2, 2, 1, 1, which are also the 8th partition of 6.
Illustration of initial terms:
The sequence represents a table of partitions (see below):
--------------------------------------------------------
.            Binary                        Partitions
n  k  T(n,k) number  Mirror   Diagram       (A026792)
.                                          1 2 3 4 5 6
--------------------------------------------------------
.                             _
1  1     1       1    1        |           1,
.                             _ _
1  1     2      10    01      _  |           2,
2  2     3      11    11       | |         1,1,
.                             _ _ _
3  1     4     100    001     _ _  |           3,
3  2     6     110    011     _  | |         2,1,
3  3     7     111    111      | | |       1,1,1,
.                             _ _ _ _
4  1     8    1000    0001    _ _    |           4,
4  2    10    1010    0101    _ _|_  |         2,2,
4  3    12    1100    0011    _ _  | |         3,1,
4  4    14    1110    0111    _  | | |       2,1,1,
4  5    15    1111    1111     | | | |     1,1,1,1,
.                             _ _ _ _ _
5  1    16   10000    00001   _ _ _    |           5,
5  2    20   10100    00101   _ _ _|_  |         3,2,
5  3    24   11000    00011   _ _    | |         4,1,
5  4    26   11010    01011   _ _|_  | |       2,2,1,
5  5    28   11100    00111   _ _  | | |       3,1,1,
5  6    30   11110    01111   _  | | | |     2,1,1,1,
5  7    31   11111    11111    | | | | |   1,1,1,1,1,
.                             _ _ _ _ _ _
6  1    32  100000    000001  _ _ _      |           6
6  2    36  100100    001001  _ _ _|_    |         3,3,
6  3    40  101000    000101  _ _    |   |         4,2,
6  4    42  101010    010101  _ _|_ _|_  |       2,2,2,
6  5    48  110000    000011  _ _ _    | |         5,1,
6  6    52  110100    001011  _ _ _|_  | |       3,2,1,
6  7    56  111000    000111  _ _    | | |       4,1,1,
6  8    58  111010    010111  _ _|_  | | |     2,2,1,1,
6  9    60  111100    001111  _ _  | | | |     3,1,1,1,
6  10   62  111110    011111  _  | | | | |   2,1,1,1,1,
6  11   63  111111    111111   | | | | | | 1,1,1,1,1,1,
.
Triangle begins:
  1;
  2,   3;
  4,   6,  7;
  8,  10, 12, 14, 15;
  16, 20, 24, 26, 28, 30, 31;
  32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63;
  ...
From _Gus Wiseman_, Apr 01 2020: (Start)
Using the encoding of A066099, this sequence ranks all finite nonempty multisets, as follows.
   1: {1}
   2: {2}
   3: {1,1}
   4: {3}
   6: {1,2}
   7: {1,1,1}
   8: {4}
  10: {2,2}
  12: {1,3}
  14: {1,1,2}
  15: {1,1,1,1}
  16: {5}
  20: {2,3}
  24: {1,4}
  26: {1,2,2}
  28: {1,1,3}
  30: {1,1,1,2}
  31: {1,1,1,1,1}
(End)
		

Crossrefs

Column 1 is A000079. Row n has length A000041(n). Right border gives A000225.
The case covering an initial interval is A333379 or A333380.
All of the following pertain to compositions in the order of A066099.
- The weakly increasing version is this sequence.
- The weakly decreasing version is A114994.
- The strictly increasing version is A333255.
- The strictly decreasing version is A333256.
- The unequal version is A233564.
- The equal version is A272919.
- The case covering an initial interval is A333217.
- Initial intervals are ranked by A164894.
- Reversed initial intervals are ranked by A246534.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],LessEqual@@stc[#]&] (* Gus Wiseman, Apr 01 2020 *)

Formula

Conjecture: a(A000070(m) - k) = 2^m - A228354(k) for m > 0, 0 < k <= A000041(m). - Mikhail Kurkov, Oct 20 2023

A333218 Numbers k such that the k-th composition in standard order is a permutation (of an initial interval).

Original entry on oeis.org

0, 1, 5, 6, 37, 38, 41, 44, 50, 52, 549, 550, 553, 556, 562, 564, 581, 582, 593, 600, 610, 616, 649, 652, 657, 664, 708, 712, 786, 788, 802, 808, 836, 840, 16933, 16934, 16937, 16940, 16946, 16948, 16965, 16966, 16977, 16984, 16994, 17000, 17033, 17036, 17041
Offset: 1

Views

Author

Gus Wiseman, Mar 16 2020

Keywords

Comments

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.

Examples

			The sequence of terms together with their corresponding compositions begins:
        0: ()             593: (3,2,4,1)      16937: (5,4,2,3,1)
        1: (1)            600: (3,2,1,4)      16940: (5,4,2,1,3)
        5: (2,1)          610: (3,1,4,2)      16946: (5,4,1,3,2)
        6: (1,2)          616: (3,1,2,4)      16948: (5,4,1,2,3)
       37: (3,2,1)        649: (2,4,3,1)      16965: (5,3,4,2,1)
       38: (3,1,2)        652: (2,4,1,3)      16966: (5,3,4,1,2)
       41: (2,3,1)        657: (2,3,4,1)      16977: (5,3,2,4,1)
       44: (2,1,3)        664: (2,3,1,4)      16984: (5,3,2,1,4)
       50: (1,3,2)        708: (2,1,4,3)      16994: (5,3,1,4,2)
       52: (1,2,3)        712: (2,1,3,4)      17000: (5,3,1,2,4)
      549: (4,3,2,1)      786: (1,4,3,2)      17033: (5,2,4,3,1)
      550: (4,3,1,2)      788: (1,4,2,3)      17036: (5,2,4,1,3)
      553: (4,2,3,1)      802: (1,3,4,2)      17041: (5,2,3,4,1)
      556: (4,2,1,3)      808: (1,3,2,4)      17048: (5,2,3,1,4)
      562: (4,1,3,2)      836: (1,2,4,3)      17092: (5,2,1,4,3)
      564: (4,1,2,3)      840: (1,2,3,4)      17096: (5,2,1,3,4)
      581: (3,4,2,1)    16933: (5,4,3,2,1)    17170: (5,1,4,3,2)
      582: (3,4,1,2)    16934: (5,4,3,1,2)    17172: (5,1,4,2,3)
		

Crossrefs

A superset of A164894.
Also a superset of A246534.
Not requiring the parts to be distinct gives A333217.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],#==0||UnsameQ@@stc[#]&&Max@@stc[#]==Length[stc[#]]&]

A124769 Number of strictly decreasing runs for compositions in standard order.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
A composition of n is a finite sequence of positive integers summing to n. 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. a(n) is the number of maximal strictly decreasing runs in this composition. Alternatively, a(n) is one plus the number of weak ascents in the same composition. For example, the strictly decreasing runs of the 1234567th composition are ((3,2,1),(2),(2,1),(2),(5,1),(1),(1)), so a(1234567) = 7. The 6 weak ascents together with the strict descents are: 3 > 2 > 1 <= 2 <= 2 > 1 <= 2 <= 5 > 1 <= 1 <= 1. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; the strictly increasing runs are 2,1; 1; so a(11) = 2.
The table starts:
  0
  1
  1 2
  1 1 2 3
  1 1 2 2 2 2 3 4
  1 1 1 2 2 2 2 3 2 2 3 3 3 3 4 5
  1 1 1 2 2 1 2 3 2 2 3 3 2 2 3 4 2 2 2 3 3 3 3 4 3 3 4 4 4 4 5 6
		

Crossrefs

Cf. A066099, A124764, A011782 (row lengths).
Compositions of n with k weak ascents are A333213.
Positions of ones are A333256.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Partial sums from the right are A048793 (triangle).
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769 (this sequence).
- Reversed initial intervals A164894.
- Weakly increasing compositions are A225620.
- Reverse is A228351 (triangle).
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are A333489.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Split[stc[n],Greater]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

Formula

a(0) = 0, a(n) = A124764(n) + 1 for n > 0.

A333255 Numbers k such that the k-th composition in standard order is strictly increasing.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 12, 16, 20, 24, 32, 40, 48, 52, 64, 72, 80, 96, 104, 128, 144, 160, 192, 200, 208, 256, 272, 288, 320, 328, 384, 400, 416, 512, 544, 576, 640, 656, 768, 784, 800, 832, 840, 1024, 1056, 1088, 1152, 1280, 1296, 1312, 1536, 1568, 1600, 1664, 1680
Offset: 1

Views

Author

Gus Wiseman, Mar 20 2020

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. 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.

Examples

			The sequence of positive terms together with the corresponding compositions begins:
     1: (1)         128: (8)         656: (2,3,5)
     2: (2)         144: (3,5)       768: (1,9)
     4: (3)         160: (2,6)       784: (1,4,5)
     6: (1,2)       192: (1,7)       800: (1,3,6)
     8: (4)         200: (1,3,4)     832: (1,2,7)
    12: (1,3)       208: (1,2,5)     840: (1,2,3,4)
    16: (5)         256: (9)        1024: (11)
    20: (2,3)       272: (4,5)      1056: (5,6)
    24: (1,4)       288: (3,6)      1088: (4,7)
    32: (6)         320: (2,7)      1152: (3,8)
    40: (2,4)       328: (2,3,4)    1280: (2,9)
    48: (1,5)       384: (1,8)      1296: (2,4,5)
    52: (1,2,3)     400: (1,3,5)    1312: (2,3,6)
    64: (7)         416: (1,2,6)    1536: (1,10)
    72: (3,4)       512: (10)       1568: (1,4,6)
    80: (2,5)       544: (4,6)      1600: (1,3,7)
    96: (1,6)       576: (3,7)      1664: (1,2,8)
   104: (1,2,4)     640: (2,8)      1680: (1,2,3,5)
		

Crossrefs

Strictly increasing runs are counted by A124768.
The normal case is A164894.
The weakly decreasing version is A114994.
The weakly increasing version is A225620.
The unequal version is A233564.
The equal version is A272919.
The strictly decreasing version is A333256.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],Less@@stc[#]&]

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
A composition of n is a finite sequence of positive integers summing to n. 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. - Gus Wiseman, Apr 03 2020

Examples

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020
		

Crossrefs

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

Formula

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)

A246534 a(n) = Sum_{k=1..n} 2^(T(k)-1), where T(k)=k(k+1)/2 = A000217(k).

Original entry on oeis.org

0, 1, 5, 37, 549, 16933, 1065509, 135283237, 34495021605, 17626681066021, 18032025190548005, 36911520172609651237, 151152638972001256489509, 1238091191924352276155613733, 20283647694843594776223406899749, 664634281540152780046679753547072037
Offset: 0

Views

Author

M. F. Hasler, Aug 28 2014

Keywords

Comments

Similar to A181388, this occurs as binary encoding of a straight n-omino lying on the y-axis, when the grid points of the first quadrant (N x N, N={0,1,2,...}) are given the weight 2^k, with k=0, 1,2, 3,4,5, ... filled in by antidiagonals.
Numbers k such that the k-th composition in standard order (row k of A066099) is a reversed initial interval. - Gus Wiseman, Apr 02 2020

Examples

			Label the cells of an infinite square matrix with 0,1,2,3,... along antidiagonals:
  0 1 3 6 10 ...
  2 4 7 ...
  5 8 ...
  9 ...
  ....
Now any subset of these cells can be represented by the sum of 2 raised to the power written in the given cells. In particular, the subset consisting of the first cell in the first 1, 2, 3, ... rows is represented by 2^0, 2^0+2^2, 2^0+2^2+2^5, ...
		

Crossrefs

The version for prime (rather than binary) indices is A002110.
The non-strict generalization is A114994.
The non-reversed version is A164894.
Intersection of A333256 and A333217.
Partial sums of A036442.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    Select[Range[0,1000],normQ[stc[#]]&&Greater@@stc[#]&] (* Gus Wiseman, Apr 02 2020 *)
  • PARI
    t=0;vector(20,n,t+=2^(n*(n+1)/2-1)) \\ yields the vector starting with a[1]=1
    
  • PARI
    t=0;vector(20,n,if(n>1,t+=2^(n*(n-1)/2-1))) \\ yields the vector starting with 0
    
  • Python
    a = 0
    for n in range(1,17): print(a, end =', '); a += 1<<(n-1)*(n+2)//2 # Ya-Ping Lu, Jan 23 2024
Showing 1-10 of 24 results. Next